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{"common":{"save":0,"post_id":"8059","level":3,"total":10,"point":10,"point_extra":0},"segment":[{"id":"5893","post_id":"8059","mon_id":"1159285","chapter_id":"1159380","question":"<p>Cho hàm s\u1ed1 <span class=\"math-tex\">$f(x)\\ne0$<\/span>, liên t\u1ee5c trên \u0111o\u1ea1n [1; 2] và th\u1ecfa mãn <span class=\"math-tex\">$f(1)=\\dfrac{1}{3};\\space x^2.f^\\prime(x)=f^2(x)$<\/span>, <span class=\"math-tex\">$\\forall x\\in[1; 2]$<\/span>. Tính tích phân <span class=\"math-tex\">$I=\\int\\limits_1^2(2x+1)^2f(x)dx$<\/span>.<\/p>","options":["<strong>A.<\/strong> <span class=\"math-tex\">$I=\\dfrac{7}{6}$<\/span>","<strong>B.<\/strong> <span class=\"math-tex\">$I=\\dfrac{5}{6}$<\/span>","<strong>C.<\/strong> <span class=\"math-tex\">$I=\\dfrac{37}{6}$<\/span>","<strong>D.<\/strong> <span class=\"math-tex\">$I=\\dfrac{1}{6}$<\/span>"],"correct":"3","level":"3","hint":"<p>S\u1eed d\u1ee5ng ph\u01b0\u01a1ng pháp <span class=\"math-tex\">$\\dfrac{f^\\prime(x)}{h[f(x)]}=g(x)$<\/span> ⇔ <span class=\"math-tex\">$\\int\\dfrac{f^\\prime(x)}{h[f(x)]}.dx=\\int g(x).dx$<\/span> ⇔ <span class=\"math-tex\">$\\int\\dfrac{d[f(x)]}{h[f(x)]}=\\int g(x).dx$<\/span><\/p>","answer":"<p>Ch\u1ecdn <span style=\"color:#16a085;\"><strong>C.<\/strong> <span class=\"math-tex\">$I=\\dfrac{37}{6}$<\/span><\/span><\/p><p>Ta có: <span class=\"math-tex\">$x^2.f^\\prime(x)=f^2(x)$<\/span> ⇒ <span class=\"math-tex\">$\\dfrac{f^\\prime(x)}{f^2(x)}=\\dfrac{1}{x^2}$<\/span> ⇒ <span class=\"math-tex\">$\\bigg(-\\dfrac{1}{f(x)}\\bigg)^\\prime=\\dfrac{1}{x^2}$<\/span><\/p><p>⇒ <span class=\"math-tex\">$-\\dfrac{1}{f(x)}=\\int\\dfrac{1}{x^2}dx$<\/span> ⇒ <span class=\"math-tex\">$\\dfrac{1}{f(x)}=-\\int\\dfrac{1}{x^2}dx$<\/span> ⇒ <span class=\"math-tex\">$\\dfrac{1}{f(x)}=\\dfrac{1}{x}+C$<\/span><\/p><p>Mà <span class=\"math-tex\">$f(1)=\\dfrac{1}{3}$<\/span> ⇒ 3 = 1 + C ⇒ C = 2. Suy ra <span class=\"math-tex\">$\\dfrac{1}{f(x)}=\\dfrac{1}{x}+2$<\/span>.<\/p><p>Do \u0111ó <span class=\"math-tex\">$f(x)=\\dfrac{x}{2x+1}$<\/span>.<\/p><p>V\u1eady <span class=\"math-tex\">$I=\\int\\limits_1^2(2x+1)^2f(x)dx$<\/span> = <span class=\"math-tex\">$\\int\\limits_1^2(2x+1)^2.\\dfrac{x}{2x+1}dx$<\/span> = <span class=\"math-tex\">$\\int\\limits_1^2(2x^2+x)dx$<\/span> = <span class=\"math-tex\">$\\bigg(\\dfrac{2}{3}x^3+\\dfrac{1}{2}x^2\\bigg) \\bigg |^2_1$<\/span> = <span class=\"math-tex\">$\\dfrac{37}{6}$<\/span><\/p>","type":"choose","extra_type":"classic","user_id":"131","test":"0","date":"2024-11-19 08:38:08","option_type":"math","len":0},{"id":"5897","post_id":"8059","mon_id":"1159285","chapter_id":"1159380","question":"<p>Cho hàm s\u1ed1 <span class=\"math-tex\">$f(x)$<\/span> \u0111\u1ed3ng bi\u1ebfn, có \u0111\u1ea1o hàm trên \u0111o\u1ea1n [1; 4] và th\u1ecfa mãn <span class=\"math-tex\">$x+2x.f(x)=\\big[f^\\prime(x)\\big]^2$<\/span>, <span class=\"math-tex\">$\\forall x\\in [1; 4]$<\/span>. Bi\u1ebft <span class=\"math-tex\">$f(1) =\\dfrac{3}{2}$<\/span>, tính <span class=\"math-tex\">$I=\\int\\limits_1^4f(x)dx$<\/span>.