Bài tập trung bình bài Đa thức một biến. Cộng trừ đa thức một biến. Nghiệm của đa thức một biến

Home Lớp 7 Toán lớp 7 - Sách cũ Đa thức một biến. Cộng trừ đa thức một biến. Nghiệm của đa thức một biến

Đa thức một biến. Cộng trừ đa thức một biến. Nghiệm của đa thức một biến

{"segment":[{"time":24,"part":[{"title":"L\u1ef1a ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang nh\u1ea5t","title_trans":"","temp":"multiple_choice","correct":[[2]],"list":[{"point":5,"ques":" Thu g\u1ecdn r\u1ed3i s\u1eafp x\u1ebfp \u0111a th\u1ee9c sau theo l\u0169y th\u1eeba gi\u1ea3m d\u1ea7n c\u1ee7a bi\u1ebfn: <br\/> $1-6{{x}^{7}}+5{{x}^{4}}-2+13{{x}^{5}}-8{{x}^{7}}$ <\/span>","select":["A. $-1+5x^4+13x^5-14x^7$","B. $-14{{x}^{7}}+13{{x}^{5}}+5{{x}^{4}}-1$","C. $-2{{x}^{7}}+13{{x}^{5}}+5{{x}^{4}}+1$","D. $2{{x}^{7}}+13{{x}^{5}}+5{{x}^{4}}-3$"],"hint":"","explain":" Ta c\u00f3: <br\/> $\\begin{align} & 1-6{{x}^{7}}+5{{x}^{4}}-2+13{{x}^{5}}-8{{x}^{7}} \\\\ & =\\left( -6{{x}^{7}}-8{{x}^{7}} \\right)+13{{x}^{5}}+5{{x}^{4}}+\\left( 1-2 \\right) \\\\ & =-14{{x}^{7}}+13{{x}^{5}}+5{{x}^{4}}-1 \\\\ \\end{align}$ <br\/> V\u1eady \u0111\u00e1p \u00e1n l\u00e0 B.<\/span><\/span>","column":2}]}],"id_ques":1221},{"time":24,"part":[{"title":"\u0110i\u1ec1n k\u1ebft qu\u1ea3 \u0111\u00fang nh\u1ea5t v\u00e0o \u00f4 tr\u1ed1ng","title_trans":"","temp":"fill_the_blank","correct":[[["3"],["2"],["3"],["-8"]]],"list":[{"point":5,"width":30,"content":"","type_input":"","ques":" Thu g\u1ecdn r\u1ed3i s\u1eafp x\u1ebfp \u0111a th\u1ee9c sau theo l\u0169y th\u1eeba gi\u1ea3m d\u1ea7n c\u1ee7a bi\u1ebfn: <br\/> $A=4{{x}^{5}}+3x-2{{x}^{2}}-{{x}^{5}}+4{{x}^{2}}-8$ <br\/> <br\/> \u0110\u00e1p \u00e1n:<\/b> $A=\\FormInput[40][bl_elm_true basic_elm basic_blank basic_blank_part]{}{{x}^{5}}+\\FormInput[40][bl_elm_true basic_elm basic_blank basic_blank_part]{}{{x}^{2}}+\\FormInput[40][bl_elm_true basic_elm basic_blank basic_blank_part]{}x+\\FormInput[40][bl_elm_true basic_elm basic_blank basic_blank_part]{}$<\/span> ","explain":"Ta c\u00f3: <br\/> $\\begin{align} & A=4{{x}^{5}}+3x-2{{x}^{2}}-{{x}^{5}}+4{{x}^{2}}-8 \\\\ & \\,\\,\\,\\,\\,\\,=\\left( 4{{x}^{5}}-{{x}^{5}} \\right)+\\left( -2{{x}^{2}}+4{{x}^{2}} \\right)+3x-8 \\\\ & \\,\\,\\,\\,\\,\\,=3{{x}^{5}}+2{{x}^{2}}+3x-8 \\\\ \\end{align}$ <br\/> V\u1eady c\u00e1c s\u1ed1 c\u1ea7n \u0111i\u1ec1n v\u00e0o c\u00e1c \u00f4 tr\u1ed1ng l\u1ea7n l\u01b0\u1ee3t l\u00e0 $3;2;3;-8$.<\/span><\/span>"}]}],"id_ques":1222},{"time":24,"part":[{"title":"Kh\u1eb3ng \u0111\u1ecbnh sau \u0111\u00e2y \u0110\u00fang hay Sai","title_trans":"","temp":"multiple_choice","correct":[[2]],"list":[{"point":5,"ques":" B\u1eadc c\u1ee7a \u0111a th\u1ee9c $2y-3y^2+3y^2-2y$ l\u00e0 $2$","select":["\u0110\u00fang","Sai"],"explain":" Thu g\u1ecdn \u0111a th\u1ee9c: <br\/> $2y-3y^2+3y^2-2y=(-3y^2+3y^2)+(2y-2y)=0$ <br\/> $0$ l\u00e0 \u0111a th\u1ee9c kh\u00f4ng c\u00f3 b\u1eadc. <br\/> Kh\u1eb3ng \u0111\u1ecbnh c\u1ee7a \u0111\u1ec1 b\u00e0i l\u00e0 sai. <br\/> V\u1eady \u0111\u00e1p \u00e1n l\u00e0 Sai.