{"common":{"save":0,"post_id":"7093","level":1,"total":10,"point":10,"point_extra":0},"segment":[{"id":"5271","post_id":"7093","mon_id":"1158924","chapter_id":"1159163","question":"<p>K\u1ebft qu\u1ea3 c\u1ee7a <span class=\"math-tex\">$\\lim\\dfrac{3^n-4.2^{n-1}-3}{3.2^n+4^n}$<\/span> là:<\/p>","options":["<strong>A.<\/strong> <span class=\"math-tex\">$+\\infty$<\/span>","<strong>B.<\/strong> <span class=\"math-tex\">$-\\infty$<\/span>","<strong>C.<\/strong> 0","<strong>D.<\/strong> 1"],"correct":"3","level":"1","hint":"","answer":"<p>Ch\u1ecdn <span style=\"color:#16a085;\"><strong>C.<\/strong> 0<\/span><\/p><p><span class=\"math-tex\">$\\lim\\dfrac{3^n-4.2^{n-1}-3}{3.2^n+4^n}$<\/span> = <span class=\"math-tex\">$\\lim\\dfrac{3^n-2.2^n-3}{3.2^n+4^n}$<\/span> <\/p><p>= <span class=\"math-tex\">$\\lim\\dfrac{\\bigg(\\dfrac{3}{4}\\bigg)^n-2.\\bigg(\\dfrac{1}{2}\\bigg)^n-3.\\bigg(\\dfrac{1}{4}\\bigg)^n}{3.\\bigg(\\dfrac{1}{2}\\bigg)^n+1}$<\/span> = 0<\/p>","type":"choose","extra_type":"shape2","user_id":"131","test":"0","date":"2024-03-03 01:10:25","option_type":"math","len":0},{"id":"5272","post_id":"7093","mon_id":"1158924","chapter_id":"1159163","question":"<p>Giá tr\u1ecb c\u1ee7a <span class=\"math-tex\">$\\lim(\\sqrt{n^2-1}-\\sqrt{3n^2+2})$<\/span> là:<\/p>","options":["<strong>A.<\/strong> <span class=\"math-tex\">$+\\infty$<\/span>","<strong>B.<\/strong> <span class=\"math-tex\">$-\\infty$<\/span>","<strong>C.<\/strong> 0","<strong>D.<\/strong> 1"],"correct":"2","level":"1","hint":"","answer":"<p>Ch\u1ecdn <span style=\"color:#16a085;\"><strong>B.<\/strong> <span class=\"math-tex\">$-\\infty$<\/span><\/span><\/p><p><span class=\"math-tex\">$\\lim(\\sqrt{n^2-1}-\\sqrt{3n^2+2})$<\/span> = <span class=\"math-tex\">$\\lim n\\bigg(\\sqrt{1-\\dfrac{1}{n^2}}-\\sqrt{3+\\dfrac{2}{n^2}}\\bigg)=-\\infty$<\/span><\/p><p>Vì <span class=\"math-tex\">$\\lim n=+\\infty$<\/span> và <span class=\"math-tex\">$\\lim \\bigg(\\sqrt{1-\\dfrac{1}{n^2}}-\\sqrt{3+\\dfrac{2}{n^2}}\\bigg)=1-\\sqrt{3}<0$<\/span><\/p>","type":"choose","extra_type":"shape1","user_id":"131","test":"0","date":"2024-03-03 01:14:37","option_type":"math","len":0},{"id":"5273","post_id":"7093","mon_id":"1158924","chapter_id":"1159163","question":"<p>Giá tr\u1ecb c\u1ee7a <span class=\"math-tex\">$\\lim (3^n-5^n)$<\/span> là:<\/p>","options":["<strong>A.<\/strong> <span class=\"math-tex\">$-\\infty$<\/span>","<strong>B.<\/strong> <span class=\"math-tex\">$+\\infty$<\/span>","<strong>C.<\/strong> 2","<strong>D.<\/strong> –2"],"correct":"1","level":"1","hint":"","answer":"<p>Ch\u1ecdn <span style=\"color:#16a085;\"><strong>A.<\/strong> <span class=\"math-tex\">$-\\infty$<\/span><\/span><\/p><p><span class=\"math-tex\">$\\lim (3^n-5^n)$<\/span> = <span class=\"math-tex\">$\\lim 5^n\\bigg[\\bigg(\\dfrac{3}{5}\\bigg)^n-1\\bigg]$<\/span> = <span class=\"math-tex\">$-\\infty$<\/span><\/p><p>vì <span class=\"math-tex\">$\\lim 5^n=+\\infty$<\/span> và <span class=\"math-tex\">$\\lim \\bigg[\\bigg(\\dfrac{3}{5}\\bigg)^n-1\\bigg]=-1<0$<\/span><\/p>","type":"choose","extra_type":"shape2","user_id":"131","test":"0","date":"2024-03-03 01:18:54","option_type":"math","len":0}]}
