Chú ý: Để đảm bảo quyền lợi và bảo vệ tài khoản của mình
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{"common":{"save":0,"post_id":"8306","level":2,"total":10,"point":10,"point_extra":0},"segment":[{"id":"6139","post_id":"8306","mon_id":"1159285","chapter_id":"1159382","question":"<p>Trong không gian v\u1edbi h\u1ec7 tr\u1ee5c t\u1ecda \u0111\u1ed9 Oxyz, cho hai \u0111i\u1ec3m <span class=\"math-tex\">$A(1;2;3)$<\/span>, <span class=\"math-tex\">$B(5;4;-1)$<\/span>. Ph\u01b0\u01a1ng trình m\u1eb7t c\u1ea7u \u0111\u01b0\u1eddng kính AB<br \/>là<\/p>","options":["<strong>A.<\/strong> <span class=\"math-tex\">$(x-3)^2+(y-3)^2+(z-1)^2=9$<\/span>","<strong>B.<\/strong> <span class=\"math-tex\">$(x-3)^2+(y-3)^2+(z-1)^2=6$<\/span>","<strong>C.<\/strong> <span class=\"math-tex\">$(x+3)^2+(y+3)^2+(z+1)^2=9$<\/span>","<strong>D.<\/strong> <span class=\"math-tex\">$(x-3)^2+(y-3)^2+(z-1)^2=36$<\/span>"],"correct":"1","level":"2","hint":"<p>Vi\u1ebft ph\u01b0\u01a1ng trình m\u1eb7t c\u1ea7u (S) có \u0111\u01b0\u1eddng kính AB, v\u1edbi A, B cho tr\u01b0\u1edbc.<\/p><p>(S) có tâm I là trung \u0111i\u1ec3m c\u1ee7a \u0111o\u1ea1n th\u1eb3ng AB và bán kính <span class=\"math-tex\">$R=\\dfrac{1}{2}AB$<\/span>.<\/p>","answer":"<p>Ch\u1ecdn <span style=\"color:#16a085;\"><strong>A.<\/strong> <span class=\"math-tex\">$(x-3)^2+(y-3)^2+(z-1)^2=9$<\/span><\/span><\/p><p>+ G\u1ecdi I là trung \u0111i\u1ec3m c\u1ee7a AB thì <span class=\"math-tex\">$I(3;3;1)$<\/span>.<\/p><p><span class=\"math-tex\">$\\overrightarrow{AB}=(4;2;-4)$<\/span> ⇒ <span class=\"math-tex\">$AB=\\sqrt{16+4+16}=6$<\/span><\/p><p>+ M\u1eb7t c\u1ea7u \u0111\u01b0\u1eddng kính AB có tâm <span class=\"math-tex\">$I(3;3;1)$<\/span>, bán kính <span class=\"math-tex\">$R=\\dfrac{1}{2}AB=3$<\/span> có ph\u01b0\u01a1ng trình là:<\/p><p><span class=\"math-tex\">$(x-3)^2+(y-3)^2+(z-1)^2=9$<\/span>.<\/p>","type":"choose","extra_type":"classic","user_id":"131","test":"0","date":"2025-01-05 16:48:32","option_type":"math","len":0},{"id":"6141","post_id":"8306","mon_id":"1159285","chapter_id":"1159382","question":"<p>Trong không gian v\u1edbi h\u1ec7 t\u1ecda \u0111\u1ed9 Oxyz, m\u1eb7t c\u1ea7u (S) qua b\u1ed1n \u0111i\u1ec3m <span class=\"math-tex\">$A(3;3;0)$<\/span>, <span class=\"math-tex\">$B(3;0;3)$<\/span>, <span class=\"math-tex\">$C(0;3;3)$<\/span>, <span class=\"math-tex\">$D(3;3;3)$<\/span>. Ph\u01b0\u01a1ng trình m\u1eb7t c\u1ea7u (S) là:<\/p>","options":["<strong>A.