Chú ý: Để đảm bảo quyền lợi và bảo vệ tài khoản của mình
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{"common":{"save":0,"post_id":"7091","level":2,"total":10,"point":10,"point_extra":0},"segment":[{"id":"5302","post_id":"7091","mon_id":"1158924","chapter_id":"1159163","question":"<p>Xét tính liên t\u1ee5c c\u1ee7a hàm s\u1ed1 <span class=\"math-tex\">$f(x)=\\begin{cases}\\dfrac{x-1}{\\sqrt{2-x}-1}&\\text{ khi }x < 1\\\\ -2x&\\text{ khi }x \\ge 1\\end{cases}$<\/span> .<br \/>Kh\u1eb3ng \u0111\u1ecbnh nào d\u01b0\u1edbi \u0111ây là \u0111úng?<\/p>","options":["<strong>A.<\/strong> <span class=\"math-tex\">$f(x)$<\/span> không liên t\u1ee5c trên R.","<strong>B.<\/strong> <span class=\"math-tex\">$f(x)$<\/span> không liên t\u1ee5c trên (0 ; 2).","<strong>C.<\/strong> <span class=\"math-tex\">$f(x)$<\/span> gián \u0111o\u1ea1n t\u1ea1i x = 1.","<strong>D.<\/strong> <span class=\"math-tex\">$f(x)$<\/span> liên t\u1ee5c trên R."],"correct":"4","level":"2","hint":"","answer":"<p>Ta có <span class=\"math-tex\">$f(1)=-2$<\/span> ; <span class=\"math-tex\">$\\displaystyle\\lim_{x\\to1^+}f(x)=\\displaystyle\\lim_{x\\to1^+}(-2x)=-2$<\/span> ; <span class=\"math-tex\">$\\displaystyle\\lim_{x\\to1^-}f(x)=\\displaystyle\\lim_{x\\to1^-}\\dfrac{x-1}{\\sqrt{2-x}-1}=\\displaystyle\\lim_{x\\to1^-}[-(\\sqrt{2-x}+1)]=-2$<\/span> .<\/p><p>Suy ra <span class=\"math-tex\">$f(x)$<\/span> liên t\u1ee5c t\u1ea1i x = 1.<\/p><p>V\u1eady hàm s\u1ed1 liên t\u1ee5c trên R.<\/p><p>Ch\u1ecdn <span style=\"color:#16a085;\"><strong>D.<\/strong> <span class=\"math-tex\">$f(x)$<\/span> liên t\u1ee5c trên R.<\/span><\/p>","type":"choose","extra_type":"classic","user_id":"131","test":"0","date":"2024-03-03 03:32:30","option_type":"math","len":0},{"id":"5306","post_id":"7091","mon_id":"1158924","chapter_id":"1159163","question":"<p>Giá tr\u1ecb nh\u1ecf nh\u1ea5t c\u1ee7a a \u0111\u1ec3 hàm s\u1ed1 <span class=\"math-tex\">$f(x)=\\begin{cases}\\dfrac{x^2-5x+6}{\\sqrt{4x-3}-x}&\\text{ khi }{x>3}\\\\{1-a^2x}&\\text{ khi }{x\\le3}\\end{cases}$<\/span> liên t\u1ee5c t\u1ea1i x = 3 là<\/p>","options":["<strong>A.<\/strong> <span class=\"math-tex\">$-\\dfrac{2}{\\sqrt{3}}$<\/span>","<strong>B.<\/strong> <span class=\"math-tex\">$\\dfrac{2}{\\sqrt{3}}$<\/span>","<strong>C.<\/strong> <span class=\"math-tex\">$-\\dfrac{4}{3}$<\/span>","<strong>D.<\/strong> <span class=\"math-tex\">$\\dfrac{4}{3}$<\/span>"],"correct":"1","level":"2","hint":"","answer":"<p>\u0110i\u1ec1u ki\u1ec7n bài toán tr\u1edf thành <span class=\"math-tex\">$\\displaystyle\\lim_{x\\to3^+}f(x)=\\displaystyle\\lim_{x\\to3^-}f(x)=f(3)$<\/span> (*).