C\u00e1c em tham kh\u1ea3o \u0110\u1ec1 thi gi\u1eefa h\u1ecdc k\u00ec 1 m\u00f4n To\u00e1n l\u1edbp 11 n\u0103m 2015 THPT Tr\u1ea7n Nh\u00e2n T\u00f4ng \u0111\u01b0\u1ee3c t\u1ed5 ch\u1ee9c ng\u00e0y 25 – 10 – 2015:<\/p>\n
KI\u1ec2M TRA 1 TI\u1ebeT CH\u01af\u01a0NG I<\/strong><\/p>\n \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\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0M\u00d4N: \u0110\u1ea0I S\u1ed0 11-C\u01a0 B\u1ea2N<\/strong><\/p>\n \u00a0TH\u1edcI GIAN: 60 PH\u00daT\u00a0<\/strong><\/p>\n \u0110\u1ec1 1.<\/strong><\/p>\n C\u00c2U 1: ( 2 \u0111i\u1ec3m) Gi\u1ea3i c\u00e1c ph\u01b0\u01a1ng tr\u00ecnh sau:<\/strong><\/p>\n <\/p>\n C\u00c2U 2: (\u00a0 4 \u0111i\u1ec3m) Gi\u1ea3i c\u00e1c ph\u01b0\u01a1ng tr\u00ecnh sau:<\/strong><\/p>\n <\/p>\n C\u00c2U 3: \u00a0( 3\u0111i\u1ec3m ) <\/strong>Cho h\u00ecnh ch\u00f3p S.ABCD.Trong tam gi\u00e1c SBC l\u1ea5y m\u1ed9t \u0111i\u1ec3m M trong tam gi\u00e1c SCD l\u1ea5y m\u1ed9t \u0111i\u1ec3m N.<\/p>\n a. T\u00ecm giao \u0111i\u1ec3m c\u1ee7a \u0111\u01b0\u1eddng th\u1eb3ng MN v\u1edbi m\u1eb7t ph\u1eb3ng(SAC)<\/p>\n b. T\u00ecm giao \u0111i\u1ec3m c\u1ee7a c\u1ea1nh SC v\u1edbi m\u1eb7t ph\u1eb3ng (AMN)<\/p>\n c. T\u00ecm thi\u1ebft di\u1ec7n c\u1ee7a m\u1eb7t ph\u1eb3ng (AMN) v\u1edbi h\u00ecnh ch\u00f3p S.ABCD<\/p>\n C\u00c2U 4: ( 1\u0111i\u1ec3m )<\/strong>T\u00ecm m \u0111\u1ec3 ph\u01b0\u01a1ng tr\u00ecnh sau\u00a0m sinx \u2013 (m -1) cosx = 3 \u2013 2m c\u00f3 nghi\u1ec7m tr\u00ean \u00a0<\/p>\n ——————————-<\/p>\n \u0110\u1ec1 2<\/strong><\/p>\n C\u00c2U 1: (2 \u0111i\u1ec3m) Gi\u1ea3i c\u00e1c ph\u01b0\u01a1ng tr\u00ecnh sau:<\/strong><\/p>\n <\/p>\n C\u00c2U 2: (4 \u0111i\u1ec3m) Gi\u1ea3i c\u00e1c ph\u01b0\u01a1ng tr\u00ecnh sau:<\/strong><\/p>\n <\/p>\n C\u00c2U 3:\u00a0 ( 3\u0111i\u1ec3m ) <\/strong>Cho h\u00ecnh ch\u00f3p S.ABCD. G\u1ecdi M l\u00e0 1 \u0111i\u1ec3m thu\u1ed9c mi\u1ec1n trong c\u1ee7a tam gi\u00e1c SCD<\/p>\n a) T\u00ecm (SBM) (SAC).<\/p>\n b) T\u00ecm BM (SAC).<\/p>\n c) T\u00ecm thi\u1ebft di\u1ec7n c\u1ee7a h\u00ecnh ch\u00f3p c\u1eaft b\u1edfi mp (ABM)<\/p>\n C\u00c2U 4:\u00a0 ( 1\u0111i\u1ec3m ) <\/strong>T\u00ecm m \u0111\u1ec3 ph\u01b0\u01a1ng tr\u00ecnh 2 sinx + m cosx = 1 \u2013 m (1) c\u00f3 nghi\u1ec7m thu\u1ed9c \u00a0<\/p>\n <\/p>\n","protected":false},"excerpt":{"rendered":" C\u00e1c em tham kh\u1ea3o \u0110\u1ec1 thi gi\u1eefa h\u1ecdc k\u00ec 1 m\u00f4n To\u00e1n l\u1edbp 11 n\u0103m 2015 THPT Tr\u1ea7n Nh\u00e2n T\u00f4ng \u0111\u01b0\u1ee3c t\u1ed5 ch\u1ee9c ng\u00e0y 25 – 10 – 2015: KI\u1ec2M TRA 1 TI\u1ebeT CH\u01af\u01a0NG I \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\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0M\u00d4N: \u0110\u1ea0I S\u1ed0 11-C\u01a0 B\u1ea2N \u00a0TH\u1edcI GIAN: 60 PH\u00daT\u00a0 \u0110\u1ec1 1. C\u00c2U 1: ( 2 \u0111i\u1ec3m) Gi\u1ea3i c\u00e1c ph\u01b0\u01a1ng tr\u00ecnh […]<\/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":[171],"tags":[],"yoast_head":"\n