Chú ý: Để đảm bảo quyền lợi và bảo vệ tài khoản của mình
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{"common":{"save":0,"post_id":"7035","level":3,"total":10,"point":10,"point_extra":0},"segment":[{"id":"4711","post_id":"7035","mon_id":"1158924","chapter_id":"1159141","question":"<p>N\u1ebfu ba góc A, B, C c\u1ee7a tam giác ABC th\u1ecfa mãn <span class=\"math-tex\">$sinA=\\dfrac{sinB+sinC}{cosB+cosC}$<\/span> thì tam giác này<\/p>","options":["<strong>A.<\/strong> vuông t\u1ea1i A","<strong>B.<\/strong> vuông t\u1ea1i B","<strong>C.<\/strong> vuông t\u1ea1i C","<strong>D.<\/strong> cân t\u1ea1i A"],"correct":"1","level":"3","hint":"","answer":"<p><span class=\"math-tex\">$sinA=\\dfrac{sinB+sinC}{cosB+cosC}$<\/span> ⇔ <span class=\"math-tex\">$sinA=\u200b\u200b\\dfrac{2sin\\dfrac{B+C}{2}.cos\\dfrac{B-C}{2}}{2cos\\dfrac{B+C}{2}.cos\\dfrac{B-C}{2}}$<\/span> <\/p><p>⇔ <span class=\"math-tex\">$sinA=\\dfrac{sin\\dfrac{B+C}{2}}{cos\\dfrac{B+C}{2}}$<\/span> ⇔ <span class=\"math-tex\">$2sin\\dfrac{A}{2}.cos\\dfrac{A}{2}=\\dfrac{sin\\dfrac{180^0-A}{2}}{cos\\dfrac{180^0-A}{2}}$<\/span> <\/p><p>⇔ <span class=\"math-tex\">$2sin\\dfrac{A}{2}.cos\\dfrac{A}{2}=\\dfrac{sin\\bigg(90^0-\\dfrac{A}{2}\\bigg)}{cos\\big(90^0-\\dfrac{A}{2}\\bigg)}$<\/span> ⇔ <span class=\"math-tex\">$2sin\\dfrac{A}{2}.cos\\dfrac{A}{2}=\\dfrac{cos\\dfrac{A}{2}}{sin\\dfrac{A}{2}}$<\/span> <\/p><p> <span class=\"math-tex\">$\\Leftrightarrow 2sin\\dfrac{A}{2}=\\dfrac{1}{sin\\dfrac{A}{2}}$<\/span> <\/p><p>⇔ <span class=\"math-tex\">$1-2sin^2\\dfrac{A}{2}=0$<\/span> ⇔ <span class=\"math-tex\">$cosA=0$<\/span> ⇒ <span class=\"math-tex\">$\\widehat{A}$<\/span> = 90° .<\/p><p>V\u1eady tam giác ABC vuông t\u1ea1i A.<\/p><p>Ch\u1ecdn <span style=\"color:#16a085;\"><strong>A.<\/strong> vuông t\u1ea1i A.<\/span><\/p>","type":"choose","extra_type":"classic","user_id":"131","test":"0","date":"2023-11-16 01:18:51","option_type":"txt","len":2},{"id":"4713","post_id":"7035","mon_id":"1158924","chapter_id":"1159141","question":"<p>\u0110\u1eb3ng th\u1ee9c nào sau \u0111ây là \u0111úng?<\/p>","options":["<strong>A.<\/strong> <span class=\"math-tex\">$sin^4a+cos^4a=\\dfrac{3+cos4a}{4}$<\/span>","<strong>B.<\/strong> <span class=\"math-tex\">$sin^4a+cos^4a=\\dfrac{1+cos4a}{2}$<\/span>","<strong>C.<\/strong> <span class=\"math-tex\">$sin^4a+cos^4a=\\dfrac{3-cos4a}{4}$<\/span>","<strong>D.<\/strong> <span class=\"math-tex\">$sin^4a+cos^4a=\\dfrac{2-2sin2a}{2}$<\/span>"],"correct":"1","level":"3","hint":"","answer":"<p><span class=\"math-tex\">$sin^4a+cos^4a$<\/span> <span class=\"math-tex\">$=(sin^2a+cos^2a)^2-2sin^2a.cos^2a$<\/span> <span class=\"math-tex\">$=1-\\dfrac{1}{2}sin^22a$<\/span><\/p><p> <span class=\"math-tex\">$=1-\\dfrac{1}{2}.