{"id":544,"date":"2026-05-05T04:51:21","date_gmt":"2026-05-05T04:51:21","guid":{"rendered":"https:\/\/79.buffdemo.com\/index.php\/2026\/05\/05\/de-thi-thu-tot-nghiep-thpt-mon-toan-2024-2025-truong-thpt-le-thanh-tong-tp-hcm-lan-2\/"},"modified":"2026-05-05T07:24:53","modified_gmt":"2026-05-05T07:24:53","slug":"de-thi-thu-tot-nghiep-thpt-mon-toan-2024-2025-truong-thpt-le-thanh-tong-tp-hcm-lan-2","status":"publish","type":"post","link":"https:\/\/79.buffdemo.com\/index.php\/2026\/05\/05\/de-thi-thu-tot-nghiep-thpt-mon-toan-2024-2025-truong-thpt-le-thanh-tong-tp-hcm-lan-2\/","title":{"rendered":"\u0110\u1ec1 thi th\u1eed t\u1ed1t nghi\u1ec7p THPT m\u00f4n To\u00e1n 2024 \u2013 2025 tr\u01b0\u1eddng THPT L\u00ea Th\u00e1nh T\u00f4ng TP HCM l\u1ea7n 2"},"content":{"rendered":"

\u0110\u1ec1 thi th\u1eed t\u1ed1t nghi\u1ec7p THPT m\u00f4n To\u00e1n 2024 \u2013 2025 tr\u01b0\u1eddng THPT L\u00ea Th\u00e1nh T\u00f4ng TP HCM l\u1ea7n 2<\/strong> l\u00e0 t\u00e0i li\u1ec7u m\u00f4n To\u00e1n d\u00e0nh cho c\u00e1c b\u1ea1n h\u1ecdc sinh 12 chu\u1ea9n b\u1ecb \u00f4n thi thpt. N\u1ed9i dung \u0111\u1ec1 thi \u0111\u01b0\u1ee3c bi\u00ean so\u1ea1n b\u00e1m s\u00e1t c\u1ea5u tr\u00fac chung c\u1ee7a b\u1ed9 v\u00e0 ph\u00e2n b\u1ed5 \u0111\u1ec1u c\u00e1c ki\u1ebfn th\u1ee9c m\u00f4n To\u00e1n \u0111\u00e3 \u0111\u01b0\u1ee3c h\u1ecdc. H\u00e3y c\u00f9ng T\u00e0i li\u1ec7u \u00d4n Thi<\/a> t\u00ecm hi\u1ec3u th\u00eam v\u1ec1 n\u1ed9i dung c\u1ee7a \u0111\u1ec1 thi th\u1eed t\u1ed1t nghi\u1ec7p n\u00e0y nh\u00e9.<\/p>\n

N\u1ed9i dung \u0110\u1ec1 thi th\u1eed t\u1ed1t nghi\u1ec7p THPT m\u00f4n To\u00e1n 2024 \u2013 2025 tr\u01b0\u1eddng THPT L\u00ea Th\u00e1nh T\u00f4ng TP HCM l\u1ea7n 2<\/strong><\/h2>\n

N\u1ed9i dung \u0110\u1ec1 thi th\u1eed t\u1ed1t nghi\u1ec7p THPT m\u00f4n To\u00e1n 2024 \u2013 2025 tr\u01b0\u1eddng THPT L\u00ea Th\u00e1nh T\u00f4ng TP HCM l\u1ea7n 2<\/strong> \u0111\u01b0\u1ee3c t\u1ed5ng h\u1ee3p t\u1eeb nhi\u1ec1u d\u1ea1ng c\u00e2u h\u1ecfi kh\u00e1c nhau v\u00e0 chia th\u00e0nh 3 ph\u1ea7n ch\u00ednh. Chi ti\u1ebft n\u1ed9i dung nh\u01b0 sau:<\/p>\n

Ph\u1ea7n 1 \u2013 C\u00e2u tr\u1eafc nghi\u1ec7m nhi\u1ec1u ph\u01b0\u01a1ng \u00e1n l\u1ef1a ch\u1ecdn<\/strong><\/p>\n

C\u00f3 t\u1ea5t c\u1ea3 12 c\u00e2u h\u1ecfi, v\u1edbi 4 \u0111\u00e1p \u00e1n A, B, C, D. Th\u00ed sinh c\u1ea7n \u0111\u1ecdc k\u0129 \u0111\u1ec1 v\u00e0 ch\u1ecdn ra \u0111\u00e1p \u00e1n ch\u00ednh x\u00e1c nh\u1ea5t. Tham kh\u1ea3o c\u00e1c c\u00e2u h\u1ecfi d\u01b0\u1edbi \u0111\u00e2y:<\/p>\n

C\u00e2u 1<\/strong>: Cho h\u00e0m s\u1ed1 y = log\u2083(x\u00b2 – 2x + 3) . H\u00e0m s\u1ed1 \u0111\u1ed3ng bi\u1ebfn tr\u00ean kho\u1ea3ng n\u00e0o sau \u0111\u00e2y?<\/p>\n

A. (-\u221e; 1) \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 B. (-1; +\u221e) \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 C. (-\u221e; -1) \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0D. (1; +\u221e) <\/p>\n

