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Home Lớp 6 Toán lớp 6 sách cũ Kiểm tra hình chương I - Lớp 6

{"segment":[{"part":[{"title":"L\u1ef1a ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang nh\u1ea5t","title_trans":"","temp":"multiple_choice","correct":[[2]],"list":[{"point":10,"ques":" \u0110i\u1ec3m I l\u00e0 trung \u0111i\u1ec3m c\u1ee7a \u0111o\u1ea1n th\u1eb3ng MN n\u1ebfu:","select":["A. \u0110i\u1ec3m I c\u00e1ch \u0111\u1ec1u hai \u0111i\u1ec3m M, N","B. \u0110i\u1ec3m I n\u1eb1m gi\u1eefa hai \u0111i\u1ec3m M, N v\u00e0 I c\u00e1ch \u0111\u1ec1u hai \u0111i\u1ec3m M, N ","C. \u0110i\u1ec3m I n\u1eb1m gi\u1eefa hai \u0111i\u1ec3m M v\u00e0 N","D. C\u1ea3 3 \u0111\u00e1p \u00e1n tr\u00ean \u0111\u1ec1u \u0111\u00fang "],"hint":"","explain":"B\u00e0i gi\u1ea3i:<\/span> <br\/> \u0110i\u1ec3m I l\u00e0 trung \u0111i\u1ec3m c\u1ee7a \u0111o\u1ea1n th\u1eb3ng MN n\u1ebfu: <br\/> +) \u0110i\u1ec3m I c\u00e1ch \u0111\u1ec1u hai \u0111i\u1ec3m M, N <br\/> +) \u0110i\u1ec3m I n\u1eb1m gi\u1eefa hai \u0111i\u1ec3m M v\u00e0 N <br\/> V\u1eady \u0111\u00e1p \u00e1n \u0111\u00fang l\u00e0: B <\/span> ","column":2}],"id_ques":2071},{"title":"L\u1ef1a ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang nh\u1ea5t","title_trans":"","temp":"multiple_choice","correct":[[3]],"list":[{"point":10,"ques":"N\u1ebfu \u0111i\u1ec3m I n\u1eb1m gi\u1eefa hai \u0111i\u1ec3m A v\u00e0 B th\u00ec:","select":["A. IA + AB = IB","B. IB + AB = IA ","C. IA + IB = AB","D. \u0110\u00e1p \u00e1n kh\u00e1c "],"hint":"","explain":"B\u00e0i gi\u1ea3i:<\/span> <br\/> \u0110i\u1ec3m I n\u1eb1m gi\u1eefa hai \u0111i\u1ec3m A v\u00e0 B n\u1ebfu: <br\/> IA + IB = AB <br\/> V\u1eady \u0111\u00e1p \u00e1n \u0111\u00fang l\u00e0: C <\/span> ","column":2}],"id_ques":2072},{"title":"L\u1ef1a ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang nh\u1ea5t","title_trans":"","temp":"multiple_choice","correct":[[3]],"list":[{"point":10,"ques":"Cho \u0111o\u1ea1n th\u1eb3ng AB = 6cm. \u0110i\u1ec3m M l\u00e0 trung \u0111i\u1ec3m c\u1ee7a \u0111o\u1ea1n th\u1eb3ng AB th\u00ec:","select":["A. AM = 4cm","B. AM = 2,5cm ","C. AM = 3cm","D. AM = 2cm "],"hint":"","explain":"B\u00e0i gi\u1ea3i:<\/span> <br\/> V\u00ec \u0111i\u1ec3m M l\u00e0 trung \u0111i\u1ec3m c\u1ee7a \u0111o\u1ea1n th\u1eb3ng AB n\u00ean: <br\/> $AM = MB = \\dfrac{AB}{2} = \\dfrac{6}{2} = 3$ (cm) <br\/> V\u1eady \u0111\u00e1p \u00e1n \u0111\u00fang l\u00e0: C. AM = 3cm <\/span> ","column":2}],"id_ques":2073},{"title":"L\u1ef1a ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang