{"common":{"save":0,"post_id":"7114","level":3,"total":10,"point":10,"point_extra":0},"segment":[{"id":"5918","post_id":"7114","mon_id":"1158924","chapter_id":"1159187","question":"<p>Cho h&agrave;m s\u1ed1&nbsp;<span class=\"math-tex\">$f(x)=\\begin{cases}\\dfrac{3-x^2}{2}\\text{ khi }x&lt;1\\\\\\dfrac{1}{x}\\text{ khi }x\\ge1\\end{cases}$<\/span>. Nh\u1eadn x&eacute;t&nbsp;n&agrave;o d\u01b0\u1edbi \u0111&acirc;y l&agrave; <strong>sai<\/strong>?<\/p>","options":["<strong>A.<\/strong> H&agrave;m s\u1ed1 li&ecirc;n t\u1ee5c t\u1ea1i x = 1.","<strong>B.<\/strong> H&agrave;m s\u1ed1 c&oacute; \u0111\u1ea1o h&agrave;m t\u1ea1i x = 1.","<strong>C.<\/strong> H&agrave;m s\u1ed1 li&ecirc;n t\u1ee5c v&agrave; c&oacute; \u0111\u1ea1o h&agrave;m t\u1ea1i x = 1.","<strong>D.<\/strong> H&agrave;m s\u1ed1 kh&ocirc;ng c&oacute; \u0111\u1ea1o h&agrave;m t\u1ea1i x = 1."],"correct":"4","level":"3","hint":"","answer":"<p>C&acirc;u <strong>sai <\/strong>l&agrave;<span style=\"color:#16a085;\">&nbsp;<strong>D.<\/strong> H&agrave;m s\u1ed1 kh&ocirc;ng c&oacute; \u0111\u1ea1o h&agrave;m t\u1ea1i x = 1.<\/span><\/p><p><span class=\"math-tex\">$\\displaystyle\\lim_{x\\to1^-}f(x)=\\displaystyle\\lim_{x\\to1^-}\\dfrac{3-x^2}{2}=1$<\/span>&nbsp;v&agrave;&nbsp;<span class=\"math-tex\">$\\displaystyle\\lim_{x\\to1^+}f(x)=\\displaystyle\\lim_{x\\to1^+}\\dfrac{1}{x}=1$<\/span>.<br \/>Do \u0111&oacute; h&agrave;m s\u1ed1 li&ecirc;n t\u1ee5c t\u1ea1i x = 1.<\/p><p><span class=\"math-tex\">$\\displaystyle\\lim_{x\\to1^-}\\dfrac{f(x)-f(1)}{x-1}=\\displaystyle\\lim_{x\\to1^-}\\dfrac{1-x^2}{2(x-1)}=\\displaystyle\\lim_{x\\to1^-}\\dfrac{1+x}{-2}=-1$<\/span>&nbsp;v&agrave;<br \/><span class=\"math-tex\">$\\displaystyle\\lim_{x\\to1^+}\\dfrac{f(x)-f(1)}{x-1}=\\displaystyle\\lim_{x\\to1^+}\\dfrac{1-x}{x(x-1)}=\\displaystyle\\lim_{x\\to1^+}\\dfrac{-1}{x}=-1$<\/span>.<\/p><p>Do \u0111&oacute;&nbsp;h&agrave;m s\u1ed1 f(x) c&oacute; \u0111\u1ea1o h&agrave;m t\u1ea1i x = 1.<\/p><p>V\u1eady c&aacute;c nh\u1eadn x&eacute;t <strong>A<\/strong>, <strong>B<\/strong>, <strong>C<\/strong> \u0111\u1ec1u \u0111&uacute;ng. Nh\u1eadn x&eacute;t <strong>D<\/strong> sai.<\/p>","type":"choose","extra_type":"classic","user_id":"131","test":"0","date":"2024-05-29 04:26:59","option_type":"txt","len":3},{"id":"5920","post_id":"7114","mon_id":"1158924","chapter_id":"1159187","question":"<p>Cho h&agrave;m s\u1ed1&nbsp;<span class=\"math-tex\">$f(x)=\\dfrac{3x}{1+|x|}$<\/span>. T&iacute;nh&nbsp;<span class=\"math-tex\">$f^\\prime(0)$<\/span>.<\/p>","options":["<strong>A.<\/strong> 0","<strong>B.<\/strong> 3","<strong>C.<\/strong> 1","<strong>D.<\/strong> &ndash;3"],"correct":"2","level":"3","hint":"","answer":"<p>Ch\u1ecdn&nbsp;<span style=\"color:#16a085;\"><strong>B.<\/strong> 3.<\/span><\/p><p>Ta c&oacute;&nbsp;<span class=\"math-tex\">$f^\\prime(0)=\\displaystyle\\lim_{x\\to0}\\dfrac{f(x)-f(0)}{x}=\\displaystyle\\lim_{x\\to0}\\dfrac{3}{1+|x|}$<\/span>.