Chú ý: Để đảm bảo quyền lợi và bảo vệ tài khoản của mình
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{"common":{"save":0,"post_id":"7611","level":3,"total":10,"point":10,"point_extra":0},"segment":[{"id":"5917","post_id":"7611","mon_id":"1159285","chapter_id":"1159380","question":"<p>Bi\u1ebft <span class=\"math-tex\">$\\displaystyle\\int\\limits_{\\frac{\\pi}{6}}^{\\frac{\\pi}{4}}(2\\cot^2x+5)dx=\\dfrac{\\pi}{a}+b\\sqrt{3}+c$<\/span> v\u1edbi a, b, c ∈ Z. Tính P = a + b + c.<\/p>","options":["<strong>A.<\/strong> 6","<strong>B.<\/strong> –4","<strong>C.<\/strong> 4","<strong>D.<\/strong> –6"],"correct":"3","level":"3","hint":"","answer":"<p>Ch\u1ecdn <span style=\"color:#16a085;\"><strong>C.<\/strong> 4.<\/span><\/p><p><span class=\"math-tex\">$\\displaystyle\\int\\limits_{\\frac{\\pi}{6}}^{\\frac{\\pi}{4}}(2\\cot^2x+5)dx$<\/span><\/p><p><span class=\"math-tex\">$=\\displaystyle\\int\\limits_{\\frac{\\pi}{6}}^{\\frac{\\pi}{4}}\\bigg(2\\bigg(\\dfrac{1}{\\sin^2x}-1\\bigg)+5\\bigg)dx$<\/span><\/p><p><span class=\"math-tex\">$=\\displaystyle\\int\\limits_{\\frac{\\pi}{6}}^{\\frac{\\pi}{4}}\\bigg(3+\\dfrac{2}{\\sin^2x}\\bigg)dx$<\/span><\/p><p><span class=\"math-tex\">$=(3x-2\\cot x)\\;\\bigg|_{\\frac{\\pi}{6}}^{\\frac{\\pi}{4}}$<\/span><\/p><p><span class=\"math-tex\">$=\\dfrac{\\pi}{4}-2(1-\\sqrt{3})$<\/span><\/p><p><span class=\"math-tex\">$=\\dfrac{\\pi}{4}+2\\sqrt{3}-2$<\/span><\/p><p>Khi \u0111ó a = 4, b = 2, c = –2 nên P = a + b + c = 4.<\/p>","type":"choose","extra_type":"classic","user_id":"131","test":"0","date":"2024-11-20 08:44:31","option_type":"txt","len":1},{"id":"5919","post_id":"7611","mon_id":"1159285","chapter_id":"1159380","question":"<p>Bi\u1ebft <span class=\"math-tex\">$\\displaystyle\\int\\limits_{0}^{\\frac{\\pi}{2}}\\sin^2\\dfrac{x}{4}\\cos^2\\dfrac{x}{4}dx=\\dfrac{\\pi}{c}-\\dfrac{a}{b}$<\/span> v\u1edbi a, b, c ∈ Z, <span class=\"math-tex\">$\\dfrac{a}{b}$<\/span> là phân s\u1ed1 t\u1ed1i gi\u1ea3n. Khi \u0111ó giá tr\u1ecb c\u1ee7a P = a + b + c là<\/p>","options":["<strong>A.<\/strong> 17","<strong>B.<\/strong> 16","<strong>C.<\/strong> 32","<strong>D.<\/strong> 25"],"correct":"4","level":"3","hint":"","answer":"<p>Ch\u1ecdn <span style=\"color:#16a085;\"><strong>D.<\/strong> 25.<\/span><\/p><p><span class=\"math-tex\">$\\displaystyle\\int\\limits_{0}^{\\frac{\\pi}{2}}\\sin^2\\dfrac{x}{4}\\cos^2\\dfrac{x}{4}dx$<\/span><\/p><p><span class=\"math-tex\">$=\\dfrac{1}{4}\\displaystyle\\int\\limits_{0}^{\\frac{\\pi}{2}}\\sin^2\\dfrac{x}{2}dx$<\/span><\/p><p><span class=\"math-tex\">$=\\dfrac{1}{4}\\displaystyle\\int\\limits_{0}^{\\frac{\\pi}{2}}\\dfrac{1-\\cos x}{2}dx$<\/span><\/p><p><span class=\"math-tex\">$=\\dfrac{1}{8}(x-\\sin x)\\;\\bigg|_0^{\\frac{\\pi}{2}}$<\/span><\/p><p><span class=\"math-tex\">$=\\dfrac{\\pi}{16}-\\dfrac{1}{8}$<\/span><\/p><p>Khi \u0111ó a = 1, b = 8, c = 16 nên P = a + b + c = 1 + 8 + 16 = 25.<\/p>","type":"choose","extra_type":"classic","user_id":"131","test":"0","date":"2024-11-20 09:00:52","option_type":"txt","len":0},{"id":"5925","post_id":"7611","mon_id":"1159285","chapter_id":"1159380","question":"<p>Tính <span class=\"math-tex\">$I=\\displaystyle\\int\\limits_{0}^{2\\pi}\\sqrt{1-\\cos2x}dx$<\/span>.<\/p>","options":["A. <span class=\"math-tex\">$\\sqrt{3}$<\/span>","<strong>B.<\/strong> <span class=\"math-tex\">$4\\sqrt{2}$<\/span>","<strong>C.<\/strong> <span class=\"math-tex\">$2\\sqrt{3}$<\/span>","<strong>D.<\/strong> <span class=\"math-tex\">$\\sqrt{2}$<\/span>"],"correct":"2","level":"3","hint":"","answer":"<p>Ch\u1ecdn <span style=\"color:#16a085;\"><strong>B.<\/strong> <span class=\"math-tex\">$4\\sqrt{2}$<\/span>.<\/span><\/p><p>V\u1edbi x ∈ [0 ; <span class=\"math-tex\">$\\pi$<\/span>] thì sin x <span class=\"math-tex\">$\\ge$<\/span> 0 nên |sin x| = sin x.<\/p><p>V\u1edbi x ∈ [<span class=\"math-tex\">$\\pi$<\/span> ; 2<span class=\"math-tex\">$\\pi$<\/span>] thì sin x <span class=\"math-tex\">$\\le$<\/span> 0 nên |sin x| = –sin x.<\/p><p>Mà <span class=\"math-tex\">$\\sqrt{1-\\cos2x}=\\sqrt{2\\sin^2x}=\\sqrt{2}|\\sin x|$<\/span> nên <\/p><p><span class=\"math-tex\">$I=\\displaystyle\\int\\limits_{0}^{2\\pi}\\sqrt{1-\\cos2x}dx$<\/span><\/p><p><span class=\"math-tex\">$=\\sqrt{2}\\displaystyle\\int\\limits_{0}^{2\\pi}|\\sin x|dx$<\/span><\/p><p><span class=\"math-tex\">$=\\sqrt{2}\\bigg(\\displaystyle\\int\\limits_{0}^{\\pi}\\sin xdx-\\displaystyle\\int\\limits_{\\pi}^{2\\pi}\\sin xdx\\bigg)$<\/span><\/p><p><span class=\"math-tex\">$=\\sqrt{2}\\bigg(-\\cos x\\bigg|_0^\\pi+\\cos x\\bigg|_\\pi^{2\\pi}\\bigg)$<\/span><\/p><p><span class=\"math-tex\">$=4\\sqrt{2}$<\/span>.<\/p>","type":"choose","extra_type":"classic","user_id":"131","test":"0","date":"2024-11-20 09:22:56","option_type":"math","len":0}]}