<\/strong> <span class=\"math-tex\">$\\bigg(x-\\dfrac{3}{2}\\bigg)^2+\\bigg(y-\\dfrac{3}{2}\\bigg)^2+\\bigg(z-\\dfrac{3}{2}\\bigg)^2=\\dfrac{3\\sqrt{3}}{4}$<\/span>","<strong>B.<\/strong> <span class=\"math-tex\">$\\bigg(x-\\dfrac{3}{2}\\bigg)^2+\\bigg(y+\\dfrac{3}{2}\\bigg)^2+\\bigg(z-\\dfrac{3}{2}\\bigg)^2=\\dfrac{27}{4}$<\/span>","<strong>C.<\/strong> <span class=\"math-tex\">$\\bigg(x-\\dfrac{3}{2}\\bigg)^2+\\bigg(y-\\dfrac{3}{2}\\bigg)^2+\\bigg(z+\\dfrac{3}{2}\\bigg)^2=\\dfrac{27}{4}$<\/span>","<strong>D.<\/strong> <span class=\"math-tex\">$\\bigg(x-\\dfrac{3}{2}\\bigg)^2+\\bigg(y-\\dfrac{3}{2}\\bigg)^2+\\bigg(z-\\dfrac{3}{2}\\bigg)^2=\\dfrac{27}{4}$<\/span>"],"correct":"4","level":"2","hint":"","answer":"<p>Ch\u1ecdn <span style=\"color:#16a085;\"><strong>D.<\/strong> <span class=\"math-tex\">$\\bigg(x-\\dfrac{3}{2}\\bigg)^2+\\bigg(y-\\dfrac{3}{2}\\bigg)^2+\\bigg(z-\\dfrac{3}{2}\\bigg)^2=\\dfrac{27}{4}$<\/span><\/span><\/p><p>G\u1ecdi ph\u01b0\u01a1ng trình m\u1eb7t c\u1ea7u là <span class=\"math-tex\">$(S):x^2+y^2+z^2-2ax-2by-2cz+d=0$<\/span>.<\/p><p>Vì m\u1eb7t c\u1ea7u (S) qua b\u1ed1n \u0111i\u1ec3m <span class=\"math-tex\">$A(3;3;0)$<\/span>, <span class=\"math-tex\">$B(3;0;3)$<\/span>, <span class=\"math-tex\">$C(0;3;3)$<\/span>, <span class=\"math-tex\">$D(3;3;3)$<\/span> nên<\/p><p><span class=\"math-tex\">$\\begin{cases}18-6a-6b\\;\\;\\;\\;\\;\\;\\;\\;+d=0\\\\18-6a\\;\\;\\;\\;\\;\\;\\;\\;-6c+d=0\\\\18\\;\\;\\;\\;\\;\\;\\;\\;-6b-6c+d=0\\\\27-6a-6b-6c+d=0\\end{cases}$<\/span> ⇔ <span class=\"math-tex\">$\\begin{cases}-6a-6b\\;\\;\\;\\;\\;\\;\\;\\;+d=-18\\\\-6a\\;\\;\\;\\;\\;\\;\\;\\;-6c+d=-18\\\\\\;\\;\\;\\;\\;\\;\\;\\;-6b-6c+d=-18\\\\-6a-6b-6c+d=-27\\end{cases}$<\/span> ⇔ <span class=\"math-tex\">$\\begin{cases}a=\\dfrac{3}{2}\\\\b=\\dfrac{3}{2}\\\\c=\\dfrac{3}{2}\\\\d=0\\end{cases}$<\/span><\/p><p>Suy ra tâm <span class=\"math-tex\">$I\\bigg(\\dfrac{3}{2};\\dfrac{3}{2};\\dfrac{3}{2}\\bigg)$<\/span> và bán kính <span class=\"math-tex\">$R=\\sqrt{\\bigg(\\dfrac{3}{2}\\bigg)^2+\\bigg(\\dfrac{3}{2}\\bigg)^2+\\bigg(\\dfrac{3}{2}\\bigg)^2}=\\dfrac{3\\sqrt{3}}{2}$<\/span>.