{"common":{"save":0,"post_id":"7033","level":1,"total":10,"point":10,"point_extra":0},"segment":[{"id":"4673","post_id":"7033","mon_id":"1158924","chapter_id":"1159141","question":"<p>\u0110\u1eb3ng th\u1ee9c nào d\u01b0\u1edbi \u0111ây \u0111úng?<\/p>","options":["<strong>A.<\/strong> <span class=\"math-tex\">$sin\\bigg(a+\\dfrac{\\pi}{6}\\bigg)=sina+\\dfrac{1}{2}$<\/span>","<strong>B.<\/strong> <span class=\"math-tex\">$sin\\bigg(a+\\dfrac{\\pi}{6}\\bigg)=\\dfrac{1}{2}sina+\\dfrac{\\sqrt{3}}{2}cosa$<\/span>","<strong>C.<\/strong> <span class=\"math-tex\">$sin\\bigg(a+\\dfrac{\\pi}{6}\\bigg)=\\dfrac{\\sqrt{3}}{2}sina-\\dfrac{1}{2}cosa$<\/span>","<strong>D.<\/strong> <span class=\"math-tex\">$sin\\bigg(a+\\dfrac{\\pi}{6}\\bigg)=\\dfrac{\\sqrt{3}}{2}sina+\\dfrac{1}{2}cosa$<\/span>"],"correct":"4","level":"1","hint":"","answer":"<p>Áp d\u1ee5ng công th\u1ee9c c\u1ed9ng <span class=\"math-tex\">$sin\\bigg(a+\\dfrac{\\pi}{6}\\bigg)$<\/span> <span class=\"math-tex\">$=sina.cos\\dfrac{\\pi}{6}+cosa.sin\\dfrac{\\pi}{6}$<\/span> <span class=\"math-tex\">$=\\dfrac{\\sqrt{3}}{2}sina+\\dfrac{1}{2}cosa$<\/span><\/p><p>Ch\u1ecdn <span style=\"color:#16a085;\"><strong>D.<\/strong> <span class=\"math-tex\">$sin\\bigg(a+\\dfrac{\\pi}{6}\\bigg)=\\dfrac{\\sqrt{3}}{2}sina+\\dfrac{1}{2}cosa$<\/span> .<\/span><\/p>","type":"choose","extra_type":"classic","user_id":"131","test":"0","date":"2023-11-12 02:16:57","option_type":"math","len":0},{"id":"4676","post_id":"7033","mon_id":"1158924","chapter_id":"1159141","question":"<p>Cho tan x = 2. Giá tr\u1ecb c\u1ee7a bi\u1ec3u th\u1ee9c <br \/>A = <span class=\"math-tex\">$\\dfrac{sin\\bigg(x+\\dfrac{\\pi}{4}\\bigg)}{sin^3\\bigg(x-\\dfrac{\\pi}{4}\\bigg)}$<\/span> b\u1eb1ng<\/p>","options":["<strong>A.<\/strong> <span class=\"math-tex\">$\\dfrac{6}{5}$<\/span>","<strong>B.<\/strong> 30","<strong>C.<\/strong> <span class=\"math-tex\">$\\dfrac{6}{7}$<\/span>","<strong>D.<\/strong> <span class=\"math-tex\">$\\dfrac{3}{5}$<\/span>"],"correct":"2","level":"1","hint":"","answer":"<p>T\u1eeb gi\u1ea3 thi\u1ebft tan x = 2 ⇒ sin x = 2 cos x .<\/p><p> <span class=\"math-tex\">$A=\\dfrac{sinxcos\\dfrac{\\pi}{4}+cosxsin\\dfrac{\\pi}{4}}{\\bigg[sinxcos\\dfrac{\\pi}{4}-cosxsin\\dfrac{\\pi}{4}\\bigg]^3}$<\/span> <span class=\"math-tex\">$= \\dfrac{\\dfrac{\\sqrt{2}}{2}(sinx+cosx)}{\\bigg[\\dfrac{\\sqrt{2}}{2}(sinx-cosx)\\bigg]^3}$<\/span> <span class=\"math-tex\">$=\\dfrac{2(sinx+cosx)}{(sinx-cosx)^3}$<\/span> <\/p><p><span class=\"math-tex\">$=\\dfrac{2(2cosx+cosx)}{(2cosx-cosx)^3}$<\/span> <span class=\"math-tex\">$=\\dfrac{6}{cos^2x}$<\/span> <span class=\"math-tex\">$=6(tan^2x+1)=30$<\/span> .<\/p><p>Ch\u1ecdn <span style=\"color:#16a085;\"><strong>B.<\/strong> 30.<\/span><\/p>","type":"choose","extra_type":"classic","user_id":"131","test":"0","date":"2023-11-12 02:28:39","option_type":"math","len":0},{"id":"4678","post_id":"7033","mon_id":"1158924","chapter_id":"1159141","question":"<p>Cho tan(a + b) = 2 ; tan(a – b) = –3. Giá tr\u1ecb c\u1ee7a bi\u1ec3u th\u1ee9c tan 2b b\u1eb1ng<\/p>","options":["<strong>A.<\/strong> <span class=\"math-tex\">$\\dfrac{5}{7}$<\/span>","<strong>B.<\/strong> <span class=\"math-tex\">$-\\dfrac{1}{7}$<\/span>","<strong>C.<\/strong> <span class=\"math-tex\">$-1$<\/span>","<strong>D.<\/strong> <span class=\"math-tex\">$1$<\/span>"],"correct":"3","level":"1","hint":"","answer":"<p>Ta có tan 2b = tan [(a + b) – (a – b)] <span class=\"math-tex\">$=\\dfrac{tan(a+b)-tan(a-b)}{1+tan(a+b)tan(a-b)}$<\/span> <span class=\"math-tex\">$=\\dfrac{2+3}{1+2.(-3)}=-1$<\/span><\/p><p>Ch\u1ecdn <span style=\"color:#16a085;\"><strong>C.<\/strong> <span class=\"math-tex\">$-1$<\/span> .<\/span><\/p>","type":"choose","extra_type":"classic","user_id":"131","test":"0","date":"2023-11-12 02:33:53","option_type":"math","len":0}]}