<\/span> <br\/> Sai l\u1ea7m c\u1ea7n tr\u00e1nh:<\/b> Tr\u01b0\u1edbc khi t\u00ecm b\u1eadc c\u1ee7a m\u1ed9t \u0111a th\u1ee9c, ta \u0111i thu g\u1ecdn \u0111a th\u1ee9c \u0111\u00f3 r\u1ed3i m\u1edbi t\u00ecm b\u1eadc c\u1ee7a n\u00f3. <\/i><\/span>","column":2}]}],"id_ques":1223},{"time":24,"part":[{"title":"L\u1ef1a ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang nh\u1ea5t","title_trans":"","temp":"multiple_choice","correct":[[1]],"list":[{"point":5,"ques":" B\u1eadc c\u1ee7a \u0111a th\u1ee9c $3x^2-2x+7+2x-3x^2-6$ l\u00e0:","select":["A. $0$","B. $1$","C. $2$","D. Kh\u00f4ng c\u00f3 b\u1eadc"],"hint":"","explain":" Ta c\u00f3: <br\/> $3x^2-2x+7+2x-3x^2-6$ <br\/> $=(3x^2-3x^2)+(-2x+2x)+(7-6)$ <br\/> $=0+0+1=1$ <br\/> $1$ l\u00e0 \u0111a th\u1ee9c b\u1eadc kh\u00f4ng. <br\/> V\u1eady \u0111\u00e1p \u00e1n l\u00e0 A.<\/span><\/span>","column":2}]}],"id_ques":1224},{"time":24,"part":[{"title":"\u0110i\u1ec1n k\u1ebft qu\u1ea3 \u0111\u00fang nh\u1ea5t v\u00e0o \u00f4 tr\u1ed1ng","title_trans":"","temp":"fill_the_blank","correct":[[["-1"],["7"]]],"list":[{"point":5,"width":40,"content":"","type_input":"","ques":" Cho \u0111a th\u1ee9c: $B=-6{{x}^{3}}+5x-1+2{{x}^{2}}+6{{x}^{3}}-2x+5{{x}^{2}}$ <br\/> H\u1ec7 s\u1ed1 t\u1ef1 do c\u1ee7a \u0111a th\u1ee9c $B$ l\u00e0 _input_ <br\/> H\u1ec7 s\u1ed1 cao nh\u1ea5t c\u1ee7a \u0111a th\u1ee9c $B$ l\u00e0 _input_ ","explain":"H\u01b0\u1edbng d\u1eabn:<\/span><br\/>Thu g\u1ecdn \u0111a th\u1ee9c $B$ r\u1ed3i t\u00ecm h\u1ec7 s\u1ed1 t\u1ef1 do v\u00e0 h\u1ec7 s\u1ed1 cao nh\u1ea5t c\u1ee7a n\u00f3. <br\/> + H\u1ec7 s\u1ed1 t\u1ef1 do l\u00e0 h\u1ec7 s\u1ed1 c\u1ee7a l\u0169y th\u1eeba b\u1eadc $0$ c\u1ee7a bi\u1ebfn <br\/> + H\u1ec7 s\u1ed1 cao nh\u1ea5t l\u00e0 h\u1ec7 s\u1ed1 c\u1ee7a l\u0169y th\u1eeba cao nh\u1ea5t c\u1ee7a bi\u1ebfn<br\/>B\u00e0i gi\u1ea3i:<\/span><br\/>Ta c\u00f3: <br\/> $\\begin{align} & B=-6{{x}^{3}}+5x-1+2{{x}^{2}}+6{{x}^{3}}-2x+5{{x}^{2}} \\\\ & \\,\\,\\,\\,\\,=\\left( -6{{x}^{3}}+6{{x}^{3}} \\right)+\\left( 2{{x}^{2}}+5{{x}^{2}} \\right)+\\left( 5x-2x \\right)-1 \\\\ & \\,\\,\\,\\,=0+7{{x}^{2}}+3x-1 \\\\ & \\,\\,\\,\\,=7{{x}^{2}}+3x-1 \\\\ \\end{align}$ <br\/> H\u1ec7 s\u1ed1 t\u1ef1 do c\u1ee7a $B$ l\u00e0 $-1$ <br\/> L\u0169y th\u1eeba cao nh\u1ea5t c\u1ee7a bi\u1ebfn l\u00e0 $2$ n\u00ean h\u1ec7 s\u1ed1 $7$ c\u1ee7a l\u0169y th\u1eeba cao nh\u1ea5t l\u00e0 h\u1ec7 s\u1ed1 cao nh\u1ea5t. <br\/> V\u1eady c\u00e1c s\u1ed1 c\u1ea7n \u0111i\u1ec1n v\u00e0o c\u00e1c \u00f4 tr\u1ed1ng l\u1ea7n l\u01b0\u1ee3t l\u00e0 $-1;7$.<\/span><\/span>"}]}],"id_ques":1225},{"time":24,"part":[{"title":"\u0110i\u1ec1n k\u1ebft qu\u1ea3 \u0111\u00fang nh\u1ea5t v\u00e0o \u00f4 tr\u1ed1ng","title_trans":"","temp":"fill_the_blank","correct":[[["4"]]],"list":[{"point":5,"width":40,"content":"","type_input":"","ques":" X\u00e1c \u0111\u1ecbnh h\u1ec7 s\u1ed1 $a$ c\u1ee7a \u0111a th\u1ee9c $P(x)=ax+2$ bi\u1ebft r\u1eb1ng $P(-1)=-2$ <br\/> \u0110\u00e1p \u00e1n:<\/b> $a=\\FormInput[40][bl_elm_true basic_elm basic_blank