2<\/sup> \u2013 3x + 1, m\u1ec7nh \u0111\u1ec1 n\u00e0o sau \u0111\u00e2y l\u00e0 \u0111\u00fang?<\/p>\nA. H\u00e0m s\u1ed1 lu\u00f4n lu\u00f4n ngh\u1ecbch bi\u1ebfn\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 B. H\u00e0m s\u1ed1 lu\u00f4n lu\u00f4n \u0111\u1ed3ng bi\u1ebfn<\/p>\n
C. H\u00e0m s\u1ed1 \u0111\u1ea1t c\u1ef1c \u0111\u1ea1i t\u1ea1i x = 1 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 D. H\u00e0m s\u1ed1 \u0111\u1ea1t c\u1ef1c ti\u1ec3u t\u1ea1i x = 1.<\/p>\n
C\u00e2u 13:<\/strong> Cho h\u00e0m s\u1ed1 y =-x4<\/sup>-2x2<\/sup>-1. S\u1ed1 giao \u0111i\u1ec3m c\u1ee7a \u0111\u1ed3 th\u1ecb h\u00e0m s\u1ed1 v\u1edbi tr\u1ee5c Ox b\u1eb1ng<\/p>\nA. 1\u00a0\u00a0\u00a0 \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 B. 2\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 C. 3 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0D. 4<\/p>\n
C\u00e2u 18. <\/strong>M\u1ed9t kh\u1ed1i l\u1eb7ng tr\u1ee5\u00a0 tam gi\u00e1c c\u00f3 c\u00e1c c\u1ea1nh \u0111\u00e1y b\u1eb1ng\u00a0 13 ,37,30 v\u00e0 di\u1ec7n t\u00edch xung quanh b\u1eb1ng 480. Khi \u0111\u00f3 th\u1ec3 t\u00edch kh\u1ed1i l\u0103ng tr\u1ee5 l\u00e0<\/p>\nA. 2010 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0B. 1080\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 C. 1010 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0D. 2040<\/p>\n
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<\/p>\n
Theo Tuyensinh247<\/strong><\/p>\n <\/p>\n","protected":false},"excerpt":{"rendered":"
Tr\u01b0\u01a1\u0300ng THPT Hu\u0300ng V\u01b0\u01a1ng t\u00f4\u0309 ch\u01b0\u0301c thi gi\u01b0\u0303a ho\u0323c ky\u0300 m\u00f4n Toa\u0301n cho ca\u0301c em ho\u0323c sinh l\u01a1\u0301p 12, ca\u0301c em theo do\u0303i \u0111\u00ea\u0300 thi chi ti\u00ea\u0301t d\u01b0\u01a1\u0301i \u0111\u00e2y: C\u00e2u 1: Cho h\u00e0m s\u1ed1 y = \u2013x3 + 3×2 \u2013 3x + 1, m\u1ec7nh \u0111\u1ec1 n\u00e0o sau \u0111\u00e2y l\u00e0 \u0111\u00fang? A. H\u00e0m s\u1ed1 lu\u00f4n lu\u00f4n […]<\/p>\n","protected":false},"author":2,"featured_media":0,"comment_status":"closed","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"_mi_skip_tracking":false,"tdm_status":"","tdm_grid_status":""},"categories":[167],"tags":[],"yoast_head":"\n
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