nh\u1ea5t","title_trans":"","temp":"multiple_choice","correct":[[1]],"list":[{"point":10,"ques":"N\u1ebfu MI + AI = AM th\u00ec:","select":["A. \u0110i\u1ec3m I n\u1eb1m gi\u1eefa hai \u0111i\u1ec3m A v\u00e0 M","B. \u0110i\u1ec3m A n\u1eb1m gi\u1eefa hai \u0111i\u1ec3m M v\u00e0 I ","C. \u0110i\u1ec3m M n\u1eb1m gi\u1eefa hai \u0111i\u1ec3m A v\u00e0 I","D. Kh\u00f4ng c\u00f3 \u0111i\u1ec3m n\u00e0o n\u1eb1m gi\u1eefa hai \u0111i\u1ec3m c\u00f2n l\u1ea1i"],"hint":"","explain":"B\u00e0i gi\u1ea3i:<\/span> <br\/> N\u1ebfu MI + AI = AM th\u00ec 3 \u0111i\u1ec3m A, I, M th\u1eb3ng h\u00e0ng <br\/> V\u00e0 \u0111i\u1ec3m I n\u1eb1m gi\u1eefa hai \u0111i\u1ec3m A v\u00e0 M <br\/> V\u1eady \u0111\u00e1p \u00e1n \u0111\u00fang l\u00e0: A <\/span> ","column":2}],"id_ques":2074},{"title":"L\u1ef1a ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang nh\u1ea5t","title_trans":"","temp":"multiple_choice","correct":[[3]],"list":[{"point":10,"ques":"Cho tia Ax, Ay l\u00e0 hai tia \u0111\u1ed1i nhau. L\u1ea5y \u0111i\u1ec3m N tr\u00ean tia Ax, M tr\u00ean tia Ay ta c\u00f3:","select":["A. \u0110i\u1ec3m M n\u1eb1m gi\u1eefa hai \u0111i\u1ec3m A v\u00e0 N","B. \u0110i\u1ec3m N n\u1eb1m gi\u1eefa hai \u0111i\u1ec3m A v\u00e0 M ","C. \u0110i\u1ec3m A n\u1eb1m gi\u1eefa hai \u0111i\u1ec3m M v\u00e0 N","D. Kh\u00f4ng c\u00f3 \u0111i\u1ec3m n\u00e0o n\u1eb1m gi\u1eefa hai \u0111i\u1ec3m c\u00f2n l\u1ea1i"],"hint":"","explain":"B\u00e0i gi\u1ea3i:<\/span> <br\/> V\u00ec Ax v\u00e0 Ay l\u00e0 hai tia \u0111\u1ed1i nhau (1) <br\/> N $\\in$ Ax, M $\\in$ Ay (2) <br\/> T\u1eeb (1) v\u00e0 (2) $\\Rightarrow$ \u0110i\u1ec3m A n\u1eb1m gi\u1eefa hai \u0111i\u1ec3m M v\u00e0 N <br\/> V\u1eady \u0111\u00e1p \u00e1n \u0111\u00fang l\u00e0: C <\/span> ","column":2}],"id_ques":2075},{"title":"L\u1ef1a ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang nh\u1ea5t","title_trans":"","temp":"multiple_choice","correct":[[2]],"list":[{"point":10,"ques":"Cho h\u00ecnh v\u1ebd: <br\/> <center><img src='https://www.luyenthi123.com/file/luyenthi123/lop6/toan/hinhhoc/kiemtra1/lv3/img\/H618_D101.png' \/><\/center> <br\/> Khi \u0111\u00f3:","select":["A. Tia BA v\u00e0 tia CA tr\u00f9ng nhau","B. Tia CA v\u00e0 tia CD \u0111\u1ed1i nhau ","C. Tia AB v\u00e0 tia BA \u0111\u1ed1i nhau","D. Tia BA v\u00e0 tia CD \u0111\u1ed1i nhau"],"hint":"","explain":"B\u00e0i gi\u1ea3i:<\/span> <br\/> +) V\u00ec tia BA v\u00e0 tia CA kh\u00f4ng chung g\u1ed1c <br\/> N\u00ean tia BA v\u00e0 tia CA kh\u00f4ng \u0111\u1ed1i nhau (lo\u1ea1i) <\/b> <br\/> +) Tia