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{"common":{"save":0,"post_id":"7052","level":2,"total":10,"point":10,"point_extra":0},"segment":[{"id":"4871","post_id":"7052","mon_id":"1158924","chapter_id":"1159147","question":"<p>Cho c\u1ea5p s\u1ed1 nhân <span class=\"math-tex\">$(u_n)$<\/span> có công b\u1ed9i âm và <span class=\"math-tex\">$u_2=\\dfrac{1}{4}$<\/span> , <span class=\"math-tex\">$u_4=16$<\/span> . Giá tr\u1ecb c\u1ee7a <span class=\"math-tex\">$u_1$<\/span> là<\/p>","options":["<strong>A.<\/strong> <span class=\"math-tex\">$u_1=-2$<\/span>","<strong>B.<\/strong> <span class=\"math-tex\">$u_1=-\\dfrac{1}{32}$<\/span>","<strong>C.<\/strong> <span class=\"math-tex\">$u_1=\\dfrac{1}{32}$<\/span>","<strong>D.<\/strong> <span class=\"math-tex\">$u_1=2$<\/span>"],"correct":"2","level":"2","hint":"","answer":"<p>Theo tính ch\u1ea5t c\u1ee7a c\u1ea5p s\u1ed1 nhân v\u1edbi <span class=\"math-tex\">$k \\ge2$<\/span> thì <span class=\"math-tex\">$u^2_k=u_{k-1}.u_{k+1}$<\/span>, ta suy ra<\/p><p><span class=\"math-tex\">$u^2_3=u_2.u_4=\\dfrac{1}{4}.16=4$<\/span> ⇔ <span class=\"math-tex\">$u_3=2$<\/span> ho\u1eb7c <span class=\"math-tex\">$u_3=-2$<\/span> .<\/p><p>Vì c\u1ea5p s\u1ed1 nhân có công b\u1ed9i âm nên <span class=\"math-tex\">$u_3=-2$<\/span> . G\u1ecdi q là công b\u1ed9i ta \u0111\u01b0\u1ee3c<\/p><p><span class=\"math-tex\">$q=\\dfrac{u_4}{u_3}=\\dfrac{16}{-2}=-8$<\/span> . <\/p><p>T\u1eeb \u0111ó <span class=\"math-tex\">$u_1=u_2:q=\\dfrac{1}{4}:(-8)=-\\dfrac{1}{32}$<\/span> .<\/p><p>Ch\u1ecdn <span style=\"color:#16a085;\"><strong>B.<\/strong> <span class=\"math-tex\">$u_1=-\\dfrac{1}{32}$<\/span> .<\/span><\/p>","type":"choose","extra_type":"classic","user_id":"131","test":"0","date":"2023-12-06 02:35:44","option_type":"math","len":0},{"id":"4873","post_id":"7052","mon_id":"1158924","chapter_id":"1159147","question":"<p>Cho c\u1ea5p s\u1ed1 nhân <span class=\"math-tex\">$(u_n)$<\/span> có t\u1ed5ng n s\u1ed1 h\u1ea1ng \u0111\u1ea7u tiên là <span class=\"math-tex\">$S_n=6^n-1$<\/span> . S\u1ed1 h\u1ea1ng th\u1ee9 6 c\u1ee7a c\u1ea5p s\u1ed1 nhân là<\/p>","options":["<strong>A.<\/strong> 30888","<strong>B.<\/strong> 88830","<strong>C.<\/strong> 83880","<strong>D.<\/strong> 38880"],"correct":"4","level":"2","hint":"","answer":"<p>C\u1ea5p s\u1ed1 nhân <span class=\"math-tex\">$(u_n)$<\/span> có s\u1ed1 h\u1ea1ng \u0111\u1ea7u <span class=\"math-tex\">$u_1$<\/span> và công b\u1ed9i <span class=\"math-tex\">$q$<\/span> .<\/p><p>Do <span class=\"math-tex\">$S_n=6^n-1$<\/span> nên <span class=\"math-tex\">$q \\ne 1$<\/span> . Khi \u0111ó <span class=\"math-tex\">$S_n=\\dfrac{u_1.(1-q^n)}{1-q}=6^n-1$<\/span> .<\/p><p>Ta có <span class=\"math-tex\">$S_1=\\dfrac{u_1.