<\/p>","options":["<strong>A.<\/strong> <span class=\"math-tex\">$I=\\dfrac{1186}{45}$<\/span>","<strong>B.<\/strong> <span class=\"math-tex\">$I=\\dfrac{1186}{9}$<\/span>","<strong>C.<\/strong> <span class=\"math-tex\">$I=\\dfrac{1186}{5}$<\/span>","<strong>D.<\/strong> <span class=\"math-tex\">$I=\\dfrac{1186}{41}$<\/span>"],"correct":"1","level":"3","hint":"<p>S\u1eed d\u1ee5ng ph\u01b0\u01a1ng pháp <span class=\"math-tex\">$\\dfrac{f^\\prime(x)}{h[f(x)]}=g(x)$<\/span> ⇔ <span class=\"math-tex\">$\\int\\dfrac{f^\\prime(x)}{h[f(x)]}.dx=\\int g(x).dx$<\/span> ⇔ <span class=\"math-tex\">$\\int\\dfrac{d[f(x)]}{h[f(x)]}=\\int g(x).dx$<\/span><\/p>","answer":"<p>Ch\u1ecdn <span style=\"color:#16a085;\"><strong>A.<\/strong> <span class=\"math-tex\">$I=\\dfrac{1186}{45}$<\/span><\/span><\/p><p>Do <span class=\"math-tex\">$f(x)$<\/span> \u0111\u1ed3ng bi\u1ebfn trên \u0111o\u1ea1n [1; 4] nên <span class=\"math-tex\">$f^\\prime(x)\\ge0,\\forall x\\in[1;4]$<\/span>.<\/p><p>Ta có: <span class=\"math-tex\">$x+2x.f(x)=\\big[f^\\prime(x)\\big]^2$<\/span> ⇔ <span class=\"math-tex\">$x(1+2f(x))=\\big[f^\\prime(x)\\big]^2$<\/span>, do <span class=\"math-tex\">$x\\in[1;4];\\space f^\\prime(x)\\ge0,\\forall x\\in[1;4]$<\/span><\/p><p>⇒ <span class=\"math-tex\">$f(x)>\\dfrac{-1}{2}$<\/span> và <span class=\"math-tex\">$f^\\prime(x)=\\sqrt{x}.\\sqrt{1+2f(x)}$<\/span><\/p><p>⇔ <span class=\"math-tex\">$\\dfrac{f^\\prime(x)}{\\sqrt{1+2f(x)}}=\\sqrt{x}$<\/span> <\/p><p>⇔ <span class=\"math-tex\">$\\Big(\\sqrt{1+2f(x)}\\Big)^\\prime=\\sqrt{x}$<\/span> <\/p><p>⇒ <span class=\"math-tex\">$\\sqrt{1+2f(x)}=\\int\\sqrt{x}dx$<\/span> ⇔ <span class=\"math-tex\">$\\sqrt{1+2f(x)}=\\dfrac{2}{3}x\\sqrt{x}+c$<\/span><\/p><p>Vì <span class=\"math-tex\">$f(1) =\\dfrac{3}{2}$<\/span> nên <span class=\"math-tex\">$\\sqrt{1+2.\\dfrac{3}{2}}=\\dfrac{2}{3}+c$<\/span> ⇒ <span class=\"math-tex\">$c=\\dfrac{4}{3}$<\/span>. <\/p><p>Suy ra <span class=\"math-tex\">$\\sqrt{1+2f(x)}=\\dfrac{2}{3}x\\sqrt{x}+\\dfrac{4}{3}$<\/span> ⇔ <span class=\"math-tex\">$1+2f(x)=\\bigg(\\dfrac{2}{3}x\\sqrt{x}+\\dfrac{4}{3}\\bigg)^2$<\/span><\/p><p>⇔ <span class=\"math-tex\">$f(x)=\\dfrac{2}{9}x^3+\\dfrac{8}{9}x^{^\\frac{3}{2}}+\\dfrac{7}{18}$<\/span><\/p><p>Khi \u0111ó <span class=\"math-tex\">$I=\\int\\limits_1^4\\bigg(\\dfrac{2}{9}x^3+\\dfrac{8}{9}x^{^\\frac{3}{2}}+\\dfrac{7}{18}\\bigg)dx$<\/span> = <span class=\"math-tex\">$\\bigg(\\dfrac{1}{18}x^4+\\dfrac{16}{45}x^{^\\frac{5}{2}}+\\dfrac{7}{18}x\\bigg)\\bigg|^4_1$<\/span> = <span class=\"math-tex\">$\\dfrac{1186}{45}$<\/span>.