{"id":550,"date":"2026-05-05T04:51:41","date_gmt":"2026-05-05T04:51:41","guid":{"rendered":"https:\/\/79.buffdemo.com\/index.php\/2026\/05\/05\/de-thi-thu-tot-nghiep-thpt-mon-toan-2024-2025-cum-chuyen-mon-so-3-dak-lak\/"},"modified":"2026-05-05T07:25:16","modified_gmt":"2026-05-05T07:25:16","slug":"de-thi-thu-tot-nghiep-thpt-mon-toan-2024-2025-cum-chuyen-mon-so-3-dak-lak","status":"publish","type":"post","link":"https:\/\/79.buffdemo.com\/index.php\/2026\/05\/05\/de-thi-thu-tot-nghiep-thpt-mon-toan-2024-2025-cum-chuyen-mon-so-3-dak-lak\/","title":{"rendered":"\u0110\u1ec1 thi th\u1eed t\u1ed1t nghi\u1ec7p THPT m\u00f4n To\u00e1n 2024 \u2013 2025 C\u1ee5m chuy\u00ean m\u00f4n s\u1ed1 3 \u0110\u1eafk L\u1eafk"},"content":{"rendered":"

T\u1ed5ng h\u1ee3p \u0110\u1ec1 thi th\u1eed t\u1ed1t nghi\u1ec7p THPT m\u00f4n To\u00e1n 2024 \u2013 2025 C\u1ee5m chuy\u00ean m\u00f4n s\u1ed1 3 \u0110\u1eafk L\u1eafk<\/strong> \u2013 t\u00e0i li\u1ec7u \u00f4n thi m\u00f4n To\u00e1n quan tr\u1ecdng d\u00e0nh cho c\u00e1c b\u1ea1n h\u1ecdc sinh 12. \u0110\u1ec1 thi \u0111\u00e3 bao qu\u00e1t c\u00e1c ki\u1ebfn th\u1ee9c tr\u1ecdng t\u00e2m, c\u00e2u h\u1ecfi c\u00f3 \u0111\u1ed9 ph\u00e2n h\u00f3a t\u1eeb d\u1ec5 \u0111\u1ebfn n\u00e2ng cao. \u0110\u1eb7c bi\u1ec7t \u0111\u1ec1 thi \u0111\u00e3 k\u00e8m \u0111\u00e1p \u00e1n sau m\u1ed7i ph\u1ea7n gi\u00fap c\u00e1c b\u1ea1n d\u1ec5 d\u00e0ng tra c\u1ee9u v\u00e0 s\u1eeda b\u00e0i sau khi luy\u1ec7n t\u1eadp. C\u00f9ng T\u00e0i li\u1ec7u \u00d4n Thi<\/a> t\u00ecm hi\u1ec3u s\u00e2u h\u01a1n v\u1ec1 n\u1ed9i dung c\u1ee7a \u0111\u1ec1 qua b\u00e0i vi\u1ebft d\u01b0\u1edbi \u0111\u00e2y.<\/p>\n

N\u1ed9i dung \u0110\u1ec1 thi th\u1eed t\u1ed1t nghi\u1ec7p THPT m\u00f4n To\u00e1n 2024 \u2013 2025 C\u1ee5m chuy\u00ean m\u00f4n s\u1ed1 3 \u0110\u1eafk L\u1eafk<\/strong><\/h2>\n

\u0110\u1ec1 thi th\u1eed t\u1ed1t nghi\u1ec7p THPT m\u00f4n To\u00e1n 2024 \u2013 2025 C\u1ee5m chuy\u00ean m\u00f4n s\u1ed1 3 \u0110\u1eafk L\u1eafk<\/strong> \u0111\u01b0\u1ee3c bi\u1ec7n so\u1ea1n theo c\u1ea5u tr\u00fac m\u1edbi chu\u1ea9n c\u1ee7a b\u1ed9 Gi\u00e1o d\u1ee5c g\u1ed3m t\u1ea5t c\u1ea3 22 c\u00e2u chia l\u00e0m 3 ph\u1ea7n ch\u00ednh. M\u1ed7i ph\u1ea7n c\u00f3 ma tr\u1eadn c\u00e2u h\u1ecfi t\u1eeb nh\u1eadn bi\u1ebft \u0111\u1ebfn v\u1eadn d\u1ee5ng cao nh\u1eb1m \u0111\u00e1nh \u0111\u00fang n\u0103ng l\u1ef1c c\u1ee7a c\u00e1c b\u1ea1n. H\u00e3y c\u00f9ng t\u00ecm hi\u1ec3u s\u00e2u h\u01a1n v\u1ec1 n\u1ed9i dung ch\u00ednh c\u1ee7a \u0111\u1ec1 nh\u01b0 sau:<\/p>\n

