{"common":{"save":0,"post_id":"7222","level":3,"total":10,"point":10,"point_extra":0},"segment":[{"id":"5322","post_id":"7222","mon_id":"1158924","chapter_id":"1159163","question":"<p>Trong các m\u1ec7nh \u0111\u1ec1 sau, m\u1ec7nh \u0111\u1ec1 nào <strong>sai<\/strong>?<\/p>","options":["<strong>A.<\/strong> <span class=\"math-tex\">$\\lim_\\limits{x\\to-\\infty}(\\sqrt{x^2-x+1}+x-2)=\\dfrac{-3}{2}$<\/span>","<strong>B.<\/strong> <span class=\"math-tex\">$\\lim_\\limits{x\\to+\\infty}(\\sqrt{x^2-x+1}+x-2)=+\\infty $<\/span>","<strong>C.<\/strong> <span class=\"math-tex\">$\\lim_\\limits{x\\to-1^-}\\dfrac{3x+2}{x+1}=-\\infty $<\/span>","<strong>D.<\/strong> <span class=\"math-tex\">$\\lim_\\limits{x\\to-1^+}\\dfrac{3x+2}{x+1}=-\\infty $<\/span>"],"correct":"3","level":"3","hint":"","answer":"<p>Ch\u1ecdn <span style=\"color:#16a085;\"><strong>C.<\/strong> <span class=\"math-tex\">$\\lim_\\limits{x\\to-1^-}\\dfrac{3x+2}{x+1}=-\\infty $<\/span><\/span><\/p><p><strong>A.<\/strong> <span class=\"math-tex\">$\\lim_\\limits{x\\to-\\infty}(\\sqrt{x^2-x+1}+x-2)=\\dfrac{-3}{2}$<\/span> <span style=\"color:#16a085;\">\u0111úng<\/span><\/p><p>vì <span class=\"math-tex\">$\\lim_\\limits{x\\to-\\infty}(\\sqrt{x^2-x+1}+x-2)= \\lim_\\limits{x\\to-\\infty}\\dfrac{3x-3}{\\sqrt{x^2-x+1}-x+2}$<\/span><\/p><p><span class=\"math-tex\">$= \\lim_\\limits{x\\to-\\infty}\\dfrac{3-\\dfrac{3}{x}}{-\\sqrt{1-\\dfrac{1}{x}+\\dfrac{1}{x^2}}-1+\\dfrac{2}{x}} =\\dfrac{-3}{2}$<\/span><\/p><p><strong>B.<\/strong> <span class=\"math-tex\">$\\lim_\\limits{x\\to+\\infty}(\\sqrt{x^2-x+1}+x-2)=+\\infty $<\/span> <span style=\"color:#16a085;\">\u0111úng<\/span><\/p><p>vì <span class=\"math-tex\">$\\lim_\\limits{x\\to+\\infty}(\\sqrt{x^2-x+1}+x-2)= \\lim_\\limits{x\\to+\\infty}\\dfrac{3x-3}{\\sqrt{x^2-x+1}-x+2}$<\/span><\/p><p><span class=\"math-tex\">$= \\lim_\\limits{x\\to+\\infty}\\dfrac{3-\\dfrac{3}{x}}{\\sqrt{1-\\dfrac{1}{x}+\\dfrac{1}{x^2}}-1+\\dfrac{2}{x}} = \\lim_\\limits{x\\to+\\infty}\\dfrac{3}{0}=+\\infty $<\/span><\/p><p><strong>C.<\/strong> <span class=\"math-tex\">$\\lim_\\limits{x\\to-1^-}\\dfrac{3x+2}{x+1}=-\\infty $<\/span> <span style=\"color:#e74c3c;\">sai <\/span><\/p><p><span style=\"color:#e74c3c;\"><\/span>vì <span class=\"math-tex\">$\\lim_\\limits{x\\to-1^-}(3x+2)=-1<0$<\/span>; <span class=\"math-tex\">$\\lim_\\limits{x\\to-1^-}(x+1)=0$<\/span> và x + 1 < 0 v\u1edbi m\u1ecdi x < –1<\/p><p>nên <span class=\"math-tex\">$\\lim_\\limits{x\\to-1^-}\\dfrac{3x+2}{x+1}=+\\infty $<\/span><\/p><p><strong>D.