<\/p><p>V\u1eady ph\u01b0\u01a1ng trình m\u1eb7t c\u1ea7u <span class=\"math-tex\">$\\bigg(x-\\dfrac{3}{2}\\bigg)^2+\\bigg(y-\\dfrac{3}{2}\\bigg)^2+\\bigg(z-\\dfrac{3}{2}\\bigg)^2=\\dfrac{27}{4}$<\/span>.<\/p>","type":"choose","extra_type":"classic","user_id":"131","test":"0","date":"2025-01-05 17:01:06","option_type":"math","len":0},{"id":"6143","post_id":"8306","mon_id":"1159285","chapter_id":"1159382","question":"<p>Trong không gian t\u1ecda \u0111\u1ed9 Oxyz, m\u1eb7t c\u1ea7u (S) \u0111i qua \u0111i\u1ec3m O và c\u1eaft các tia Ox, Oy, Oz l\u1ea7n l\u01b0\u1ee3t t\u1ea1i các \u0111i\u1ec3m A, B, C khác O th\u1ecfa mãn tam giác ABC có tr\u1ecdng tâm là \u0111i\u1ec3m G(–6; –12; 18). T\u1ecda \u0111\u1ed9 tâm c\u1ee7a m\u1eb7t c\u1ea7u (S) là:<\/p>","options":["<strong>A.<\/strong> (9; 18; –27)","<strong>B.<\/strong> (–3; –6; 9)","<strong>C.<\/strong> (3; 6; –9)","<strong>D.<\/strong> (–9; –18; 27)"],"correct":"4","level":"2","hint":"","answer":"<p>Ch\u1ecdn <span style=\"color:#16a085;\"><strong>D.<\/strong> (–9; –18; 27)<\/span><\/p><p>G\u1ecdi t\u1ecda \u0111\u1ed9 các \u0111i\u1ec3m trên ba tia Ox, Oy, Oz l\u1ea7n l\u01b0\u1ee3t là <span class=\"math-tex\">$A(a;0;0),B(0;b;0),C(0;0;c)$<\/span> v\u1edbi <span class=\"math-tex\">$a.b.c>0$<\/span>.<\/p><p>Vì G(–6; –12; 18) là tr\u1ecdng tâm tam giác ABC nên <span class=\"math-tex\">$\\begin{cases}\\dfrac{a}{3}=-6\\\\\\dfrac{b}{3}=-12\\\\\\dfrac{c}{3}=18\\end{cases}$<\/span> ⇔ <span class=\"math-tex\">$\\begin{cases}a=-18\\\\b=-36\\\\c=54\\end{cases}$<\/span>.<\/p><p>G\u1ecdi ph\u01b0\u01a1ng trình m\u1eb7t c\u1ea7u <span class=\"math-tex\">$(S):x^2+y^2+z^2-2mx-2ny-2pz+q=0$<\/span>. <\/p><p>Vì (S) qua các \u0111i\u1ec3m O, A, B, C nên ta có h\u1ec7: <\/p><p><span class=\"math-tex\">$\\begin{cases}q=0\\\\36m+q=-18^2\\\\72n+q=-36^2\\\\-108p+q=-54^2\\end{cases}$<\/span> ⇔ <span class=\"math-tex\">$\\begin{cases}m=-9\\\\n=-18\\\\p=27\\\\q=0\\end{cases}$<\/span><\/p><p>V\u1eady t\u1ecda \u0111\u1ed9 tâm m\u1eb7t c\u1ea7u (S) là (–9; –18; 27).<\/p>","type":"choose","extra_type":"classic","user_id":"131","test":"0","date":"2025-01-06 08:56:02","option_type":"txt","len":2}]}