{"id":686,"date":"2026-05-05T04:59:14","date_gmt":"2026-05-05T04:59:14","guid":{"rendered":"https:\/\/79.buffdemo.com\/index.php\/2026\/05\/05\/100-cau-trac-nghiem-dung-sai-anh-giao-kid\/"},"modified":"2026-05-05T07:33:57","modified_gmt":"2026-05-05T07:33:57","slug":"100-cau-trac-nghiem-dung-sai-anh-giao-kid","status":"publish","type":"post","link":"https:\/\/79.buffdemo.com\/index.php\/2026\/05\/05\/100-cau-trac-nghiem-dung-sai-anh-giao-kid\/","title":{"rendered":"100 C\u00e2u tr\u1eafc nghi\u1ec7m \u0111\u00fang sai Anh Gi\u00e1o Kid"},"content":{"rendered":"

100 C\u00e2u tr\u1eafc nghi\u1ec7m \u0111\u00fang sai Anh Gi\u00e1o Kid <\/strong>t\u1ed5ng h\u1ee3p c\u00e1c c\u00e2u h\u1ecfi tr\u1eafc nghi\u1ec7m c\u1ee7ng c\u1ed1 ki\u1ebfn th\u1ee9c to\u00e1n 12 th\u00f4ng qua c\u00e1c c\u00e2u h\u1ecfi \u0111\u00fang\/sai. T\u00e0i li\u1ec7u n\u00e0y bao g\u1ed3m nhi\u1ec1u d\u1ea1ng b\u00e0i t\u1eadp \u0111a d\u1ea1ng, t\u1eeb h\u00e0m s\u1ed1, x\u00e1c su\u1ea5t, h\u00ecnh h\u1ecdc kh\u00f4ng gian \u0111\u1ebfn c\u00e1c b\u00e0i to\u00e1n th\u1ef1c t\u1ebf, \u0111\u01b0\u1ee3c tr\u00ecnh b\u00e0y m\u1ed9t c\u00e1ch khoa h\u1ecdc, r\u00f5 r\u00e0ng, gi\u00fap h\u1ecdc sinh d\u1ec5 d\u00e0ng ti\u1ebfp c\u1eadn v\u00e0 ki\u1ec3m tra l\u1ea1i ki\u1ebfn th\u1ee9c c\u1ee7a m\u00ecnh m\u1ed9t c\u00e1ch hi\u1ec7u qu\u1ea3. H\u00e3y\u00a0c\u00f9ng Tailieuonthi edu<\/strong><\/a> t\u00ecm hi\u1ec3u chi ti\u1ebft qua b\u00e0i vi\u1ebft sau.<\/p>\n

N\u1ed9i dung 100 C\u00e2u tr\u1eafc nghi\u1ec7m \u0111\u00fang sai Anh Gi\u00e1o Kid<\/strong><\/h2>\n

100 C\u00e2u tr\u1eafc nghi\u1ec7m \u0111\u00fang sai Anh Gi\u00e1o Kid<\/strong> \u0111\u01b0\u1ee3c chia th\u00e0nh 10 \u0110\u1ec1 thi. M\u1ed7i \u0110\u1ec1 thi bao g\u1ed3m 10 c\u00e2u h\u1ecfi tr\u1eafc nghi\u1ec7m \u0111\u00fang\/sai. T\u00e0i li\u1ec7u to\u00e1n 12<\/a> n\u00e0y t\u1eadp trung ch\u1ee7 y\u1ebfu v\u00e0o c\u00e1c ph\u1ea7n Gi\u1ea3i t\u00edch (H\u00e0m s\u1ed1, Nguy\u00ean h\u00e0m \u2013 T\u00edch ph\u00e2n) v\u00e0 H\u00ecnh h\u1ecdc kh\u00f4ng gian Oxyz, c\u00f9ng v\u1edbi c\u00e1c b\u00e0i to\u00e1n X\u00e1c su\u1ea5t \u2013 Th\u1ed1ng k\u00ea \u1ee9ng d\u1ee5ng th\u1ef1c t\u1ebf. \u0110\u00e2y \u0111\u1ec1u l\u00e0 nh\u1eefng ph\u1ea7n ki\u1ebfn th\u1ee9c tr\u1ecdng t\u00e2m trong ch\u01b0\u01a1ng tr\u00ecnh To\u00e1n 12, \u0111\u1eb7c bi\u1ec7t l\u00e0 cho c\u00e1c k\u1ef3 thi THPT Qu\u1ed1c gia. D\u01b0\u1edbi \u0111\u00e2y l\u00e0 m\u1ed9t s\u1ed1 c\u00e2u h\u1ecfi tham kh\u1ea3o \u0111\u01b0\u1ee3c tr\u00edch xu\u1ea5t t\u1eeb b\u1ed9 t\u00e0i li\u1ec7u:<\/p>\n