basic_blank_part]{}$<\/span> ","explain":"H\u01b0\u1edbng d\u1eabn:<\/span><br\/> $P(-1)=-2$ t\u1ee9c l\u00e0 gi\u00e1 tr\u1ecb c\u1ee7a \u0111a th\u1ee9c $P$ t\u1ea1i $x=-1$ l\u00e0 $-2$ <br\/> Do \u0111\u00f3, thay $x=-1$ v\u00e0o \u0111a th\u1ee9c $P$ v\u00e0 cho gi\u00e1 tr\u1ecb c\u1ee7a \u0111a th\u1ee9c \u0111\u00f3 b\u1eb1ng $-2$ \u0111\u1ec3 t\u00ecm $a$ <br\/>B\u00e0i gi\u1ea3i:<\/span><br\/>Ta c\u00f3: <br\/> $\\begin{align} P\\left( -1 \\right)&=-2 \\\\ a\\,.\\left( -1 \\right)+2 &=-2 \\\\ -a+2 &=-2 \\\\ -a &=-4 \\\\ a &=4 \\\\ \\end{align}$ <br\/> V\u1eady s\u1ed1 c\u1ea7n \u0111i\u1ec1n v\u00e0o \u00f4 tr\u1ed1ng l\u00e0 $4$.<\/span><\/span>"}]}],"id_ques":1226},{"time":24,"part":[{"title":"L\u1ef1a ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang nh\u1ea5t","title_trans":"","temp":"multiple_choice","correct":[[4]],"list":[{"point":5,"ques":" Cho $P\\left( x \\right)=100{{x}^{100}}+99{{x}^{99}}+98{{x}^{98}}+\\cdots +2{{x}^{2}}+x$ <br\/> Ta t\u00ednh \u0111\u01b0\u1ee3c $P(1)$ b\u1eb1ng: <\/span>","select":["A. $4000$","B. $4950$","C. $5000$","D. $5050$"],"explain":"H\u01b0\u1edbng d\u1eabn:<\/span><br\/> Thay $x=1$ v\u00e0o \u0111a th\u1ee9c $P$ r\u1ed3i t\u00ednh. <br\/> C\u00f4ng th\u1ee9c t\u00ednh t\u1ed5ng $n$ s\u1ed1 t\u1ef1 nhi\u00ean li\u00ean ti\u1ebfp: $\\dfrac{(\\text{s\u1ed1 \u0111\u1ea7u +s\u1ed1 cu\u1ed1i}).n}{2}$ <br\/> B\u00e0i gi\u1ea3i:<\/span><br\/>Ta c\u00f3: <br\/> $\\begin{align} & P\\left( x \\right)=100{{x}^{100}}+99{{x}^{99}}+98{{x}^{98}}+\\cdots +2{{x}^{2}}+x \\\\ & \\Rightarrow P\\left( 1 \\right)={{100.1}^{100}}+{{99.1}^{99}}+{{98.1}^{98}}+\\cdots +{{2.1}^{2}}+1 \\\\ & \\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,=100+99+98+\\cdots +2+1 \\\\ & \\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,=\\dfrac{\\left( 100+1 \\right).100}{2} \\\\ & \\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,=101.50 \\\\ & \\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,=5050 \\\\ \\end{align}$ <br\/> V\u1eady \u0111\u00e1p \u00e1n l\u00e0 D.<\/span><\/span>","column":4}]}],"id_ques":1227},{"time":24,"part":[{"title":"L\u1ef1a ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang nh\u1ea5t","title_trans":"","temp":"multiple_choice","correct":[[3]],"list":[{"point":5,"ques":" Cho c\u00e1c \u0111a th\u1ee9c: <br\/> $A=3x-2{{x}^{2}}-2+6{{x}^{3}}$ <br\/> $ B=3{{x}^{2}}-x-2{{x}^{3}}+4 $ <br\/> $ C=1+4{{x}^{3}}-2x$ <br\/> T\u1ed5ng $A+B+C$ l\u00e0: <\/span>","select":["A. $12{{x}^{3}}-{{x}^{2}}+3$","B. $8{{x}^{3}}+{{x}^{2}}+6x+3$","C. $8{{x}^{3}}+{{x}^{2}}+3$","D. $12{{x}^{3}}-{{x}^{2}}+6x+3$"],"explain":"Ta c\u00f3: <br\/> $\\begin{align}& A+B+C=\\left( 3x-2{{x}^{2}}-2+6{{x}^{3}} \\right)+\\left( 3{{x}^{2}}-x-2{{x}^{3}}+4 \\right)+\\left( 1+4{{x}^{3}}-2x \\right) \\\\ & \\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,=3x-2{{x}^{2}}-2+6{{x}^{3}}+3{{x}^{2}}-x-2{{x}^{3}}+4+1+4{{x}^{3}}-2x \\\\ & \\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,=\\left( 6{{x}^{3}}-2{{x}^{3}}+4{{x}^{3}} \\right)+\\left( -2{{x}^{2}}+3{{x}^{2}} \\right)+\\left( 3x-x-2x \\right)+\\left( -2+4+1 \\right) \\\\ & \\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,=8{{x}^{3}}+{{x}^{2}}+3 \\\\ \\end{align}$ <br\/> V\u1eady \u0111\u00e1p \u00e1n l\u00e0 C.