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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":"7522","level":2,"total":10,"point":10,"point_extra":0},"segment":[{"id":"5399","post_id":"7522","mon_id":"1159285","chapter_id":"1159288","question":"<p>Cho hàm s\u1ed1 <span class=\"math-tex\">$y=\\dfrac{x^2+2x+3}{\\sqrt{x^4-3x^2+2}}$<\/span>. \u0110\u1ed3 th\u1ecb hàm s\u1ed1 \u0111ã cho có bao nhiêu \u0111\u01b0\u1eddng ti\u1ec7m c\u1eadn?<\/p>","options":["<strong>A.<\/strong> 4","<strong>B.<\/strong> 5","<strong>C.<\/strong> 3","<strong>D.<\/strong> 6"],"correct":"1","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>A.<\/strong> 4<\/span><\/p><p>\u0110i\u1ec1u ki\u1ec7n: x ∈ <span class=\"math-tex\">$(-\\infty;-\\sqrt{2})\\cup(-1;1)\\cup(\\sqrt{2};+\\infty)$<\/span>.<\/p><p><span class=\"math-tex\">$\\displaystyle\\lim_{x\\to\\pm\\infty}\\dfrac{x^2+2x+3}{\\sqrt{x^4-3x^2+2}} =\\displaystyle\\lim_{x\\to\\pm\\infty}\\dfrac{1+\\dfrac{2}{x}+\\dfrac{3}{x^2}}{\\sqrt{1-\\dfrac{3}{x^2}+\\dfrac{2}{x^4}}}=1$<\/span><\/p><p>suy ra <span style=\"color:#16a085;\"><strong>y = 1<\/strong> là \u0111\u01b0\u1eddng ti\u1ec7m c\u1eadn ngang c\u1ee7a \u0111\u1ed3 th\u1ecb hàm s\u1ed1.<\/span><\/p><p>Có <span class=\"math-tex\">$\\displaystyle\\lim_{x\\to1^-}\\dfrac{x^2+2x+3}{\\sqrt{x^4-3x^2+2}} =+\\infty$<\/span> nên <span style=\"color:#16a085;\"><strong>x = 1<\/strong> là \u0111\u01b0\u1eddng ti\u1ec7m c\u1eadn \u0111\u1ee9ng.<\/span><\/p><p><span class=\"math-tex\">$\\displaystyle\\lim_{x\\to-1^+}\\dfrac{x^2+2x+3}{\\sqrt{x^4-3x^2+2}} =\\displaystyle\\lim_{x\\to-1^+}\\dfrac{(x+1)(x+2)}{\\sqrt{(x+1)(x-1)(x+\\sqrt{2})(x-\\sqrt{2})}} $<\/span><\/p><p><span class=\"math-tex\">$=\\displaystyle\\lim_{x\\to-1^+}\\dfrac{\\sqrt{x+1}(x+2)}{\\sqrt{(x-1)(x+\\sqrt{2})(x-\\sqrt{2})}} =0$<\/span><\/p><p>suy ra \u0111\u01b0\u1eddng th\u1eb3ng x = –1 <strong>không <\/strong>là \u0111\u01b0\u1eddng ti\u1ec7m c\u1eadn \u0111\u1ee9ng.<\/p><p>Có <span class=\"math-tex\">$\\displaystyle\\lim_{x\\to\\sqrt{2}^+}\\dfrac{x^2+2x+3}{\\sqrt{x^4-3x^2+2}} =+\\infty $<\/span> nên <span style=\"color:#16a085;\"><strong>x = <span class=\"math-tex\">$\\sqrt{2}$<\/span> <\/strong>là \u0111\u01b0\u1eddng ti\u1ec7m c\u1eadn \u0111\u1ee9ng.<\/span><\/p><p>Có <span class=\"math-tex\">$\\displaystyle\\lim_{x\\to-\\sqrt{2}^-}\\dfrac{x^2+2x+3}{\\sqrt{x^4-3x^2+2}} =+\\infty $<\/span> nên <span style=\"color:#16a085;\"><strong>x = <span class=\"math-tex\">$-\\sqrt{2}$<\/span> <\/strong>là \u0111\u01b0\u1eddng ti\u1ec7m c\u1eadn \u0111\u1ee9ng.<\/span><\/p><p>V\u1eady \u0111\u1ed3 th\u1ecb hàm s\u1ed1 có <span style=\"color:#16a085;\"><strong>4 \u0111\u01b0\u1eddng ti\u1ec7m c\u1eadn<\/strong><\/span> (1 ti\u1ec7m c\u1eadn ngang, 3 ti\u1ec7m c\u1eadn \u0111\u1ee9ng).<\/p>","type":"choose","extra_type":"shape1","user_id":"131","test":"0","date":"2024-06-30 01:16:13","option_type":"txt","len":0},{"id":"5401","post_id":"7522","mon_id":"1159285","chapter_id":"1159288","question":"<p>Cho hàm s\u1ed1 <span class=\"math-tex\">$y=\\dfrac{2x^2-3x-1}{x-2}$<\/span>. Ti\u1ec7m c\u1eadn xiên c\u1ee7a \u0111\u1ed3 th\u1ecb hàm s\u1ed1 là \u0111\u01b0\u1eddng th\u1eb3ng<\/p>","options":["<strong>A.