Ch\u1ecdn c\u00e2u \u0111\u00fang\/sai:<\/p>\n

C\u00e2u 1 \u2013 \u0110\u1ec1 1:<\/strong> Cho h\u00e0m s\u1ed1 $y = \\frac{2x+1}{x-1}$ c\u00f3 \u0111\u1ed3 th\u1ecb $(C)$.<\/p>\n

a) \u0110\u1ed3 th\u1ecb $(C)$ c\u00f3 ti\u1ec7m c\u1eadn \u0111\u1ee9ng $x=1$ v\u00e0 ti\u1ec7m c\u1eadn ngang $y=1$.
\nb) H\u00e0m s\u1ed1 \u0111\u00e3 cho ngh\u1ecbch bi\u1ebfn tr\u00ean c\u00e1c kho\u1ea3ng x\u00e1c \u0111\u1ecbnh c\u1ee7a n\u00f3.
\nc) C\u00f3 \u0111\u00fang 3 \u0111i\u1ec3m thu\u1ed9c \u0111\u1ed3 th\u1ecb $(C)$ m\u00e0 t\u1ecda \u0111\u1ed9 c\u1ee7a ch\u00fang l\u00e0 nh\u1eefng s\u1ed1 nguy\u00ean.
\nd) G\u1ecdi $M$ l\u00e0 \u0111i\u1ec3m n\u1eb1m tr\u00ean \u0111\u1ed3 th\u1ecb $(C)$ v\u00e0 $H, K$ t\u01b0\u01a1ng \u1ee9ng l\u00e0 h\u00ecnh chi\u1ebfu vu\u00f4ng g\u00f3c c\u1ee7a $M$ tr\u00ean c\u00e1c tr\u1ee5c $Ox$ v\u00e0 $Oy$. C\u00f3 2 \u0111i\u1ec3m $M$ c\u00f3 ho\u00e0nh \u0111\u1ed9 d\u01b0\u01a1ng th\u1ecfa m\u00e3n t\u1ee9 gi\u00e1c $MHOK$ c\u00f3 di\u1ec7n t\u00edch b\u1eb1ng $2$.<\/p>\n

C\u00e2u 3 \u2013 \u0110\u1ec1 1:<\/strong> M\u1ed9t m\u00e1y bay di chuy\u1ec3n ra \u0111\u1ebfn \u0111\u01b0\u1eddng b\u0103ng v\u00e0 b\u1eaft \u0111\u1ea7u ch\u1ea1y \u0111\u00e0 \u0111\u1ec3 c\u1ea5t c\u00e1nh. Gi\u1ea3 s\u1eed v\u1eadn t\u1ed1c c\u1ee7a m\u00e1y bay khi ch\u1ea1y \u0111\u00e0 \u0111\u01b0\u1ee3c cho b\u1edfi $v(t) = 5 + 3t \\, (m\/s)$, v\u1edbi $t$ l\u00e0 th\u1eddi gian k\u1ec3 t\u1eeb khi m\u00e1y bay b\u1eaft \u0111\u1ea7u ch\u1ea1y \u0111\u00e0. Sau 32 gi\u00e2y th\u00ec m\u00e1y bay c\u1ea5t c\u00e1nh tr\u00ean \u0111\u01b0\u1eddng b\u0103ng. G\u1ecdi $s(t)$ l\u00e0 qu\u00e3ng \u0111\u01b0\u1eddng m\u00e1y bay \u0111i chuy\u1ec3n \u0111\u01b0\u1ee3c sau $t$ gi\u00e2y k\u1ec3 t\u1eeb l\u00fac b\u1eaft \u0111\u1ea7u ch\u1ea1y \u0111\u00e0.<\/p>\n

a) $v(t) = s'(t)$.
\nb) $s(t) = \\frac{3}{2}t^2 + 5t + 5$.
\nc) Qu\u00e3ng \u0111\u01b0\u1eddng m\u00e1y bay \u0111i chuy\u1ec3n \u0111\u01b0\u1ee3c sau 4 gi\u00e2y k\u1ec3 t\u1eeb khi b\u1eaft \u0111\u1ea7u ch\u1ea1y \u0111\u00e0 l\u00e0 49 m\u00e9t.
\nd) Qu\u00e3ng \u0111\u01b0\u1eddng m\u00e1y bay \u0111\u00e3 di chuy\u1ec3n t\u1eeb khi b\u1eaft \u0111\u1ea7u ch\u1ea1y \u0111\u00e0 \u0111\u1ebfn khi r\u1eddi \u0111\u01b0\u1eddng b\u0103ng l\u00e0 1696 m\u00e9t.<\/p>\n