<\/span><\/span>","column":2}]}],"id_ques":1228},{"time":24,"part":[{"title":"L\u1ef1a ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang nh\u1ea5t","title_trans":"","temp":"multiple_choice","correct":[[1]],"list":[{"point":5,"ques":" Cho c\u00e1c \u0111a th\u1ee9c: <br\/> $A=3x-2{{x}^{2}}-2+6{{x}^{3}}$ <br\/> $ B=3{{x}^{2}}-x-2{{x}^{3}}+4 $ <br\/> $ C=1+4{{x}^{3}}-2x$ <br\/> Hi\u1ec7u $A-B-C$ l\u00e0: <\/span>","select":["A. $4{{x}^{3}}-5{{x}^{2}}+6x-7$","B. $8{{x}^{3}}+{{x}^{2}}+6x-7$","C. $4{{x}^{3}}+5{{x}^{2}}+6x+7$","D. $12{{x}^{3}}-5{{x}^{2}}+5x+3$"],"explain":"Ta c\u00f3: <br\/> $\\begin{align} & A-B-C=\\left( 3x-2{{x}^{2}}-2+6{{x}^{3}} \\right)-\\left( 3{{x}^{2}}-x-2{{x}^{3}}+4 \\right)-\\left( 1+4{{x}^{3}}-2x \\right) \\\\ & \\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,=3x-2{{x}^{2}}-2+6{{x}^{3}}-3{{x}^{2}}+x+2{{x}^{3}}-4-1-4{{x}^{3}}+2x \\\\ & \\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,=\\left( 6{{x}^{3}}+2{{x}^{3}}-4{{x}^{3}} \\right)+\\left( -2{{x}^{2}}-3{{x}^{2}} \\right)+\\left( 3x+x+2x \\right)+\\left( -2-4-1 \\right) \\\\ & \\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,=4{{x}^{3}}-5{{x}^{2}}+6x-7 \\\\ \\end{align}$ <br\/> V\u1eady \u0111\u00e1p \u00e1n l\u00e0 A.<\/span><\/span>","column":2}]}],"id_ques":1229},{"time":24,"part":[{"title":"\u0110i\u1ec1n k\u1ebft qu\u1ea3 \u0111\u00fang nh\u1ea5t v\u00e0o \u00f4 tr\u1ed1ng","title_trans":"","temp":"fill_the_blank","correct":[[["-5"]]],"list":[{"point":5,"width":40,"content":"","type_input":"","ques":" T\u00ecm $x,$ bi\u1ebft $ \\left( 5{{x}^{3}}-4{{x}^{2}}+2x-1 \\right)+\\left( 3-x+4{{x}^{2}}-5{{x}^{3}} \\right) =-3$ <br\/> \u0110\u00e1p \u00e1n: <\/b> $x=\\FormInput[40][bl_elm_true basic_elm basic_blank basic_blank_part]{}$","explain":"Ta c\u00f3: <br\/> $\\begin{align} \\left( 5{{x}^{3}}-4{{x}^{2}}+2x-1 \\right)+\\left( 3-x+4{{x}^{2}}-5{{x}^{3}} \\right) &=-3 \\\\ 5{{x}^{3}}-4{{x}^{2}}+2x-1+3-x+4{{x}^{2}}-5{{x}^{3}} &=-3 \\\\ \\left( 5{{x}^{3}}-5{{x}^{3}} \\right)+\\left( -4{{x}^{2}}+4{{x}^{2}} \\right)+\\left( 2x-x \\right)+\\left( -1+3 \\right) &=-3 \\\\ 0+0+x+2 &=-3 \\\\ x+2 &=-3 \\\\ x &=-5 \\\\ \\end{align}$ <br\/> V\u1eady s\u1ed1 c\u1ea7n \u0111i\u1ec1n v\u00e0o \u00f4 tr\u1ed1ng l\u00e0 $-5$.<\/span><\/span>"}]}],"id_ques":1230},{"time":24,"part":[{"title":"L\u1ef1a ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang nh\u1ea5t","title_trans":"","temp":"multiple_choice","correct":[[3]],"list":[{"point":5,"ques":" Cho c\u00e1c \u0111a th\u1ee9c: <br\/> $M=-2{{x}^{4}}-7x+\\dfrac{1}{2}-6{{x}^{4}}+2{{x}^{2}}-x $ <br\/> $ N=3{{x}^{3}}-{{x}^{4}}-5{{x}^{2}}+{{x}^{3}}-6x+\\dfrac{3}{4}$ <br\/> T\u00ednh $M+N$ v\u00e0 t\u00ecm h\u1ec7 s\u1ed1 cao nh\u1ea5t c\u1ee7a \u0111a th\u1ee9c thu \u0111\u01b0\u1ee3c. <br\/> H\u1ec7 s\u1ed1 cao nh\u1ea5t c\u1ee7a \u0111a th\u1ee9c