CA v\u00e0 tia CD c\u00f3 chung g\u1ed1c C v\u00e0 hai tia t\u1ea1o th\u00e0nh m\u1ed9t \u0111\u01b0\u1eddng th\u1eb3ng <br\/> N\u00ean tia CA v\u00e0 tia CD \u0111\u1ed1i nhau (th\u1ecfa m\u00e3n) <\/b> <br\/> +) Tia AB v\u00e0 tia BA kh\u00f4ng chung g\u1ed1c <br\/> N\u00ean tia AB v\u00e0 tia BA kh\u00f4ng \u0111\u1ed1i nhau (lo\u1ea1i) <\/b> <br\/> +) Tia BA v\u00e0 tia CD kh\u00f4ng chung g\u1ed1c <br\/> N\u00ean tia BA v\u00e0 tia CD kh\u00f4ng \u0111\u1ed1i nhau (lo\u1ea1i) <\/b> <br\/> V\u1eady \u0111\u00e1p \u00e1n \u0111\u00fang l\u00e0: B <\/span> ","column":2}],"id_ques":2076},{"title":"L\u1ef1a ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang nh\u1ea5t","title_trans":"","temp":"multiple_choice","correct":[[2]],"list":[{"point":10,"ques":"Cho 3 \u0111i\u1ec3m A, B, C th\u1eb3ng h\u00e0ng, \u0111i\u1ec3m C n\u1eb1m gi\u1eefa hai \u0111i\u1ec3m A v\u00e0 B. Kh\u1eb3ng \u0111\u1ecbnh n\u00e0o sau \u0111\u00e2y \u0111\u00fang:","select":["A. Hai \u0111i\u1ec3m A v\u00e0 B n\u1eb1m c\u00f9ng ph\u00eda \u0111\u1ed1i v\u1edbi \u0111i\u1ec3m C","B. Hai \u0111i\u1ec3m A v\u00e0 B n\u1eb1m kh\u00e1c ph\u00eda \u0111\u1ed1i v\u1edbi \u0111i\u1ec3m C","C. Hai \u0111i\u1ec3m A v\u00e0 C n\u1eb1m kh\u00e1c ph\u00eda \u0111\u1ed1i v\u1edbi \u0111i\u1ec3m B","D. Hai \u0111i\u1ec3m B v\u00e0 C n\u1eb1m kh\u00e1c ph\u00eda \u0111\u1ed1i v\u1edbi \u0111i\u1ec3m A"],"hint":"","explain":"B\u00e0i gi\u1ea3i:<\/span> <br\/> <center><img src='https://www.luyenthi123.com/file/luyenthi123/lop6/toan/hinhhoc/kiemtra1/lv3/img\/H618_D102.png' \/><\/center> <br\/> V\u00ec \u0111i\u1ec3m C n\u1eb1m gi\u1eefa hai \u0111i\u1ec3m A v\u00e0 B n\u00ean: <br\/> +) Hai \u0111i\u1ec3m A v\u00e0 B n\u1eb1m kh\u00e1c ph\u00eda \u0111\u1ed1i v\u1edbi \u0111i\u1ec3m C <br\/> +) Hai \u0111i\u1ec3m A v\u00e0 C n\u1eb1m c\u00f9ng ph\u00eda \u0111\u1ed1i v\u1edbi \u0111i\u1ec3m B <br\/> +) Hai \u0111i\u1ec3m B v\u00e0 C n\u1eb1m c\u00f9ng ph\u00eda \u0111\u1ed1i v\u1edbi \u0111i\u1ec3m A <br\/> V\u1eady \u0111\u00e1p \u00e1n \u0111\u00fang l\u00e0: B <\/span> ","column":2}],"id_ques":2077},{"title":"\u0110i\u1ec1n s\u1ed1 th\u00edch h\u1ee3p v\u00e0o ch\u1ed7 tr\u1ed1ng","title_trans":"","temp":"fill_the_blank","correct":[[["20"]]],"list":[{"point":10,"width":80,"content":"","type_input":"","ques":"Cho n \u0111i\u1ec3m trong \u0111\u00f3 c\u00f3 12 \u0111i\u1ec3m th\u1eb3ng h\u00e0ng, ngo\u00e0i ra kh\u00f4ng c\u00f3 3 \u0111i\u1ec3m th\u1eb3ng h\u00e0ng. Qua hai \u0111i\u1ec3m k\u1ebb \u0111\u01b0\u1ee3c m\u1ed9t \u0111\u01b0\u1eddng