{"common":{"save":0,"post_id":"8305","level":1,"total":10,"point":10,"point_extra":0},"segment":[{"id":"6132","post_id":"8305","mon_id":"1159285","chapter_id":"1159382","question":"<p>Cho \u0111i\u1ec3m M n\u1eb1m ngoài m\u1eb7t c\u1ea7u S(O; R). Kh\u1eb3ng \u0111\u1ecbnh nào d\u01b0\u1edbi \u0111ây <strong>\u0111úng<\/strong>?<\/p>","options":["<strong>A.<\/strong> OM < R","<strong>B.<\/strong> OM = R","<strong>C.<\/strong> OM > R","<strong>D.<\/strong> OM <span class=\"math-tex\">$\\le$<\/span> R"],"correct":"3","level":"1","hint":"<p>\u2219 N\u1ebfu OM = R thì M n\u1eb1m trên m\u1eb7t c\u1ea7u.<br \/>\u2219 N\u1ebfu OM < R thì M n\u1eb1m trong m\u1eb7t c\u1ea7u.<br \/>\u2219 N\u1ebfu OM > R thì M n\u1eb1m ngoài m\u1eb7t c\u1ea7u.<\/p>","answer":"<p>Ch\u1ecdn <span style=\"color:#16a085;\"><strong>C.<\/strong> OM > R<\/span><\/p><p>M n\u1eb1m ngoài m\u1eb7t c\u1ea7u S(O; R) ⇔ OM > R.<\/p>","type":"choose","extra_type":"classic","user_id":"131","test":"0","date":"2025-01-05 16:25:22","option_type":"math","len":0},{"id":"6133","post_id":"8305","mon_id":"1159285","chapter_id":"1159382","question":"<p>Trong không gian Oxyz, cho m\u1eb7t c\u1ea7u <span class=\"math-tex\">$(S):x^2+(y-2)^2+(z+1)^2=6$<\/span>. \u0110\u01b0\u1eddng kính c\u1ee7a (S) b\u1eb1ng:<\/p>","options":["<strong>A.<\/strong> <span class=\"math-tex\">$\\sqrt{6}$<\/span>","<strong>B.<\/strong> 12","<strong>C.<\/strong> 3","<strong>D.<\/strong> <span class=\"math-tex\">$2\\sqrt{6}$<\/span>"],"correct":"4","level":"1","hint":"<p>Ph\u01b0\u01a1ng trình m\u1eb7t c\u1ea7u (S) có d\u1ea1ng <span class=\"math-tex\">$(x-a)^2+(y-b)^2+(z-c)^2=R^2$<\/span> thì m\u1eb7t c\u1ea7u có tâm I(a; b; c) và có bán kính R.<\/p>","answer":"<p>Ch\u1ecdn <span style=\"color:#16a085;\"><strong>D.<\/strong> <span class=\"math-tex\">$2\\sqrt{6}$<\/span><\/span><\/p><p>Ta có bán kính c\u1ee7a (S) là <span class=\"math-tex\">$2\\sqrt{6}$<\/span> nên \u0111\u01b0\u1eddng kính c\u1ee7a (S) b\u1eb1ng <span class=\"math-tex\">$2\\sqrt{6}$<\/span>.<\/p>","type":"choose","extra_type":"classic","user_id":"131","test":"0","date":"2025-01-05 16:28:37","option_type":"math","len":0},{"id":"6135","post_id":"8305","mon_id":"1159285","chapter_id":"1159382","question":"<p>M\u1eb7t c\u1ea7u <span class=\"math-tex\">$(S):3x^2+3y^2+3z^2-6x+12y+2=0$<\/span> có bán kính b\u1eb1ng:<\/p>","options":["<strong>A.<\/strong> <span class=\"math-tex\">$\\sqrt{\\dfrac{13}{3}}$<\/span>","<strong>B.<\/strong> <span class=\"math-tex\">$\\dfrac{2\\sqrt{7}}{3}$<\/span>","<strong>C.<\/strong> <span class=\"math-tex\">$\\dfrac{\\sqrt{21}}{3}$<\/span>","<strong>D.<\/strong> <span class=\"math-tex\">$\\dfrac{\\sqrt{7}}{3}$<\/span>"],"correct":"1","level":"1","hint":"<p>Ph\u01b0\u01a1ng trình <span class=\"math-tex\">$x^2+y^2+z^2-2ax-2by-2cz+d=0$<\/span> v\u1edbi <span class=\"math-tex\">$a^2+b^2+c^2-d>0$<\/span> là ph\u01b0\u01a1ng trình c\u1ee7a m\u1eb7t c\u1ea7u tâm I(a; b; c) và bán kính <span class=\"math-tex\">$R=\\sqrt{a^2+b^2+c^2-d}$<\/span>.<\/p>","answer":"<p>Ch\u1ecdn <span style=\"color:#16a085;\"><strong>A.<\/strong> <span class=\"math-tex\">$\\sqrt{\\dfrac{13}{3}}$<\/span><\/span><\/p><p><span class=\"math-tex\">$(S):3x^2+3y^2+3z^2-6x+12y+2=0$<\/span> ⇔ <span class=\"math-tex\">$(S):x^2+y^2+z^2-2x+4y+\\dfrac{2}{3}=0$<\/span> có tâm <span class=\"math-tex\">$I(1;-2;0)$<\/span> và bán kính <span class=\"math-tex\">$R=\\sqrt{\\dfrac{13}{3}}$<\/span>.<\/p>","type":"choose","extra_type":"classic","user_id":"131","test":"0","date":"2025-01-05 16:36:07","option_type":"math","len":0}]}