\\dfrac{1-cos4a}{2}$<\/span> <span class=\"math-tex\">$=1-\\dfrac{1-cos4a}{4}$<\/span> <span class=\"math-tex\">$=\\dfrac{3+cos4a}{4}$<\/span> .<\/p><p>Ch\u1ecdn <span style=\"color:#16a085;\"><strong>A.<\/strong> <span class=\"math-tex\">$sin^4a+cos^4a=\\dfrac{3+cos4a}{4}$<\/span> .<\/span><\/p>","type":"choose","extra_type":"classic","user_id":"131","test":"0","date":"2023-11-16 01:33:02","option_type":"math","len":0},{"id":"4717","post_id":"7035","mon_id":"1158924","chapter_id":"1159141","question":"<p>Trong các công th\u1ee9c sau, công th\u1ee9c nào là sai?<\/p>","options":["<strong>A.<\/strong> <span class=\"math-tex\">$1-sin2x=(sinx-cosx)^2$<\/span>","<strong>B.<\/strong> <span class=\"math-tex\">$1+sin2x=2sin^2\\bigg(x+\\dfrac{\\pi}{4}\\bigg)$<\/span>","<strong>C.<\/strong> <span class=\"math-tex\">$sin2x=(sinx+cosx+1)(sinx+cosx-1)$<\/span>","<strong>D.<\/strong> <span class=\"math-tex\">$sin2x=2sin\\bigg(x+\\dfrac{\\pi}{4}\\bigg).cos\\bigg(x-\\dfrac{\\pi}{4}\\bigg)$<\/span>"],"correct":"4","level":"3","hint":"","answer":"<p><strong>A.<\/strong> <span class=\"math-tex\">$1-sin2x=sin^2x+cos^2x-2.sinx.cosx=(sinx-cosx)^2$<\/span> , do \u0111ó <strong>A<\/strong> \u0111úng.<\/p><p><strong>B.<\/strong> <span class=\"math-tex\">$2sin^2\\bigg(x+\\dfrac{\\pi}{4}\\bigg)=1-cos\\bigg(2x+\\dfrac{\\pi}{2}\\bigg)$<\/span><\/p><p><strong><span class=\"math-tex\">$=1+cos\\bigg(\\pi-2x-\\dfrac{\\pi}{2}\\bigg)=1+cos\\bigg(\\dfrac{\\pi}{2}-2x\\bigg)=1+sin2x$<\/span><\/strong> , do \u0111ó <strong>B<\/strong> \u0111úng.<\/p><p><strong>C.<\/strong> <span class=\"math-tex\">$(sinx+cosx+1)(sinx+cosx-1)=(sinx+cosx)^2-1$<\/span><\/p><p><span class=\"math-tex\">$=sin^2x+cos^2x+2sinx.cosx-1$<\/span> <span class=\"math-tex\">$=1+sin2x-1=sin2x$<\/span>, do \u0111ó <strong>C<\/strong> \u0111úng.<\/p><p><strong>D.<\/strong> <span class=\"math-tex\">$2sin\\bigg(x+\\dfrac{\\pi}{4}\\bigg).cos\\bigg(x-\\dfrac{\\pi}{4}\\bigg)$<\/span> <span class=\"math-tex\">$=sin\\dfrac{\\pi}{2}-sin2x$<\/span> <span class=\"math-tex\">$=1-sin2x$<\/span> , do \u0111ó <strong>D<\/strong> sai.<\/p><p>Ch\u1ecdn <span style=\"color:#16a085;\"><strong>D.<\/strong> <span class=\"math-tex\">$sin2x=2sin\\bigg(x+\\dfrac{\\pi}{4}\\bigg).cos\\bigg(x-\\dfrac{\\pi}{4}\\bigg)$<\/span> .<\/span><\/p>","type":"choose","extra_type":"classic","user_id":"131","test":"0","date":"2023-11-16 01:56:16","option_type":"math","len":0}]}