{"common":{"save":0,"post_id":"6431","level":2,"total":10,"point":10,"point_extra":0},"segment":[{"id":"6429","mon_id":"1158767","chapter_id":"1158829","post_id":"6431","question":"<p>Cho các \u0111i\u1ec3m <span class=\"math-tex\">$A(1;2),B(1;4),C(7;2)$<\/span> Tính s\u1ed1 \u0111o góc <span class=\"math-tex\">$A$<\/span> c\u1ee7a tam giác ABC<\/p>","options":["A. <span class=\"math-tex\">$30^o$<\/span>","B. <span class=\"math-tex\">$60^o$<\/span>","C. <span class=\"math-tex\">$90^o$<\/span>","D. <span class=\"math-tex\">$45^o$<\/span>"],"correct":"3","level":"2","hint":"","answer":"<p><span style=\"color:#27ae60;\"><strong>\u0110áp án \u0111úng là: C. <span class=\"math-tex\">$90^o$<\/span><\/strong><\/span><\/p><p>Ta có: <\/p><p><span class=\"math-tex\">$\\overrightarrow{AB}=(1-1;4-2)=(0;2)$<\/span><\/p><p><span class=\"math-tex\">$\\overrightarrow{AC}=(7-1;2-2)=(6;0)$<\/span><\/p><p><span class=\"math-tex\">$\\overrightarrow{BC}=(7-1;2-4)=(6;-2)$<\/span><\/p><p>\u0110\u1ed9 dài các c\u1ea1nh c\u1ee7a tam giác là:<\/p><p><span class=\"math-tex\">$AB=\\sqrt{0^2+2^2}=2;AC=\\sqrt{0^2+6^2}=\\sqrt{36}=6$<\/span><\/p><p><span class=\"math-tex\">$BC=\\sqrt{6^2+(-2)^2}=2\\sqrt{10}$<\/span><\/p><p>Xét \u0111\u1ed9 dài các c\u1ea1nh, ta th\u1ea5y <span class=\"math-tex\">$AB^2+AC^2=BC^2$<\/span> nên tam giác ABC là tam giác vuông t\u1ea1i A.<\/p>","type":"choose","extra_type":"classic","user_id":"147","test":"0","date":"2023-03-06 00:54:07","option_type":"math","len":0},{"id":"6430","mon_id":"1158767","chapter_id":"1158829","post_id":"6431","question":"<p>Cho tam giác <span class=\"math-tex\">$ABC$<\/span> , <span class=\"math-tex\">$m_a$<\/span>\u200b là \u0111\u1ed9 dài \u0111\u01b0\u1eddng trung tuy\u1ebfn k\u1ebb t\u1eeb \u0111\u1ec9nh A xu\u1ed1ng c\u1ea1nh BC trong tam giác ABC. <\/p><p>Ch\u1ecdn công th\u1ee9c \u0111úng:<\/p>","options":["A. <span class=\"math-tex\">$m_a^2=\\frac{b^2+c^2}{2}+\\frac{a^2}{4}$<\/span>","B. <span class=\"math-tex\">$m_a^2=\\frac{a^2+c^2}{2}-\\frac{b^2}{4}$<\/span>","C. <span class=\"math-tex\">$m_a^2=\\frac{a^2+b^2}{2}-\\frac{c^2}{4}$<\/span>","D. <span class=\"math-tex\">$m_a^2=\\frac{2c^2+2b^2-a^2}{4}$<\/span> "],"correct":"4","level":"2","hint":"<p>Áp d\u1ee5ng h\u1ec7 qu\u1ea3 c\u1ee7a \u0111\u1ecbnh lí Côsin trong tam giác ta có:<\/p><p><span class=\"math-tex\">$m_a^2=\\frac{2c^2+2b^2-a^2}{4}$<\/span><\/p>","answer":"<p><span style=\"color:#27ae60;\"><\/span>\u0110áp án :<span style=\"color:#27ae60;\"><strong> <\/strong><strong>D. <span class=\"math-tex\">$m_a^2=\\frac{2c^2+2b^2-a^2}{4}$<\/span><\/strong><\/span><\/p><p>Áp d\u1ee5ng h\u1ec7 qu\u1ea3 c\u1ee7a \u0111\u1ecbnh lí Côsin trong tam giác.<\/p>","type":"choose","extra_type":"classic","user_id":"147","test":"0","date":"2023-03-06 01:01:50","option_type":"math","len":0},{"id":"6431","mon_id":"1158767","chapter_id":"1158829","post_id":"6431","question":"<p>Tam giác <span class=\"math-tex\">$ABC$<\/span> có <span class=\"math-tex\">$AB=9cm,BC=15cm,AC=13cm$<\/span>. Khi \u0111ó \u0111\u01b0\u1eddng trung tuy\u1ebfn <span class=\"math-tex\">$AM$<\/span> c\u1ee7a tam giác có \u0111\u1ed9 dài là<\/p>","options":["A. <span class=\"math-tex\">${\\sqrt{267}\\over2}cm$<\/span> .","B. <span class=\"math-tex\">${\\sqrt{267}\\over4}cm$<\/span> .","C. <span class=\"math-tex\">${\\sqrt{267}\\over6}cm$<\/span> .","D. <span class=\"math-tex\">${\\sqrt{267}\\over8}cm$<\/span> ."],"correct":"1","level":"2","hint":"<p><span class=\"svgedit\"><svg height=\"217\" width=\"319\" xmlns:xlink=\"http:\/\/www.w3.org\/1999\/xlink\"> <g><title><\/title><rect fill=\"#fff\" height=\"219\" id=\"canvas_background\" width=\"321\" x=\"-1\" y=\"-1\"><\/rect> <g display=\"none\" height=\"100%\" id=\"canvasGrid\" overflow=\"visible\" width=\"100%\" x=\"0\" y=\"0\"> <rect fill=\"url(#gridpattern)\" height=\"100%\" stroke-width=\"0\" width=\"100%\" x=\"0\" y=\"0\"><\/rect> <\/g> <\/g> <g><title><\/title><image height=\"217\" id=\"svg_1\" width=\"319\" x=\"0\" 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y=\"0\"><\/image> <\/g> <\/svg><\/span><\/p><p>Áp d\u1ee5ng h\u1ec7 qu\u1ea3 c\u1ee7a \u0111\u1ecbnh lí côsin ta có: <\/p><p><span class=\"math-tex\">$AM^2=\\frac{2.9^2+2.13^2-15^2}{4}=\\frac{267}{4}\\Rightarrow AM=$<\/span><span class=\"math-tex\">${\\sqrt{267}\\over2}$<\/span><\/p>","answer":"<p><span style=\"color:#27ae60;\"><strong>\u0110áp án: A<\/strong><\/span><\/p>","type":"choose","extra_type":"classic","user_id":"147","test":"0","date":"2023-03-06 01:11:11","option_type":"math","len":0}]}