(1-q)}{1-q}=6-1$<\/span> ⇔ <span class=\"math-tex\">$u_1=5$<\/span> <\/p><p><span class=\"math-tex\">$S_2=\\dfrac{u_1(1-q^2)}{1-q}=6^2-1$<\/span> ⇔ <span class=\"math-tex\">$5.(1+q)=35$<\/span> ⇔ <span class=\"math-tex\">$q=6$<\/span> .<\/p><p>V\u1eady <span class=\"math-tex\">$u_6=u_1.q^5=5.6^5=38880$<\/span> .<\/p><p>Ch\u1ecdn <span style=\"color:#16a085;\"> <strong>D.<\/strong> 38880.<\/span><\/p>","type":"choose","extra_type":"classic","user_id":"131","test":"0","date":"2023-12-06 02:46:42","option_type":"txt","len":0},{"id":"4876","post_id":"7052","mon_id":"1158924","chapter_id":"1159147","question":"<p><span class=\"math-tex\">$u_n$<\/span> \u0111\u01b0\u1ee3c cho b\u1edfi công th\u1ee9c nào d\u01b0\u1edbi \u0111ây là s\u1ed1 h\u1ea1ng t\u1ed5ng quát c\u1ee7a m\u1ed9t c\u1ea5p s\u1ed1 nhân?<\/p>","options":["<strong>A.<\/strong> <span class=\"math-tex\">$u_n=\\dfrac{1}{2^{n+1}}$<\/span>","<strong>B.<\/strong> <span class=\"math-tex\">$u_n=n^2-\\dfrac{1}{2}$<\/span>","<strong>C.<\/strong> <span class=\"math-tex\">$u_n=\\dfrac{1}{2^n}-1$<\/span>","<strong>D.<\/strong> <span class=\"math-tex\">$u_n=n^2+\\dfrac{1}{2}$<\/span>"],"correct":"1","level":"2","hint":"","answer":"<p><strong>A.<\/strong> <span class=\"math-tex\">$u_n=\\dfrac{1}{2^{n+1}}=\\dfrac{1}{4}.\\bigg(\\dfrac{1}{2}\\bigg)^{n-1}$<\/span> là s\u1ed1 h\u1ea1ng t\u1ed5ng quát c\u1ee7a m\u1ed9t c\u1ea5p s\u1ed1 nhân có <span class=\"math-tex\">$u_1=\\dfrac{1}{4}$<\/span> và <span class=\"math-tex\">$q=\\dfrac{1}{2}$<\/span> .<\/p><p><strong>B.<\/strong> <span class=\"math-tex\">$u_n=n^2-\\dfrac{1}{2}$<\/span> có <span class=\"math-tex\">$u_1=\\dfrac{1}{2}$<\/span> ; <span class=\"math-tex\">$u_2=\\dfrac{7}{2}=u_1.7$<\/span> ; <span class=\"math-tex\">$u_3=\\dfrac{17}{2} \\ne u_2.7$<\/span> nên không ph\u1ea3i s\u1ed1 h\u1ea1ng t\u1ed5ng quát c\u1ee7a m\u1ed9t c\u1ea5p s\u1ed1 nhân.<\/p><p><strong>C.<\/strong> <span class=\"math-tex\">$u_n=\\dfrac{1}{2^n}-1$<\/span> có <span class=\"math-tex\">$u_1=-\\dfrac{1}{2};u_2=-\\dfrac{3}{4}=u_1.\\dfrac{3}{2};u_3=-\\dfrac{7}{8} \\ne u_2.\\dfrac{3}{2}$<\/span> nên không ph\u1ea3i s\u1ed1 h\u1ea1ng t\u1ed5ng quát c\u1ee7a m\u1ed9t c\u1ea5p s\u1ed1 nhân.<\/p><p><strong>D.<\/strong> <span class=\"math-tex\">$u_n=n^2+\\dfrac{1}{2}$<\/span> có <span class=\"math-tex\">$u_1=\\dfrac{3}{2};u_2=\\dfrac{9}{2}=u_1.3;u_3=\\dfrac{19}{2} \\ne u_2.3$<\/span> nên không ph\u1ea3i s\u1ed1 h\u1ea1ng t\u1ed5ng quát c\u1ee7a m\u1ed9t c\u1ea5p s\u1ed1 nhân.<\/p><p>Ch\u1ecdn <span style=\"color:#16a085;\"> <strong>A.<\/strong> <span class=\"math-tex\">$u_n=\\dfrac{1}{2^{n+1}}$<\/span> .<\/span><\/p>","type":"choose","extra_type":"classic","user_id":"131","test":"0","date":"2023-12-06 02:57:29","option_type":"math","len":0}]}