thu \u0111\u01b0\u1ee3c l\u00e0: <\/span>","select":["A. $4$","B. $-14$","C. $-9$","D. $-3$"],"explain":"Ta c\u00f3: <br\/> $\\begin{align}& M+N=\\left( -2{{x}^{4}}-7x+\\dfrac{1}{2}-6{{x}^{4}}+2{{x}^{2}}-x \\right)+\\left( 3{{x}^{3}}-{{x}^{4}}-5{{x}^{2}}+{{x}^{3}}-6x+\\dfrac{3}{4} \\right) \\\\ & \\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,=-2{{x}^{4}}-7x+\\dfrac{1}{2}-6{{x}^{4}}+2{{x}^{2}}-x+3{{x}^{3}}-{{x}^{4}}-5{{x}^{2}}+{{x}^{3}}-6x+\\dfrac{3}{4} \\\\ & \\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,=\\left( -2{{x}^{4}}-6{{x}^{4}}-{{x}^{4}} \\right)+\\left( 3{{x}^{3}}+{{x}^{3}} \\right)+\\left( 2{{x}^{2}}-5{{x}^{2}} \\right)+\\left( -7x-x-6x \\right)+\\left( \\dfrac{1}{2}+\\dfrac{3}{4} \\right) \\\\ & \\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,=-9{{x}^{4}}+4{{x}^{3}}-3{{x}^{2}}-14x+\\dfrac{5}{4} \\\\ \\end{align}$ <br\/> L\u0169y th\u1eeba cao nh\u1ea5t c\u1ee7a bi\u1ebfn l\u00e0 $4$ n\u00ean h\u1ec7 s\u1ed1 cao nh\u1ea5t l\u00e0 $-9$ <br\/> V\u1eady \u0111\u00e1p \u00e1n l\u00e0 C.<\/span><\/span>","column":4}]}],"id_ques":1231},{"time":24,"part":[{"title":"L\u1ef1a ch\u1ecdn \u0111\u00fang hay sai","title_trans":"","temp":"true_false","correct":[["f","f","t"]],"list":[{"point":5,"image":"","col_name":["Cho hai \u0111a th\u1ee9c: <br\/> $A=5-2{{x}^{3}}+\\dfrac{1}{4}{{x}^{2}} $ <br\/> $ B=-\\dfrac{3}{4}{{x}^{2}}+{{x}^{3}}+5+x$ ","\u0110\u00fang","Sai"],"arr_ques":[" $A+B=-3{{x}^{3}}-\\dfrac{1}{2}{{x}^{2}}+x+10$ ","H\u1ec7 s\u1ed1 cao nh\u1ea5t c\u1ee7a \u0111a th\u1ee9c t\u1ed5ng $A+B$ l\u00e0 $-3$","B\u1eadc c\u1ee7a \u0111a th\u1ee9c t\u1ed5ng $A+B$ l\u00e0 $3$"],"hint":"","explain":[" Sai, v\u00ec <br\/> $\\begin{align} & A+B=\\left( 5-2{{x}^{3}}+\\dfrac{1}{4}{{x}^{2}} \\right)+\\left( -\\dfrac{3}{4}{{x}^{2}}+{{x}^{3}}+5+x \\right) \\\\ & \\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,=5-2{{x}^{3}}+\\dfrac{1}{4}{{x}^{2}}-\\dfrac{3}{4}{{x}^{2}}+{{x}^{3}}+5+x \\\\ & \\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,=\\left( -2{{x}^{3}}+{{x}^{3}} \\right)+\\left( \\dfrac{1}{4}{{x}^{2}}-\\dfrac{3}{4}{{x}^{2}} \\right)+x+\\left( 5+5 \\right) \\\\ & \\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,=-{{x}^{3}}-\\dfrac{1}{2}{{x}^{2}}+x+10 \\\\ \\end{align}$ <\/span>"," <br\/> Sai, v\u00ec <br\/> $A+B=-{{x}^{3}}-\\dfrac{1}{2}{{x}^{2}}+x+10$ (theo c\u00e2u tr\u00ean) <br\/> Do \u0111\u00f3, h\u1ec7 s\u1ed1 cao nh\u1ea5t c\u1ee7a \u0111a th\u1ee9c t\u1ed5ng $A+B$ l\u00e0 $-1$<\/span> ","<br\/> \u0110\u00fang, v\u00ec <br\/> $A+B=-{{x}^{3}}-\\dfrac{1}{2}{{x}^{2}}+x+10$ (theo c\u00e2u tr\u00ean) <br\/> Do \u0111\u00f3, b\u1eadc c\u1ee7a \u0111a th\u1ee9c t\u1ed5ng $A+B$ l\u00e0 $3$<\/span>"]}]}],"id_ques":1232},{"time":24,"part":[{"title":"L\u1ef1a ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang nh\u1ea5t","title_trans":"","temp":"multiple_choice","correct":[[4]],"list":[{"point":5,"ques":" X\u00e1c \u0111\u1ecbnh \u0111a th\u1ee9c b\u1eadc nh\u1ea5t $P(x)=ax+b$ bi\u1ebft r\u1eb1ng $P(-1)=5$ v\u00e0 $P(-2)=7$ <\/span>","select":["A. $P(x)=6x+8 $","B. $P(x)=-3x+5$","C. $P(x)=12x+17$","D. $P(x)=-2x+3$"],"explain":"H\u01b0\u1edbng d\u1eabn:<\/span><br\/> T\u00ecm hai bi\u1ec3u th\u1ee9c li\u00ean h\u1ec7 gi\u1eefa $a$ v\u00e0 $b$ t\u1eeb $P(-1)=5$ v\u00e0 $P(-2)=7$ <br\/> T\u1eeb hai bi\u1ec3u th\u1ee9c li\u00ean h\u1ec7 \u0111\u00f3 bi\u1ebfn \u0111\u1ed5i \u0111\u1ec3 t\u00ecm ra $a$ v\u00e0 $b$ <br\/>B\u00e0i gi\u1ea3i:<\/span><br\/> Ta c\u00f3: <br\/> $\\begin{align} & P\\left( -1 \\right)=5 \\\\ & \\Rightarrow 5=a.\\left( -1 \\right)+b \\\\ & \\Rightarrow -a+b=5 \\\\ & \\Rightarrow a-b=-5\\,\\,\\,\\,\\,\\left( 1 \\right) \\\\ \\end{align}$ <br\/> V\u00e0: <br\/> $\\begin{align} & P\\left( -2 \\right)=7 \\\\ & \\Rightarrow 7=a.\\left( -2 \\right)+b \\\\ & \\Rightarrow -2a+b=7 \\\\ & \\Rightarrow 2a-b=-7\\,\\,\\,\\,\\left( 2 \\right) \\\\ \\end{align}$ <br\/> T\u1eeb (1) ta c\u00f3: $a=b-5$ <br\/> Thay $a=b-5$ v\u00e0o (2) ta \u0111\u01b0\u1ee3c: <br\/> $\\begin{align} & 2\\left( b-5 \\right)-b=-7 \\\\ & \\Rightarrow 2b-10-b=-7 \\\\ & \\Rightarrow b-10=-7 \\\\ & \\Rightarrow b=10-7=3 \\\\ & \\Rightarrow a=b-5=3-5=-2 \\\\ \\end{align}$ <br\/> V\u1eady \u0111a th\u1ee9c c\u1ea7n t\u00ecm l\u00e0 $P(x)=-2x+3$ <br\/> V\u1eady \u0111\u00e1p \u00e1n l\u00e0 D.<\/span><\/span>","column":2}]}],"id_ques":1233},{"time":24,"part":[{"title":"\u0110i\u1ec1n k\u1ebft qu\u1ea3 \u0111\u00fang nh\u1ea5t v\u00e0o \u00f4 tr\u1ed1ng","title_trans":"","temp":"fill_the_blank","correct":[[["0,2"]]],"list":[{"point":5,"width":40,"content":"","ques":"Cho \u0111a th\u1ee9c sau: <br\/> $P\\left( x \\right)=7,4x-8-\\left[ -5,6x+\\left( -2,4x+7 \\right)-11,92 \\right]$ <br\/> \u0110\u1ec3 $P(x)=0$ th\u00ec gi\u00e1 tr\u1ecb c\u1ee7a $x$ l\u00e0 _input_","explain":"H\u01b0\u1edbng d\u1eabn:<\/span><br\/>R\u00fat g\u1ecdn \u0111a th\u1ee9c $P(x)$ <br\/> Cho $P(x)=0$ r\u1ed3i t\u00ecm $x$ <br\/> B\u00e0i gi\u1ea3i:<\/span><br\/>Ta c\u00f3: <br\/> $\\begin{align} & P\\left( x \\right)=7,4x-8-\\left[ -5,6x+\\left( -2,4x+7 \\right)-11,92 \\right] \\\\ & \\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,=7,4x-8-\\left(-5,6x-2,4x+7-11,92 \\right) \\\\ & \\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,=7,4x-8-\\left(-8x-4,92 \\right) \\\\ & \\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,=7,4x-8+8x+4,92 \\\\ & \\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,=\\left( 7,4x+8x \\right)+\\left( -8+4,92 \\right) \\\\ & \\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,=15,4x-3,08 \\\\ \\end{align}$ <br\/> \u0110\u1ec3 $P(x)=0$ ta c\u00f3: <br\/> $\\begin{align} 15,4x-3,08 &=0 \\\\ 15,4x &=3,08 \\\\ x &=3,08:15,4=0,2 \\\\ \\end{align}$<\/span>"}]}],"id_ques":1234},{"time":24,"part":[{"title":"Ch\u1ecdn nh\u1eefng<\/u> \u0111\u00e1p \u00e1n \u0111\u00fang","title_trans":"Ch\u1ecdn \u0111\u01b0\u1ee3c nhi\u1ec1u \u0111\u00e1p \u00e1n","temp":"checkbox","correct":[["3","4"]],"list":[{"point":5,"img":"","ques":" Trong c\u00e1c gi\u00e1 tr\u1ecb sau, gi\u00e1 tr\u1ecb n\u00e0o l\u00e0 nghi\u1ec7m c\u1ee7a \u0111a