C\u00e2u 5 \u2013 \u0110\u1ec1 1:<\/strong> M\u1ed9t th\u00e1p trung t\u00e2m ki\u1ec3m so\u00e1t kh\u00f4ng l\u01b0u \u1edf s\u00e2n bay cao $80 \\, m$ s\u1eed d\u1ee5ng ra \u0111a c\u00f3 ph\u1ea1m vi theo d\u00f5i $500 \\, km$ \u0111\u01b0\u1ee3c \u0111\u1eb7t tr\u00ean \u0111\u1ec9nh th\u00e1p. Ch\u1ecdn h\u1ec7 tr\u1ee5c t\u1ecda \u0111\u1ed9 $Oxyz$ c\u00f3 g\u1ed1c $O$ tr\u00f9ng v\u1edbi v\u1ecb tr\u00ed ch\u00e2n th\u00e1p, m\u1eb7t ph\u1eb3ng $(Oxy)$ tr\u00f9ng v\u1edbi m\u1eb7t \u0111\u1ea5t sao cho tr\u1ee5c $Ox$ h\u01b0\u1edbng v\u1ec1 ph\u00eda t\u00e2y, tr\u1ee5c $Oy$ h\u01b0\u1edbng v\u1ec1 ph\u00eda nam, tr\u1ee5c $Oz$ h\u01b0\u1edbng th\u1eb3ng \u0111\u1ee9ng l\u00ean ph\u00eda tr\u00ean. M\u1ed9t m\u00e1y bay t\u1ea1i v\u1ecb tr\u00ed $A$ c\u00e1ch m\u1eb7t \u0111\u1ea5t $10 \\, km$, c\u00e1ch $300 \\, km$ v\u1ec1 ph\u00eda \u0111\u00f4ng v\u00e0 $200 \\, km$ v\u1ec1 ph\u00eda b\u1eafc so v\u1edbi th\u00e1p trung t\u00e2m ki\u1ec3m so\u00e1t kh\u00f4ng l\u01b0u. Trong c\u00e1c kh\u1eb3ng \u0111\u1ecbnh sau \u0111\u00e2y, kh\u1eb3ng \u0111\u1ecbnh n\u00e0o \u0111\u00fang, kh\u1eb3ng \u0111\u1ecbnh n\u00e0o sai?<\/p>\n

\"\"<\/p>\n

a) Ra \u0111a \u1edf v\u1ecb tr\u00ed c\u00f3 t\u1ecda \u0111\u1ed9 $(0;0;0)$.
\nb) Ra \u0111a c\u1ee7a trung t\u00e2m ki\u1ec3m so\u00e1t kh\u00f4ng l\u01b0u kh\u00f4ng ph\u00e1t hi\u1ec7n \u0111\u01b0\u1ee3c m\u00e1y bay t\u1ea1i v\u1ecb tr\u00ed $A$.
\nc) Kho\u1ea3ng c\u00e1ch t\u1eeb m\u00e1y bay \u0111\u1ebfn ra \u0111a l\u00e0 $360,69 \\, km$.
\nd) V\u1ecb tr\u00ed $A$ c\u00f3 t\u1ecda \u0111\u1ed9 $(300;200;10)$.<\/p>\n

T\u1ea3i 100 C\u00e2u tr\u1eafc nghi\u1ec7m \u0111\u00fang sai Anh Gi\u00e1o Kid<\/strong><\/h2>\n

100 C\u00e2u tr\u1eafc nghi\u1ec7m \u0111\u00fang sai Anh Gi\u00e1o Kid<\/strong> l\u00e0 t\u00e0i li\u1ec7u \u00f4n luy\u1ec7n h\u1eefu \u00edch v\u1edbi c\u00e1c d\u1ea1ng b\u00e0i t\u1eadp tr\u1eafc nghi\u1ec7m \u0111\u00fang\/sai, bao g\u1ed3m nhi\u1ec1u ch\u1ee7 \u0111\u1ec1 kh\u00e1c nhau nh\u01b0 h\u00e0m s\u1ed1, x\u00e1c su\u1ea5t, h\u00ecnh h\u1ecdc kh\u00f4ng gian, v\u00e0 t\u00edch ph\u00e2n. C\u00e1c b\u1ea1n c\u00f3 th\u1ec3 t\u1ea3i file n\u00e0y \u0111\u1ec3 luy\u1ec7n t\u1eadp.<\/p>\n

\"\" \"\" \"\"<\/p>\n

T\u1ea3i 100 C\u00e2u tr\u1eafc nghi\u1ec7m \u0111\u00fang sai Anh Gi\u00e1o Kid<\/a><\/p>\n

T\u00e0i li\u1ec7u li\u00ean quan:<\/strong><\/p>\n