th\u1eb3ng. Bi\u1ebft k\u1ebb \u0111\u01b0\u1ee3c t\u1ea5t c\u1ea3 125 \u0111\u01b0\u1eddng th\u1eb3ng. <br\/> V\u1eady n = _input_","hint":"D\u1ef1a v\u00e0o c\u00f4ng th\u1ee9c: S\u1ed1 \u0111\u01b0\u1eddng th\u1eb3ng = $\\dfrac{n . (n - 1)}{2}$","explain":"H\u01b0\u1edbng d\u1eabn:<\/span><br\/> B\u01b0\u1edbc 1: <\/b> T\u00ednh s\u1ed1 \u0111\u01b0\u1eddng th\u1eb3ng v\u1ebd \u0111\u01b0\u1ee3c qua n \u0111i\u1ec3m: <br\/> $\\dfrac{n . (n - 1)}{2}$ (\u0111\u01b0\u1eddng th\u1eb3ng) <br\/> B\u01b0\u1edbc 2: <\/b> T\u00ednh s\u1ed1 \u0111\u01b0\u1eddng th\u1eb3ng v\u1ebd \u0111\u01b0\u1ee3c qua 12 \u0111i\u1ec3m kh\u00f4ng th\u1eb3ng h\u00e0ng: $\\dfrac{12. (12 - 1)}{2}$ <br\/> B\u01b0\u1edbc 3: <\/b>V\u00ec qua 12 \u0111i\u1ec3m ch\u1ec9 v\u1ebd \u0111\u01b0\u1ee3c m\u1ed9t \u0111\u01b0\u1eddng th\u1eb3ng <br\/> V\u00ec t\u1ea5t c\u1ea3 v\u1ebd \u0111\u01b0\u1ee3c 125 \u0111\u01b0\u1eddng th\u1eb3ng n\u00ean ta c\u00f3: <br\/> $\\dfrac{n . (n - 1)}{2} - \\dfrac{12 . (12 - 1)}{2} + 1 = 125$ <br\/> T\u1eeb \u0111\u00f3 ta t\u00ecm \u0111\u01b0\u1ee3c n <\/span> <br\/><br\/> B\u00e0i gi\u1ea3i:<\/span><br\/> Gi\u1ea3 s\u1eed trong n \u0111i\u1ec3m kh\u00f4ng c\u00f3 3 \u0111i\u1ec3m n\u00e0o th\u1eb3ng h\u00e0ng <br\/> V\u1eady t\u1ea5t c\u1ea3 v\u1ebd \u0111\u01b0\u1ee3c s\u1ed1 \u0111\u01b0\u1eddng th\u1eb3ng l\u00e0: <br\/> $\\dfrac{n . (n - 1)}{2}$ (\u0111\u01b0\u1eddng th\u1eb3ng) <br\/> Gi\u1ea3 s\u1eed trong 12 \u0111i\u1ec3m kh\u00f4ng c\u00f3 3 \u0111i\u1ec3m n\u00e0o th\u1eb3ng h\u00e0ng th\u00ec v\u1ebd \u0111\u01b0\u1ee3c: <br\/> $\\dfrac{12 . (12 - 1)}{2} = 66$ (\u0111\u01b0\u1eddng th\u1eb3ng) <br\/> Nh\u01b0ng qua 12 \u0111i\u1ec3m th\u1eb3ng h\u00e0ng ch\u1ec9 v\u1ebd \u0111\u01b0\u1ee3c m\u1ed9t \u0111\u01b0\u1eddng th\u1eb3ng <br\/> N\u00ean qua n \u0111i\u1ec3m trong \u0111\u00f3 c\u00f3 12 \u0111i\u1ec3m th\u1eb3ng h\u00e0ng v\u1ebd \u0111\u01b0\u1ee3c: <br\/> $\\dfrac{n . (n - 1)}{2} - 66 + 1$ (\u0111\u01b0\u1eddng th\u1eb3ng) <br\/> V\u00ec k\u1ebb \u0111\u01b0\u1ee3c t\u1ea5t c\u1ea3 125 \u0111\u01b0\u1eddng th\u1eb3ng n\u00ean ta c\u00f3: <br\/>$\\begin{align*} \\dfrac{n . (n - 1)}{2} - 66 + 1 &= 125 \\\\ \\dfrac{n . (n - 1)}{2} &= 125 + 66 - 1 \\\\ \\dfrac{n . (n - 1)}{2} &= 190 \\\\ n . (n - 1) &= 190 . 2 \\\\ n . (n - 1) &= 20 . 