Ph\u1ea7n 1 \u2013 C\u00e2u h\u1ecfi tr\u1eafc nghi\u1ec7m nhi\u1ec1u ph\u01b0\u01a1ng \u00e1n<\/strong><\/p>\n

Tr\u00edch m\u1ed9t s\u1ed1 c\u00e2u h\u1ecfi trong \u0110\u1ec1 thi th\u1eed t\u1ed1t nghi\u1ec7p THPT m\u00f4n To\u00e1n 2024 \u2013 2025 C\u1ee5m chuy\u00ean m\u00f4n s\u1ed1 3 \u0110\u1eafk L\u1eafk<\/strong> nh\u01b0 sau:<\/p>\n

C\u00e2u 1<\/strong>: Trong kh\u00f4ng gian v\u1edbi h\u1ec7 t\u1ecda \u0111\u1ed9 Oxyz, cho \u0111i\u1ec3m $M(2;1;-3)$. H\u00ecnh chi\u1ebfu vu\u00f4ng g\u00f3c c\u1ee7a \u0111i\u1ec3m $M(2;1;-3)$ tr\u00ean tr\u1ee5c $Ox$ c\u00f3 t\u1ecda \u0111\u1ed9 l\u00e0:<\/p>\n

A. $(0;1;0)$\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 B. $(0;1;-3)$\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0C. $(2;0;0)$\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 D. $(0;0;-3)$<\/p>\n

C\u00e2u 2<\/strong>: Trong kh\u00f4ng gian Oxyz, cho hai vect\u01a1 a\u20d7 = (2;3;3), b\u20d7 = (3;2;-1). Khi \u0111\u00f3 t\u00edch v\u00f4 h\u01b0\u1edbng $\\vec{a} \\cdot \\vec{b}$ b\u1eb1ng:<\/p>\n

A. $\\vec{a} \\cdot \\vec{b} = 9$\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 B. $\\vec{a} \\cdot \\vec{b} = 7$\u00a0 \u00a0 \u00a0 C. $\\vec{a} \\cdot \\vec{b} = 3$\u00a0 \u00a0 \u00a0 \u00a0 D. $\\vec{a} \\cdot \\vec{b} = 15$<\/p>\n

Ph\u1ea7n 2 \u2013 C\u00e2u h\u1ecfi tr\u1eafc nghi\u1ec7m \u0111\u00fang sai
\n<\/strong>
\nV\u1edbi 4 ph\u01b0\u01a1ng \u00e1n a), b), c), d), m\u1ed7i th\u00ed sinh ch\u1ec9 c\u1ea7n ghi m\u1ec7nh \u0111\u1ec1 \u0110\u00daNG\/SAI<\/strong> \u1edf m\u1ed7i c\u00e2u. V\u00ed d\u1ee5:<\/p>\n

C\u00e2u 1<\/strong>: Gi\u1ea3 s\u1eed m\u1ed9t h\u1ea1t chuy\u1ec3n \u0111\u1ed9ng tr\u00ean m\u1ed9t tr\u1ee5c th\u1eb3ng \u0111\u1ee9ng chi\u1ec1u d\u01b0\u01a1ng h\u01b0\u1edbng l\u00ean tr\u00ean sao cho t\u1ecda \u0111\u1ed9 c\u1ee7a h\u1ea1t (\u0111\u01a1n v\u1ecb: m\u00e9t) t\u1ea1i th\u1eddi \u0111i\u1ec3m $t$ (gi\u00e2y) l\u00e0 y = t\u00b3 – 12t + 3, (t \u2265 0).\n<\/p>\n