<\/p>","type":"choose","extra_type":"classic","user_id":"131","test":"0","date":"2024-11-19 08:47:50","option_type":"math","len":0},{"id":"5903","post_id":"8059","mon_id":"1159285","chapter_id":"1159380","question":"<p>Cho hàm s\u1ed1 <span class=\"math-tex\">$f(x)$<\/span> th\u1ecfa mãn <span class=\"math-tex\">$f(x)+f^\\prime(x)=e^{-x},\\forall x\\in\\mathbb R$<\/span> và <span class=\"math-tex\">$f(0)=2$<\/span>. Tính <span class=\"math-tex\">$I=\\int\\limits_1^2\\dfrac{f(x)e^x}{x}dx$<\/span>.<\/p>","options":["<strong>A.<\/strong> <span class=\"math-tex\">$I=2\\ln2$<\/span>","<strong>B.<\/strong> <span class=\"math-tex\">$I=\\ln2$<\/span>","<strong>C.<\/strong> <span class=\"math-tex\">$I=1+\\ln2$<\/span>","<strong>D.<\/strong> <span class=\"math-tex\">$I=1+2\\ln2$<\/span>"],"correct":"4","level":"3","hint":"<p>S\u1eed d\u1ee5ng ph\u01b0\u01a1ng pháp: d\u1ea1ng <span class=\"math-tex\">$f^\\prime(x)+f(x)=h(x)$<\/span>.<\/p><p>Nhân hai v\u1ebf v\u1edbi <span class=\"math-tex\">$e^x$<\/span> ta \u0111\u01b0\u1ee3c <span class=\"math-tex\">$e^x.f^\\prime(x)+e^x.f(x)=e^x.h(x)$<\/span> ⇔ <span class=\"math-tex\">$\\Big[e^x.f(x)\\big]^\\prime=e^x.h(x)$<\/span>.<\/p><p>Suy ra <span class=\"math-tex\">$e^x.f(x)=\\int e^x.h(x)dx$<\/span>.<\/p><p>T\u1eeb \u0111ây ta d\u1ec5 dàng tính \u0111\u01b0\u1ee3c <span class=\"math-tex\">$f(x)$<\/span>.<\/p>","answer":"<p>Ch\u1ecdn <span style=\"color:#16a085;\"><strong>D.<\/strong> <span class=\"math-tex\">$I=1+2\\ln2$<\/span><\/span><\/p><p><span class=\"math-tex\">$f(x)+f^\\prime(x)=e^{-x},\\forall x\\in\\mathbb R$<\/span> ⇔ <span class=\"math-tex\">$f(x)e^x+f^\\prime(x)e^x=1$<\/span> ⇔ <span class=\"math-tex\">$\\Big[e^x.f(x)\\big]^\\prime=1$<\/span><\/p><p>Suy ra <span class=\"math-tex\">$e^x.f(x)=\\int 1dx$<\/span> ⇔ <span class=\"math-tex\">$e^x.f(x)=x+C$<\/span><\/p><p>Vì <span class=\"math-tex\">$f(0)=2$<\/span> nên <span class=\"math-tex\">$e^0.2=0+C$<\/span> ⇒ C = 2. Do \u0111ó <span class=\"math-tex\">$e^x.f(x)=x+2$<\/span>.<\/p><p>V\u1eady <span class=\"math-tex\">$I=\\int\\limits_1^2\\dfrac{f(x)e^x}{x}dx$<\/span> = <span class=\"math-tex\">$\\int\\limits_1^2\\dfrac{x+2}{x}dx$<\/span> = <span class=\"math-tex\">$\\int\\limits_1^2\\bigg(1+\\dfrac{2}{x}\\bigg)dx$<\/span> = <span class=\"math-tex\">$(x+2\\ln|x|)\\Big|^2_1$<\/span> = <span class=\"math-tex\">$1+2\\ln2$<\/span>.<\/p>","type":"choose","extra_type":"classic","user_id":"131","test":"0","date":"2024-11-19 09:17:24","option_type":"math","len":0}]}