<\/strong> <span class=\"math-tex\">$\\lim_\\limits{x\\to-1^+}\\dfrac{3x+2}{x+1}=-\\infty $<\/span> <span style=\"color:#16a085;\">\u0111úng <\/span><\/p><p><span style=\"color:#16a085;\"><\/span>vì <span class=\"math-tex\">$\\lim_\\limits{x\\to-1^-}(3x+2)=-1<0$<\/span>; <span class=\"math-tex\">$\\lim_\\limits{x\\to-1^-}(x+1)=0$<\/span> và x + 1 > 0 v\u1edbi m\u1ecdi x > –1<\/p>","type":"choose","extra_type":"classic","user_id":"131","test":"0","date":"2024-03-07 01:38:25","option_type":"math","len":0},{"id":"5325","post_id":"7222","mon_id":"1158924","chapter_id":"1159163","question":"<p>Tính gi\u1edbi h\u1ea1n <span class=\"math-tex\">$K= \\lim_\\limits{x\\to 0}\\dfrac{\\sqrt{4x+1}-1}{x^2-3x}$<\/span>.<\/p>","options":["<strong>A.<\/strong> <span class=\"math-tex\">$K=-\\dfrac{2}{3}$<\/span>","<strong>B.<\/strong> <span class=\"math-tex\">$K=\\dfrac{2}{3}$<\/span>","<strong>C.<\/strong> <span class=\"math-tex\">$K=\\dfrac{4}{3}$<\/span>","<strong>D.<\/strong> <span class=\"math-tex\">$K=0$<\/span>"],"correct":"1","level":"3","hint":"","answer":"<p>Ch\u1ecdn <span style=\"color:#16a085;\"><strong>A.<\/strong> <span class=\"math-tex\">$K=-\\dfrac{2}{3}$<\/span><\/span><\/p><p><span class=\"math-tex\">$K= \\lim_\\limits{x\\to 0}\\dfrac{\\sqrt{4x+1}-1}{x^2-3x}$<\/span> = <span class=\"math-tex\">$\\lim_\\limits{x\\to 0}\\dfrac{4x}{x(x-3)(\\sqrt{4x+1}+1)}$<\/span> <\/p><p>= <span class=\"math-tex\">$\\lim_\\limits{x\\to 0}\\dfrac{4}{(x-3)(\\sqrt{4x+1}+1)}$<\/span> = <span class=\"math-tex\">$-\\dfrac{2}{3}$<\/span><\/p>","type":"choose","extra_type":"classic","user_id":"131","test":"0","date":"2024-03-07 01:57:26","option_type":"math","len":0},{"id":"5326","post_id":"7222","mon_id":"1158924","chapter_id":"1159163","question":"<p>Cho hàm s\u1ed1 <span class=\"math-tex\">$f(x)=\\begin{cases}ax^2+bx&\\text{khi}\\space x\\ge1\\\\2x-1&\\text{khi}\\space x<1\\end{cases}$<\/span>. \u0110\u1ec3 hàm s\u1ed1 \u0111ã cho có \u0111\u1ea1o hàm t\u1ea1i x = 1 thì 2a + b b\u1eb1ng:<\/p>","options":["<strong>A.<\/strong> 5","<strong>B.<\/strong> 2","<strong>C.<\/strong> –2","<strong>D.<\/strong> –5"],"correct":"2","level":"3","hint":"","answer":"<p>Ch\u1ecdn <span style=\"color:#16a085;\"><strong>B.<\/strong> 2<\/span><\/p><p><span class=\"math-tex\">$\\lim_\\limits{x\\to1^-}\\dfrac{f(x)-f(1)}{x-1} =\\lim_\\limits{x\\to1^-}\\dfrac{2x-1-1}{x-1}=2$<\/span><\/p><p><span class=\"math-tex\">$\\lim_\\limits{x\\to1^+}\\dfrac{f(x)-f(1)}{x-1} =\\lim_\\limits{x\\to1^+}\\dfrac{ax^2+bx-a-b}{x-1} =\\lim_\\limits{x\\to1^+}\\dfrac{a(x^2-1)+b(x-1)}{x-1}$<\/span><\/p><p><span class=\"math-tex\">$=\\lim_\\limits{x\\to1^+}\\dfrac{(x-1)[a(x+1)+b]}{x-1}\\n=\\lim_\\limits{x\\to1^+}[a(x+1)+b]=2a+b$<\/span><\/p><p>Theo yêu c\u1ea7u bài toán thì<\/p><p> <span class=\"math-tex\">$\\lim_\\limits{x\\to1^-}\\dfrac{f(x)-f(1)}{x-1}\\n=\\lim_\\limits{x\\to1^+}\\dfrac{f(x)-f(1)}{x-1}$<\/span><span style=\"color:#16a085;\"><span class=\"math-tex\">$\\Leftrightarrow 2a+b=2$<\/span><\/span><\/p>","type":"choose","extra_type":"shape1","user_id":"131","test":"0","date":"2024-03-07 02:03:45","option_type":"txt","len":0}]}