<\/strong> y = 2x – 1 ","<strong>B.<\/strong> y = 2x + 1 ","<strong>C.<\/strong> y = 2x – 3 ","<strong>D.<\/strong> y = 2x + 3"],"correct":"2","level":"2","hint":"<p>S\u1eed d\u1ee5ng \u0111\u1ecbnh ngh\u0129a \u0111\u01b0\u1eddng ti\u1ec7m c\u1eadn xiên.<\/p>","answer":"<p>Ch\u1ecdn <span style=\"color:#16a085;\"><strong>B.<\/strong> y = 2x + 1 <\/span><\/p><p><span class=\"math-tex\">$y=\\dfrac{2x^2-3x-1}{x-2}$<\/span><span class=\"math-tex\">$=2x+1+\\dfrac{1}{x-2}$<\/span><\/p><p><span class=\"math-tex\">$\\lim_\\limits{x\\to+\\infty}[y-(2x+1)]= \\lim_\\limits{x\\to+\\infty}\\bigg[2x+1+\\dfrac{1}{x-2}-(2x+1)\\bigg]$<\/span><span class=\"math-tex\">$=\\lim_\\limits{x\\to+\\infty}\\dfrac{1}{x-2}=0$<\/span><\/p><p><span class=\"math-tex\">$\\lim_\\limits{x\\to-\\infty}[y-(2x+1)]= \\lim_\\limits{x\\to-\\infty}\\bigg[2x+1+\\dfrac{1}{x-2}-(2x+1)\\bigg]$<\/span><span class=\"math-tex\">$=\\lim_\\limits{x\\to-\\infty}\\dfrac{1}{x-2}=0$<\/span><\/p><p>V\u1eady \u0111\u01b0\u1eddng th\u1eb3ng y = 2x + 1 là \u0111\u01b0\u1eddng ti\u1ec7m c\u1eadn xiên c\u1ee7a \u0111\u1ed3 th\u1ecb hàm s\u1ed1.<\/p>","type":"choose","extra_type":"classic","user_id":"131","test":"0","date":"2024-06-30 01:32:04","option_type":"txt","len":2},{"id":"5402","post_id":"7522","mon_id":"1159285","chapter_id":"1159288","question":"<p>Bi\u1ebft \u0111\u1ed3 th\u1ecb hàm s\u1ed1 <span class=\"math-tex\">$y=\\dfrac{x^2-2x+2}{x-3}$<\/span> có \u0111\u01b0\u1eddng ti\u1ec7m c\u1eadn xiên là \u0111\u01b0\u1eddng th\u1eb3ng y = ax + b. Tính a² + 2b.<\/p>","options":["<strong>A.<\/strong> 2","<strong>B.<\/strong> 4","<strong>C.<\/strong> 5","<strong>D.<\/strong> 3"],"correct":"4","level":"2","hint":"<p>S\u1eed d\u1ee5ng \u0111\u1ecbnh ngh\u0129a \u0111\u01b0\u1eddng ti\u1ec7m c\u1eadn xiên.<\/p>","answer":"<p>Ch\u1ecdn <span style=\"color:#16a085;\"><strong>D.<\/strong> 3<\/span><\/p><p><span class=\"math-tex\">$y=\\dfrac{x^2-2x+2}{x-3}=x+1+\\dfrac{5}{x-3}$<\/span><\/p><p><span class=\"math-tex\">$\\lim_\\limits{x\\to+\\infty}[y-(x+1)]= \\lim_\\limits{x\\to+\\infty}\\bigg[x+1+\\dfrac{5}{x-3}-(x+1)\\bigg]$<\/span><span class=\"math-tex\">$=\\lim_\\limits{x\\to+\\infty}\\dfrac{5}{x-3}=0$<\/span><\/p><p><span class=\"math-tex\">$\\lim_\\limits{x\\to-\\infty}[y-(x+1)]= \\lim_\\limits{x\\to-\\infty}\\bigg[x+1+\\dfrac{5}{x-3}-(x+1)\\bigg]$<\/span><span class=\"math-tex\">$=\\lim_\\limits{x\\to-\\infty}\\dfrac{5}{x-3}=0$<\/span><\/p><p>Suy ra \u0111\u01b0\u1eddng th\u1eb3ng y = x + 1 là \u0111\u01b0\u1eddng ti\u1ec7m c\u1eadn xiên c\u1ee7a \u0111\u1ed3 th\u1ecb hàm s\u1ed1.<\/p><p>Suy ra a = 1, b = 1 và <span style=\"color:#16a085;\"><strong>a² + 2b = 3<\/strong><\/span>.<\/p>","type":"choose","extra_type":"sheep","user_id":"131","test":"0","date":"2024-06-30 01:34:45","option_type":"txt","len":0}]}