19 \\end{align*}$ <br\/> V\u1eady n = 20 <br\/> V\u1eady s\u1ed1 c\u1ea7n \u0111i\u1ec1n l\u00e0: 20 <\/span>"}],"id_ques":2078},{"title":"\u0110i\u1ec1n s\u1ed1 th\u00edch h\u1ee3p v\u00e0o ch\u1ed7 tr\u1ed1ng","title_trans":"","temp":"fill_the_blank","correct":[[["435"]]],"list":[{"point":10,"width":80,"content":"","type_input":"","ques":"Cho 30 \u0111i\u1ec3m trong \u0111\u00f3 c\u00f3 kh\u00f4ng c\u00f3 3 \u0111i\u1ec3m n\u00e0o th\u1eb3ng h\u00e0ng. Qua hai \u0111i\u1ec3m k\u1ebb \u0111\u01b0\u1ee3c m\u1ed9t \u0111\u01b0\u1eddng th\u1eb3ng. H\u1ecfi k\u1ebb \u0111\u01b0\u1ee3c t\u1ea5t c\u1ea3 bao nhi\u00eau \u0111\u01b0\u1eddng th\u1eb3ng. <br\/> \u0110\u00e1p \u00e1n l\u00e0: _input_ \u0111\u01b0\u1eddng th\u1eb3ng","hint":"D\u1ef1a v\u00e0o c\u00f4ng th\u1ee9c: S\u1ed1 \u0111\u01b0\u1eddng th\u1eb3ng = $\\dfrac{n . (n - 1)}{2}$","explain":" B\u00e0i gi\u1ea3i:<\/span><br\/> V\u00ec trong 30 \u0111i\u1ec3m kh\u00f4ng c\u00f3 3 \u0111i\u1ec3m n\u00e0o th\u1eb3ng h\u00e0ng n\u00ean v\u1ebd \u0111\u01b0\u1ee3c: <br\/> $\\dfrac{30 . (30 - 1)}{2} = 435$ (\u0111\u01b0\u1eddng th\u1eb3ng) <br\/> V\u1eady s\u1ed1 c\u1ea7n \u0111i\u1ec1n l\u00e0: 435 <\/span>"}],"id_ques":2079},{"title":"\u0110i\u1ec1n s\u1ed1 th\u00edch h\u1ee3p v\u00e0o ch\u1ed7 tr\u1ed1ng","title_trans":"","temp":"fill_the_blank","correct":[[["1"]]],"list":[{"point":10,"width":80,"content":"","type_input":"","ques":"Cho \u0111o\u1ea1n th\u1eb3ng AB = 6cm. L\u1ea5y hai \u0111i\u1ec3m E v\u00e0 F n\u1eb1m gi\u1eefa hai \u0111i\u1ec3m A v\u00e0 B sao cho AE + BF = 7cm. T\u00ednh \u0111\u1ed9 d\u00e0i EF. <br\/> \u0110\u00e1p \u00e1n l\u00e0: EF = _input_ cm","hint":"","explain":" B\u00e0i gi\u1ea3i:<\/span><br\/> T\u00f3m t\u1eaft: <\/b> <br\/> AB = 6cm <br\/> E, F thu\u1ed9c \u0111o\u1ea1n th\u1eb3ng AB <br\/> AE + BF = 7cm <br\/> T\u00ednh EF = ? <\/span> <br\/> <center><img src='https://www.luyenthi123.com/file/luyenthi123/lop6/toan/hinhhoc/kiemtra1/lv3/img\/H618_D103.png' \/><\/center> <br\/> V\u00ec \u0111i\u1ec3m E n\u1eb1m gi\u1eefa hai \u0111i\u1ec3m A v\u00e0 B (theo \u0111\u1ec1 b\u00e0i) <br\/> N\u00ean AE + BE = AB = 6cm (1) <br\/> M\u1eb7t kh\u00e1c AE + BF = 7cm (2) <br\/> T\u1eeb (1) v\u00e0 (2) $\\Rightarrow$ BE < BF <br\/> Tr\u00ean tia BA c\u00f3 BE < BF n\u00ean \u0111i\u1ec3m E n\u1eb1m gi\u1eefa hai \u0111i\u1ec3m B v\u00e0 F <br\/> $\\Rightarrow$ BE + EF = BF (3) <br\/> Ta c\u00f3: AE + BF = 7cm (4) <br\/> Thay (3) v\u00e0o (4) ta \u0111\u01b0\u1ee3c: <br\/> $\\begin{align*} AE + (BE + EF) &= 7cm \\\\ AE + BE + EF &= 7cm \\\\ AB + EF &= 7cm \\\\ 6cm + EF &= 7cm \\\\ EF &= 7cm - 6cm \\\\ EF &= 1cm \\end{align*}$ <br\/> V\u1eady EF = 1cm <\/span>"}],"id_ques":2080}],"id_ques":0}],"lesson":{"save":1,"level":3,"time":44}}

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