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HomeLớp 12Toán lớp 12 - Sách Kết nối tri thứcBài 3. Đường tiệm cận của đồ thị hàm sốBài tập trung bình
{"common":{"save":0,"post_id":"7521","level":2,"total":10,"point":10,"point_extra":0},"segment":[{"id":"5387","post_id":"7521","mon_id":"1159285","chapter_id":"1159288","question":"<p>Ti\u1ec7m c\u1eadn ngang c\u1ee7a \u0111\u1ed3 th\u1ecb hàm s\u1ed1 <span class=\"math-tex\">$y=\\dfrac{x-2}{x+1}$<\/span> là:<\/p>","options":["<strong>A.<\/strong> y = –2","<strong>B.<\/strong> y = 1","<strong>C.<\/strong> x = –1","<strong>D.<\/strong> x = 2"],"correct":"2","level":"2","hint":"<p>S\u1eed d\u1ee5ng \u0111\u1ecbnh ngh\u0129a \u0111\u01b0\u1eddng ti\u1ec7m c\u1eadn \u0111\u1ee9ng, \u0111\u01b0\u1eddng ti\u1ec7m c\u1eadn ngang.<\/p>","answer":"<p>Ch\u1ecdn <span style=\"color:#16a085;\"><strong>B.<\/strong> y = 1<\/span><\/p><p>Ta có: <span class=\"math-tex\">$\\lim_\\limits{x\\to+\\infty}\\dfrac{x-2}{x+1}=1$<\/span> và <span class=\"math-tex\">$\\lim_\\limits{x\\to-\\infty}\\dfrac{x-2}{x+1}=1$<\/span><\/p><p>Suy ra y = 1 là ti\u1ec7m c\u1eadn ngang c\u1ee7a \u0111\u1ed3 th\u1ecb hàm s\u1ed1.<\/p>","type":"choose","extra_type":"shape1","user_id":"131","test":"0","date":"2024-06-29 02:31:36","option_type":"txt","len":0},{"id":"5388","post_id":"7521","mon_id":"1159285","chapter_id":"1159288","question":"<p>Ti\u1ec7m c\u1eadn \u0111\u1ee9ng c\u1ee7a \u0111\u1ed3 th\u1ecb hàm s\u1ed1 <span class=\"math-tex\">$y=\\dfrac{x-1}{x-3}$<\/span> là:<\/p>","options":["<strong>A.<\/strong> x = –3","<strong>B.<\/strong> x = –1","<strong>C.<\/strong> x = 1","<strong>D.<\/strong> x = 3"],"correct":"4","level":"2","hint":"<p>S\u1eed d\u1ee5ng \u0111\u1ecbnh ngh\u0129a \u0111\u01b0\u1eddng ti\u1ec7m c\u1eadn \u0111\u1ee9ng, \u0111\u01b0\u1eddng ti\u1ec7m c\u1eadn ngang.<\/p>","answer":"<p>Ch\u1ecdn <span style=\"color:#16a085;\"><strong>D.<\/strong> x = 3<\/span><\/p><p>Ta có: <span class=\"math-tex\">$\\lim_\\limits{x\\to3^-}\\dfrac{x-1}{x-3}=-\\infty$<\/span>.<\/p><p>Suy ra ti\u1ec7m c\u1eadn \u0111\u1ee9ng là \u0111\u01b0\u1eddng th\u1eb3ng x = 3.<\/p>","type":"choose","extra_type":"shape2","user_id":"131","test":"0","date":"2024-06-29 02:41:56","option_type":"txt","len":0},{"id":"5389","post_id":"7521","mon_id":"1159285","chapter_id":"1159288","question":"<p>T\u1ed5ng s\u1ed1 ti\u1ec7m c\u1eadn \u0111\u1ee9ng và ti\u1ec7m c\u1eadn ngang c\u1ee7a \u0111\u1ed3 th\u1ecb hàm s\u1ed1 <span class=\"math-tex\">$y=\\dfrac{5x^2-4x-1}{x^2-1}$<\/span> là:<\/p>","options":["<strong>A.<\/strong> 0","<strong>B.<\/strong> 1","<strong>C.<\/strong> 2","<strong>D.<\/strong> 3"],"correct":"3","level":"2","hint":"<p>S\u1eed d\u1ee5ng \u0111\u1ecbnh ngh\u0129a \u0111\u01b0\u1eddng ti\u1ec7m c\u1eadn \u0111\u1ee9ng, \u0111\u01b0\u1eddng ti\u1ec7m c\u1eadn ngang.<\/p>","answer":"<p>Ch\u1ecdn <span style=\"color:#16a085;\"><strong>C.<\/strong> 2<\/span><\/p><p>Ta có: <span class=\"math-tex\">$\\lim_\\limits{x\\to+\\infty}\\dfrac{5x^2-4x-1}{x^2-1}$<\/span> = <span class=\"math-tex\">$\\lim_\\limits{x\\to+\\infty}\\dfrac{x^2\\bigg(5-\\dfrac{4}{x}-\\dfrac{1}{x^2}\\bigg)}{x^2\\bigg(1-\\dfrac{1}{x^2}\\bigg)}$<\/span> = <span class=\"math-tex\">$\\lim_\\limits{x\\to+\\infty}\\dfrac{5-\\dfrac{4}{x}-\\dfrac{1}{x^2}}{1-\\dfrac{1}{x^2}}=5$<\/span><\/p><p>nên \u0111\u1ed3 th\u1ecb hàm s\u1ed1 có m\u1ed9t ti\u1ec7m c\u1eadn ngang y = 5.<\/p><p>Ta có: <span class=\"math-tex\">$\\lim_\\limits{x\\to-1^+}\\dfrac{5x^2-4x-1}{x^2-1}$<\/span> = <span class=\"math-tex\">$\\lim_\\limits{x\\to-1^+}\\dfrac{5x^2-4x-1}{(x-1)(x+1)}$<\/span> = <span class=\"math-tex\">$\\lim_\\limits{x\\to-1^+}\\bigg(\\dfrac{1}{x+1}.\\dfrac{5x^2-4x-1}{x-1}\\bigg)=-\\infty$<\/span><\/p><p>vì <span class=\"math-tex\">$\\lim_\\limits{x\\to-1^+}\\dfrac{1}{x+1}=+\\infty$<\/span> và <span class=\"math-tex\">$\\lim_\\limits{x\\to-1^+}\\dfrac{5x^2-4x-1}{x-1}=-4<0$<\/span><\/p><p>Khi \u0111ó, \u0111\u1ed3 th\u1ecb hàm s\u1ed1 có m\u1ed9t ti\u1ec7m c\u1eadn \u0111\u1ee9ng x = –1.<\/p><p>T\u1ed5ng c\u1ed9ng \u0111\u1ed3 th\u1ecb hàm s\u1ed1 có 2 ti\u1ec7m c\u1eadn.<\/p>","type":"choose","extra_type":"shape1","user_id":"131","test":"0","date":"2024-06-29 02:46:37","option_type":"txt","len":0}]}