a) H\u1ea1t chuy\u1ec3n \u0111\u1ed9ng l\u00ean tr\u00ean khi $t > 2$ v\u00e0 h\u1ea1t chuy\u1ec3n \u0111\u1ed9ng xu\u1ed1ng d\u01b0\u1edbi khi $t < 2$.<\/p>\n

b) Qu\u00e3ng \u0111\u01b0\u1eddng h\u1ea1t \u0111i \u0111\u01b0\u1ee3c trong kho\u1ea3ng th\u1eddi gian $0 < t < 3$ l\u00e0 9 m.<\/p>\n

c) H\u00e0m v\u1eadn t\u1ed1c l\u00e0: $v(t) = y\u2019 = 3t^2 \u2013 12, \\quad (t \\ge 0)$ v\u00e0 h\u00e0m gia t\u1ed1c l\u00e0 $a(t) = v'(t), \\quad (t \\ge 0)$.<\/p>\n

d) H\u1ea1t t\u0103ng t\u1ed1c khi $t > 2$ v\u00e0 h\u1ea1t gi\u1ea3m t\u1ed1c $0 < t < 2$.<\/p>\n

C\u00e2u 2<\/strong>: Trong kh\u00f4ng gian Oxyz, cho tam gi\u00e1c ABC c\u00f3 A(1;2;2),B(-2;1;-3), C(1;0;-5).\n<\/p>\n

a) cos( AB\u20d7, BC\u20d7 ) = (\u221a10)\/(35).\n<\/p>\n

b) G\u1ecdi D l\u00e0 \u0111\u1ec9nh th\u1ee9 t\u01b0 c\u1ee7a h\u00ecnh b\u00ecnh h\u00e0nh ABCD . Khi \u0111\u00f3 \u0111i\u1ec3m D(0;3;-6).\n<\/p>\n

c) T\u1ecda \u0111\u1ed9 tr\u1ecdng t\u00e2m c\u1ee7a tam gi\u00e1c ABC l\u00e0 G(0;1;-2).\n<\/p>\n

d) G\u1ecdi M(a;b;c) l\u00e0 \u0111i\u1ec3m thu\u1ed9c m\u1eb7t ph\u1eb3ng (Oxz) , sao cho | MA\u20d7 + MB\u20d7 + MC\u20d7 | \u0111\u1ea1t gi\u00e1 tr\u1ecb nh\u1ecf nh\u1ea5t. Khi \u0111\u00f3 gi\u00e1 tr\u1ecb c\u1ee7a bi\u1ec3u th\u1ee9c T = a + b – c + 2024 b\u1eb1ng 2025 .<\/p>\n

Ph\u1ea7n 3 \u2013 Tr\u1eafc nghi\u1ec7m tr\u1ea3 l\u1eddi ng\u1eafn<\/strong><\/p>\n

T\u01b0\u01a1ng t\u1ef1 nh\u01b0 c\u00e1c ph\u1ea7n tr\u00ean, ph\u1ea7n 3 v\u1edbi c\u00e2u h\u1ecfi tr\u1ea3 l\u1eddi ng\u1eafn, \u0111\u1ed9 kh\u00f3 c\u1ee7a \u0111\u1ec1 d\u1eabn t\u0103ng l\u00ean v\u00e0 ph\u00e2n chia theo nhi\u1ec1u b\u1eadc. Do \u0111\u00f3, \u0111\u1ec3 n\u1eafm tr\u1ecdn s\u1ed1 \u0111i\u1ec3m c\u1ee7a ph\u1ea7n n\u00e0y, th\u00ed sinh c\u1ea7n \u00f4n t\u1eadp th\u1eadt k\u1ef9, n\u1eafm v\u1eefng ki\u1ebfn th\u1ee9c v\u00e0 gi\u1ea3i \u0111\u1ec1 th\u1eadt nhi\u1ec1u. Xem qua c\u00e1c c\u00e2u h\u1ecfi trong \u0111\u1ec1 thi th\u1eed to\u00e1n 12<\/a> d\u01b0\u1edbi \u0111\u00e2y:<\/p>\n