th\u1ee9c $P(x)=6x^2-7x-3$?","hint":"","column":2,"number_true":2,"select":["A. $2$","B. $-3$","C. $\\dfrac{3}{2}$","D. $-\\dfrac{1}{3}$"],"explain":"H\u01b0\u1edbng d\u1eabn:<\/span><br\/>Thay gi\u00e1 tr\u1ecb c\u1ee7a $x$ \u1edf m\u1ed7i \u0111\u00e1p \u00e1n \u0111\u00e3 cho v\u00e0o \u0111a th\u1ee9c $P(x)$ <br\/> Gi\u00e1 tr\u1ecb n\u00e0o th\u1ecfa m\u00e3n $P(x)=0$ th\u00ec \u0111\u00f3 l\u00e0 nghi\u1ec7m c\u1ee7a \u0111a th\u1ee9c $P(x)$<br\/> B\u00e0i gi\u1ea3i:<\/span><br\/> $\\bullet $ Thay $x=2$ v\u00e0o $P(x)$ ta c\u00f3: <br\/> $6x^2-7x-3=6.2^2-7.2-3=7 \\ne 0$ <br\/> $\\Rightarrow x=2$ kh\u00f4ng ph\u1ea3i nghi\u1ec7m c\u1ee7a \u0111a th\u1ee9c $P(x)$ <br\/> $\\bullet $ Thay $x=3$ v\u00e0o $P(x)$ ta c\u00f3: <br\/> $6x^2-7x-3=6.3^2-7.3-3=30 \\ne 0$ <br\/> $\\Rightarrow x=3$ kh\u00f4ng ph\u1ea3i nghi\u1ec7m c\u1ee7a \u0111a th\u1ee9c $P(x)$ <br\/> $\\bullet $ Thay $x=\\dfrac{3}{2}$ v\u00e0o $P(x)$ ta c\u00f3: <br\/> $6x^2-7x-3=6.{{\\left( \\dfrac{3}{2} \\right)}^{2}}-7.\\dfrac{3}{2}-3=0$ <br\/> $\\Rightarrow x=\\dfrac{3}{2}$ l\u00e0 m\u1ed9t nghi\u1ec7m c\u1ee7a \u0111a th\u1ee9c $P(x)$ <br\/> $\\bullet $ Thay $x=-\\dfrac{1}{3}$ v\u00e0o $P(x)$ ta c\u00f3: <br\/> $6x^2-7x-3=6.{{\\left( -\\dfrac{1}{3} \\right)}^{2}}-7.\\left( -\\dfrac{1}{3} \\right)-3=0 $ <br\/> $\\Rightarrow x=-\\dfrac{1}{3}$ l\u00e0 m\u1ed9t nghi\u1ec7m c\u1ee7a \u0111a th\u1ee9c $P(x)$ <br\/> V\u1eady \u0111\u00e1p \u00e1n c\u1ea7n ch\u1ecdn l\u00e0 C v\u00e0 D.<\/span><\/span>"}]}],"id_ques":1235},{"time":24,"part":[{"title":"N\u1ed1i m\u1ed9t nghi\u1ec7m c\u1ee7a \u0111a th\u1ee9c \u1edf c\u1ed9t ph\u1ea3i t\u01b0\u01a1ng \u1ee9ng v\u1edbi \u0111a th\u1ee9c \u0111\u00e3 cho \u1edf c\u1ed9t tr\u00e1i","title_trans":"","audio":"","temp":"matching","correct":[["1","3","4","2"]],"list":[{"point":5,"image":"","left":["$P(x)=2x+1$","$Q(x)=5-2x$","$H(x)=\\dfrac{1}{3}x-2$","$G(x)=7x+3$"],"right":["$-\\dfrac{1}{2}$","$-\\dfrac{3}{7}$ ","$\\dfrac{5}{2}$","$6$"],"top":100,"hint":"","explain":" Ta c\u00f3: <br\/> $\\begin{aligned} & \\begin{matrix} \\begin{aligned} & \\bullet \\,\\,\\,P\\left( x \\right)=0 \\\\ & \\Rightarrow 2x+1=0 \\\\ & \\Rightarrow 2x=-1 \\\\ & \\Rightarrow x=-\\dfrac{1}{2} \\\\ \\end{aligned} & \\begin{aligned} & \\bullet \\,Q\\left( x \\right)=0 \\\\ & \\Rightarrow 5-2x=0 \\\\ & \\Rightarrow -2x=-5 \\\\ & \\Rightarrow x=\\dfrac{5}{2} \\\\ \\end{aligned} \\\\\\end{matrix} \\\\ & \\begin{matrix} \\begin{aligned}& \\bullet \\,H\\left( x \\right)=0 \\\\ & \\Rightarrow \\dfrac{1}{3}x-2=0 \\\\ & \\Rightarrow \\dfrac{1}{3}x=2 \\\\ & \\Rightarrow x=2:\\dfrac{1}{3}=6 \\\\ \\end{aligned} & \\begin{aligned} & \\bullet \\,G\\left( x \\right)=0 \\\\ & \\Rightarrow 7x+3=0 \\\\ & \\Rightarrow 7x=-3 \\\\ & \\Rightarrow x=-\\dfrac{3}{7} \\\\ & \\\\ & \\\\ \\end{aligned} \\\\\\end{matrix} \\\\ \\end{aligned}$ <\/span> "}]}],"id_ques":1236},{"time":24,"part":[{"title":"Kh\u1eb3ng \u0111\u1ecbnh sau \u0111\u00e2y \u0110\u00fang hay Sai","title_trans":"","temp":"multiple_choice","correct":[[1]],"list":[{"point":5,"ques":" \u0110a th\u1ee9c $G(x)=x^2+1$ kh\u00f4ng c\u00f3 nghi\u1ec7m. ","select":["\u0110\u00fang","Sai"],"explain":" Ta c\u00f3: $x^2 \\ge 0 \\Rightarrow x^2+1 \\ge 1$ v\u1edbi m\u1ecdi $x$ <br\/> Do \u0111\u00f3 $G(x)=x^2+1 \\ne 0$ v\u1edbi m\u1ecdi $x$ <br\/> \u0110a th\u1ee9c $G(x)=x^2+1$ kh\u00f4ng c\u00f3 nghi\u1ec7m. <br\/> V\u1eady \u0111\u00e1p \u00e1n l\u00e0 \u0110\u00fang.