<\/p><p>M&agrave;&nbsp;<span class=\"math-tex\">$\\displaystyle\\lim_{x\\to0^+}\\dfrac{3}{1+|x|}=\\displaystyle\\lim_{x\\to0^+}\\dfrac{3}{1+x}=3$<\/span>;&nbsp;<span class=\"math-tex\">$\\displaystyle\\lim_{x\\to0^-}\\dfrac{3}{1+|x|}=\\displaystyle\\lim_{x\\to0^-}\\dfrac{3}{1-x}=3$<\/span><\/p><p>&rArr;&nbsp;<span class=\"math-tex\">$\\displaystyle\\lim_{x\\to0^+}\\dfrac{3}{1+|x|}=\\displaystyle\\lim_{x\\to0^-}\\dfrac{3}{1+|x|}=3$<\/span><\/p><p>&rArr;&nbsp;<span class=\"math-tex\">$f^\\prime(0)=\\displaystyle\\lim_{x\\to0}\\dfrac{3}{1+|x|}=3$<\/span>.<\/p>","type":"choose","extra_type":"classic","user_id":"131","test":"0","date":"2024-05-29 04:42:17","option_type":"txt","len":1},{"id":"5922","post_id":"7114","mon_id":"1158924","chapter_id":"1159187","question":"<p>Cho h&agrave;m s\u1ed1&nbsp;<span class=\"math-tex\">$f(x)=\\begin{cases}\\dfrac{\\sqrt{3x+1}-2x}{x-1}&amp;\\text{ khi }x\\ne1\\\\-\\dfrac{5}{4}&amp;\\text{ khi }x=1\\end{cases}$<\/span>. T&iacute;nh&nbsp;<span class=\"math-tex\">$f^\\prime(1)$<\/span>.<\/p>","options":["<strong>A.<\/strong> Kh&ocirc;ng t\u1ed3n t\u1ea1i","<strong>B.<\/strong> 0","<strong>C.<\/strong> <span class=\"math-tex\">$-\\dfrac{7}{50}$<\/span>","<strong>D.<\/strong> <span class=\"math-tex\">$-\\dfrac{9}{64}$<\/span>"],"correct":"4","level":"3","hint":"","answer":"<p>Ch\u1ecdn&nbsp;<span style=\"color:#16a085;\"><strong>D.<\/strong>&nbsp;<span class=\"math-tex\">$-\\dfrac{9}{64}$<\/span>.<\/span><\/p><p>Ta c&oacute;&nbsp;<span class=\"math-tex\">$\\displaystyle\\lim_{x\\to1}f(x)=\\displaystyle\\lim_{x\\to1}\\dfrac{\\sqrt{3x+1}-2x}{x-1}$<\/span><\/p><p><span class=\"math-tex\">$=\\displaystyle\\lim_{x\\to1}\\dfrac{3x+1-4x^2}{(x-1)(\\sqrt{3x+1}+2x)}$<\/span><\/p><p><span class=\"math-tex\">$=\\displaystyle\\lim_{x\\to1}\\dfrac{(x-1)(-4x-1)}{(x-1)(\\sqrt{3x+1}+2x)}$<\/span><\/p><p><span class=\"math-tex\">$=\\displaystyle\\lim_{x\\to1}\\dfrac{-4x-1}{\\sqrt{3x+1}+2x}$<\/span><\/p><p><span class=\"math-tex\">$=-\\dfrac{5}{4}$<\/span>&nbsp;<span class=\"math-tex\">$=f(1)$<\/span><\/p><p>Do \u0111&oacute; h&agrave;m s\u1ed1 li&ecirc;n t\u1ee5c t\u1ea1i x = 1.<\/p><p><span class=\"math-tex\">$f^\\prime(1)=\\displaystyle\\lim_{x\\to1}\\dfrac{f(x)-f(1)}{x-1}$<\/span>&nbsp;<\/p><p><span class=\"math-tex\">$=\\displaystyle\\lim_{x\\to1}\\dfrac{\\dfrac{\\sqrt{3x+1}-2x}{x-1}+\\dfrac{5}{4}}{x-1}$<\/span><\/p><p><span class=\"math-tex\">$=\\displaystyle\\lim_{x\\to1}\\dfrac{4\\sqrt{3x+1}-3x-5}{4(x-1)^2}$<\/span><\/p><p><span class=\"math-tex\">$=\\displaystyle\\lim_{x\\to1}\\dfrac{16(3x+1)-(3x+5)^2}{4(x-1)^2(4\\sqrt{3x+1}+3x+5)}$<\/span><\/p><p><span class=\"math-tex\">$=\\displaystyle\\lim_{x\\to1}\\dfrac{-9}{4(4\\sqrt{3x+1}+3x+5)}$<\/span><\/p><p><span class=\"math-tex\">$=-\\dfrac{9}{64}$<\/span>.<\/p>","type":"choose","extra_type":"classic","user_id":"131","test":"0","date":"2024-05-29 04:45:42","option_type":"math","len":0}]}