C\u00e2u 1<\/strong>: Cho C(x) = 16000 + 500x – 1,6x\u00b2 + 0,004x\u00b3 l\u00e0 h\u00e0m chi ph\u00ed v\u00e0 p(x) = 1700 – 7x l\u00e0 h\u00e0m c\u1ea7u. H\u00e3y t\u00ecm m\u1ee9c s\u1ea3n xu\u1ea5t (t\u00ednh theo \u0111\u01a1n v\u1ecb h\u00e0ng ho\u00e1) s\u1ebd t\u1ed1i \u0111a ho\u00e1 l\u1ee3i nhu\u1eadn.<\/p>\n

C\u00e2u 2<\/strong>: B\u1ea3ng sau \u0111\u00e2y bi\u1ec3u di\u1ec5n m\u1eabu s\u1ed1 li\u1ec7u gh\u00e9p nh\u00f3m v\u1ec1 c\u00e2n n\u1eb7ng c\u1ee7a m\u1ed9t s\u1ed1 qu\u1ea3 d\u01b0a l\u01b0\u1edbi thu ho\u1ea1ch \u0111\u01b0\u1ee3c \u1edf m\u1ed9t khu v\u01b0\u1eddn (\u0111\u01a1n v\u1ecb: gam):<\/p>\n

\"\"<\/p>\n

T\u00ecm ph\u01b0\u01a1ng sai c\u1ee7a m\u1eabu s\u1ed1 li\u1ec7u gh\u00e9p nh\u00f3m tr\u00ean (k\u1ebft qu\u1ea3 l\u00e0m tr\u00f2n \u0111\u1ebfn h\u00e0ng \u0111\u01a1n v\u1ecb).<\/p>\n

C\u00e2u 3<\/strong>: Cho h\u00ecnh ch\u00f3p $S.ABC$ c\u00f3 $SA = 3$, $SB = 4$, $SC = 5$. M\u1ed9t m\u1eb7t ph\u1eb3ng $(\\alpha)$ lu\u00f4n \u0111i qua tr\u1ecdng t\u00e2m $G$ c\u1ee7a tam gi\u00e1c $ABC$ v\u00e0 c\u1eaft c\u00e1c c\u1ea1nh $SA, SB, SC$ l\u1ea7n l\u01b0\u1ee3t t\u1ea1i $A\u2019, B\u2019, C\u2019$. T\u00ecm gi\u00e1 tr\u1ecb nh\u1ecf nh\u1ea5t c\u1ee7a bi\u1ec3u th\u1ee9c T = (1)\/(SA\u2019\u00b2) + (1)\/(SB\u2019\u00b2) + (1)\/(SC\u2019\u00b2).\n<\/p>\n

T\u1ea3i \u0110\u1ec1 thi th\u1eed t\u1ed1t nghi\u1ec7p THPT m\u00f4n To\u00e1n 2024 \u2013 2025 C\u1ee5m chuy\u00ean m\u00f4n s\u1ed1 3 \u0110\u1eafk L\u1eafk<\/strong><\/h2>\n

Th\u1eed s\u1ee9c v\u1edbi \u0110\u1ec1 thi th\u1eed t\u1ed1t nghi\u1ec7p THPT m\u00f4n To\u00e1n 2024 \u2013 2025 C\u1ee5m chuy\u00ean m\u00f4n s\u1ed1 3 \u0110\u1eafk L\u1eafk<\/strong> s\u1ebd gi\u00fap c\u00e1c b\u1ea1n bi\u1ebft \u0111\u01b0\u1ee3c n\u0103ng l\u1ef1c c\u1ee7a b\u1ea3n th\u00e2n v\u00e0 t\u00ecm ra l\u1ed9 tr\u00ecnh \u00f4n t\u1eadp hi\u1ec7u qu\u1ea3 nh\u1ea5t. H\u00e3y t\u1ea3i \u0111\u1ec1 thi v\u00e0 \u0111\u00e1p \u00e1n qua \u0111\u01b0\u1eddng link sau \u0111\u00e2y v\u00e0 b\u1eaft tay v\u00e0o gi\u1ea3i \u0111\u1ec1 ngay h\u00f4m nay nh\u00e9!<\/p>\n