<\/span> <\/span>","column":2}]}],"id_ques":1237},{"time":24,"part":[{"title":"L\u1ef1a ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang nh\u1ea5t","title_trans":"","temp":"multiple_choice","correct":[[3]],"list":[{"point":5,"ques":" S\u1ed1 nghi\u1ec7m c\u1ee7a \u0111a th\u1ee9c $H(x)=x^2-25$ l\u00e0: <\/span>","select":["A. $0 $","B. $1$","C. $2$"],"explain":" Ta c\u00f3: <br\/> $\\begin{aligned} & \\,\\,\\,\\,H\\left( x \\right)=0 \\\\ & \\Rightarrow {{x}^{2}}-25=0 \\\\ & \\Rightarrow {{x}^{2}}=25 \\\\ & \\Rightarrow \\left[ \\begin{aligned} & x=5 \\\\ & x=-5 \\\\ \\end{aligned} \\right. \\\\ \\end{aligned}$. <br\/> \u0110a th\u1ee9c $H(x)=x^2-25$ c\u00f3 hai nghi\u1ec7m l\u00e0 $5$ v\u00e0 $-5$ <br\/> V\u1eady \u0111\u00e1p \u00e1n l\u00e0 C.<\/span><\/span>","column":3}]}],"id_ques":1238},{"time":24,"part":[{"title":"Kh\u1eb3ng \u0111\u1ecbnh sau \u0111\u00e2y \u0110\u00fang hay Sai","title_trans":"","temp":"multiple_choice","correct":[[1]],"list":[{"point":5,"ques":" X\u00e9t \u0111a th\u1ee9c $P(x)=ax^2+bx+c$ <br\/> N\u1ebfu $a+b+c=0$ th\u00ec $P(x)$ c\u00f3 m\u1ed9t nghi\u1ec7m l\u00e0 $x=1$<\/span> ","select":["\u0110\u00fang","Sai"],"explain":"H\u01b0\u1edbng d\u1eabn:<\/span><br\/> Ki\u1ec3m tra xem $x=1$ c\u00f3 th\u1ecfa m\u00e3n $P(x)=0$ kh\u00f4ng? <br\/> Thay $x=1$ v\u00e0o \u0111a th\u1ee9c $P(x)$ <br\/> B\u00e0i gi\u1ea3i:<\/span><br\/>Thay $x=1$ v\u00e0o \u0111a th\u1ee9c $P(x)$ ta \u0111\u01b0\u1ee3c: <br\/> $a.1^2+b.1+c=a+b+c=0$ <br\/> Do \u0111\u00f3 $x=1$ l\u00e0 nghi\u1ec7m c\u1ee7a $P(x)$ n\u1ebfu $a+b+c=0$.<br\/> V\u1eady \u0111\u00e1p \u00e1n l\u00e0 \u0110\u00fang.<\/span> <\/span>","column":2}]}],"id_ques":1239},{"time":24,"part":[{"title":"Kh\u1eb3ng \u0111\u1ecbnh sau \u0111\u00e2y \u0110\u00fang hay Sai","title_trans":"","temp":"multiple_choice","correct":[[1]],"list":[{"point":5,"ques":" X\u00e9t \u0111a th\u1ee9c $P(x)=ax^2+bx+c$ <br\/> N\u1ebfu $a-b+c=0$ th\u00ec $P(x)$ c\u00f3 m\u1ed9t nghi\u1ec7m l\u00e0 $x=-1$<\/span> ","select":["\u0110\u00fang","Sai"],"explain":"H\u01b0\u1edbng d\u1eabn:<\/span><br\/> Ki\u1ec3m tra xem $x=-1$ c\u00f3 th\u1ecfa m\u00e3n $P(x)=0$ kh\u00f4ng? <br\/> Thay $x=-1$ v\u00e0o \u0111a th\u1ee9c $P(x)$ <br\/> B\u00e0i gi\u1ea3i:<\/span><br\/>Thay $x=-1$ v\u00e0o \u0111a th\u1ee9c $P(x)$ ta \u0111\u01b0\u1ee3c: <br\/> $a.(-1)^2+b.(-1)+c=a-b+c =0$ <br\/> Do \u0111\u00f3 $x=-1$ l\u00e0 nghi\u1ec7m c\u1ee7a $P(x)$ n\u1ebfu $a-b+c=0$.<br\/> V\u1eady \u0111\u00e1p \u00e1n l\u00e0 \u0110\u00fang.<\/span> <\/span>","column":2}]}],"id_ques":1240}],"lesson":{"save":0,"level":2}}

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