<\/p><p>Ta có <span class=\"math-tex\">$f(3)=1-3a^2$<\/span><\/p><p><span class=\"math-tex\">$\\displaystyle\\lim_{x\\to3^+}f(x)=\\displaystyle\\lim_{x\\to3^+}\\dfrac{x^2-5x+6}{\\sqrt{4x-3}-x}=\\displaystyle\\lim_{x\\to3^+}\\dfrac{(x-2)(\\sqrt{4x-3}+x)}{1-x}=-3$<\/span><\/p><p><span class=\"math-tex\">$\\displaystyle\\lim_{x\\to3^-}f(x)=\\displaystyle\\lim_{x\\to3^-}(1-a^2x)=1-3a^2$<\/span><\/p><p>Khi \u0111ó (*) ⇔ <span class=\"math-tex\">$1-3a^2=-3$<\/span> ⇔ <span class=\"math-tex\">$a=-\\dfrac{2}{\\sqrt{3}}$<\/span> ho\u1eb7c <span class=\"math-tex\">$a=\\dfrac{2}{\\sqrt{3}}$<\/span> .<\/p><p>V\u1eady giá tr\u1ecb nh\u1ecf nh\u1ea5t c\u1ee7a a là <span class=\"math-tex\">$a=-\\dfrac{2}{\\sqrt{3}}$<\/span> .<\/p><p>Ch\u1ecdn <span style=\"color:#16a085;\"><strong>A.<\/strong> <span class=\"math-tex\">$-\\dfrac{2}{\\sqrt{3}}$<\/span>.<\/span><\/p>","type":"choose","extra_type":"classic","user_id":"131","test":"0","date":"2024-03-03 03:44:04","option_type":"math","len":0},{"id":"5310","post_id":"7091","mon_id":"1158924","chapter_id":"1159163","question":"<p>Tìm giá tr\u1ecb l\u1edbn nh\u1ea5t c\u1ee7a a \u0111\u1ec3 hàm s\u1ed1 <span class=\"math-tex\">$f(x)=\\begin{cases}\\dfrac{\\sqrt[3]{3x+2}-2}{x-2}&\\text{ khi }{x>2}\\\\{a^2x-\\dfrac{7}{4}}&\\text{ khi }{x\\le2}\\end{cases}$<\/span> liên t\u1ee5c t\u1ea1i x = 2.<\/p>","options":["<strong>A.<\/strong> 3","<strong>B.<\/strong> 0","<strong>C.<\/strong> 1","<strong>D.<\/strong> 2"],"correct":"3","level":"2","hint":"","answer":"<p>Ta c\u1ea7n có <span class=\"math-tex\">$\\displaystyle\\lim_{x\\to2^+}f(x)=\\displaystyle\\lim_{x\\to2^-}f(x)=f(2)$<\/span> (*).<\/p><p><span class=\"math-tex\">$f(2)=2a^2-\\dfrac{7}{4}$<\/span><\/p><p><span class=\"math-tex\">$\\displaystyle\\lim_{x\\to2^+}f(x)=\\displaystyle\\lim_{x\\to2^+}\\dfrac{\\sqrt[3]{3x+2}-2}{x-2}$<\/span><\/p><p> <span class=\"math-tex\">$=\\displaystyle\\lim_{x\\to2^+}\\dfrac{3(x-2)}{(x-2)[(\\sqrt[3]{3x+2})^2+2\\sqrt[3]{3x+2}+4]}$<\/span><\/p><p> <span class=\"math-tex\">$=\\displaystyle\\lim_{x\\to2^+}\\dfrac{3}{(\\sqrt[3]{3x+2})^2+2\\sqrt[3]{3x+2}+4} $<\/span><\/p><p> <span class=\"math-tex\">$=\\dfrac{1}{4}$<\/span><\/p><p><span class=\"math-tex\">$\\displaystyle\\lim_{x\\to2^-}f(x)=\\displaystyle\\lim_{x\\to2^-}\\bigg(a^2x-\\dfrac{7}{4}\\bigg)=2a^2-\\dfrac{7}{4}$<\/span><\/p><p>Do \u0111ó (*) ⇔ <span class=\"math-tex\">$2a^2-\\dfrac{7}{4}=\\dfrac{1}{4}$<\/span> ⇔ a = 1 ho\u1eb7c a = –1.<\/p><p>V\u1eady giá tr\u1ecb l\u1edbn nh\u1ea5t c\u1ee7a a là a = 1.<\/p><p>Ch\u1ecdn <span style=\"color:#16a085;\"><strong>C.<\/strong> 1.<\/span><\/p>","type":"choose","extra_type":"classic","user_id":"131","test":"0","date":"2024-03-06 00:52:05","option_type":"txt","len":0}]}