C\u00e2u 2<\/strong>: M\u1ed9t nh\u00e0 ph\u00e2n t\u00edch th\u1ecb tr\u01b0\u1eddng l\u00e0m vi\u1ec7c cho m\u1ed9t c\u00f4ng ty s\u1ea3n xu\u1ea5t thi\u1ebft b\u1ecb gia d\u1ee5ng nh\u1eadn th\u1ea5y r\u1eb1ng n\u1ebfu c\u00f4ng ty s\u1ea3n xu\u1ea5t v\u00e0 b\u00e1n x chi\u1ebfc m\u00e1y xay sinh t\u1ed1 h\u00e0ng th\u00e1ng th\u00ec l\u1ee3i nhu\u1eadn thu \u0111\u01b0\u1ee3c (ngh\u00ecn \u0111\u1ed3ng) c\u00f3 th\u1ec3 \u0111\u01b0\u1ee3c t\u00ednh b\u1eb1ng c\u00f4ng th\u1ee9c P(x) = -0,3x\u00b3 + 36x\u00b2 + 1800x – 48000. \u0110\u1ec3 c\u00f3 l\u1ee3i nhu\u1eadn l\u1edbn nh\u1ea5t, c\u00f4ng ty c\u1ea7n s\u1ea3n xu\u1ea5t \u0111\u00fang bao nhi\u00eau chi\u1ebfc m\u00e1y sinh t\u1ed1 m\u1ed7i th\u00e1ng?<\/p>\n

A. 90\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 B. 100\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0C. 110\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 D. 120<\/p>\n

Ph\u1ea7n 2 \u2013 C\u00e2u tr\u1eafc nghi\u1ec7m \u0111\u00fang sai<\/strong><\/p>\n

Th\u00ed sinh tr\u1ea3 l\u1eddi c\u00e2u 1 \u0111\u1ebfn c\u00e2u 4 v\u1edbi c\u00e1c n\u1ed9i dung v\u1ec1 ph\u01b0\u01a1ng tr\u00ecnh logarit, \u0111\u1ed3 th\u1ecb h\u00e0m s\u1ed1, h\u00ecnh h\u1ecdc kh\u00f4ng gian,\u2026 Xem qua c\u00e1c c\u00e2u h\u1ecfi c\u00f3 trong \u0111\u1ec1 thi th\u1eed to\u00e1n thpt<\/a> l\u1ea7n 2 d\u01b0\u1edbi d\u00e2y:<\/p>\n

C\u00e2u 1<\/strong>: Trong m\u1ed9t cu\u1ed9c th\u1eed t\u00ean l\u1eeda, Tri\u1ec1u Ti\u00ean \u0111\u00e3 cho ph\u00f3ng m\u1ed9t qu\u1ea3 t\u00ean l\u1eeda c\u00f3 g\u1eafn \u0111\u1ea7u \u0111\u1ea1n h\u1ea1t nh\u00e2n v\u1edbi v\u1eadn t\u1ed1c v(t) = ( 1\/9000t\u00b2 + n\/100 )( m\/s ), n > 0 trong \u0111\u00f3 t (\u0111\u01a1n v\u1ecb gi\u00e2y) t\u00ednh t\u1eeb l\u00fac t\u00ean l\u1eeda Tri\u1ec1u Ti\u00ean b\u1eaft \u0111\u1ea7u ph\u00f3ng v\u00e0 d\u1ef1 t\u00ednh s\u1ebd r\u01a1i xu\u1ed1ng m\u1ed9t v\u00f9ng bi\u1ec3n. D\u1ef1 \u0111o\u00e1n 1 gi\u1edd th\u00ec bay ngang v\u00f9ng bi\u1ec3n thu\u1ed9c ch\u1ee7 quy\u1ec1n c\u1ee7a Nh\u1eadt B\u1ea3n, nay l\u1eadp t\u1ee9c Rada nh\u1eadn \u0111\u01b0\u1ee3c t\u00edn hi\u1ec7u v\u1ec1 c\u0103n c\u1ee9 qu\u00e2n \u0111\u1ed9i.
\nKhi nh\u1eadn \u0111\u01b0\u1ee3c t\u00edn hi\u1ec7u v\u00e0 g\u1eedi tin hi\u1ec7u v\u1ec1 c\u0103n c\u1ee9 qu\u00e2n \u0111\u1ed9i. Khi nh\u1eadn \u0111\u01b0\u1ee3c t\u00edn hi\u1ec7u c\u1ee7a Rada sau 30 ph\u00fat qu\u00e2n \u0111\u1ed9i Nh\u1eadt B\u1ea3n \u0111\u00e3 cho ph\u00f3ng 1 qu\u1ea3 t\u00ean l\u1eeda t\u1ea7m trung \u0111\u00e3 x\u00e1c \u0111\u1ecbnh s\u1eb5n m\u1ee5c ti\u00eau \u0111i v\u1edbi gia t\u1ed1c a(t\u2081) = ( 1\/4500t\u2081 + n\/100 ) (m\/s\u00b2), n > 0 trong \u0111\u00f3 t\u2081 \u0111\u01a1n v\u1ecb gi\u00e2y t\u00ednh t\u1eeb l\u00fac t\u00ean l\u1eeda t\u1ea7m trung b\u1eaft \u0111\u1ea7u ph\u00f3ng.<\/p>\n

a) V\u1eadn t\u1ed1c c\u1ee7a t\u00ean l\u1eeda t\u1ea7m trung \u0111\u01b0\u1ee3c bi\u1ec3u th\u1ecb d\u01b0\u1edbi h\u00e0m v(t\u2081) = ( 1\/9000t\u2081\u00b2 + n\/100 )(m\/s), n > 0\n<\/p>\n

b) K\u1ec3 t\u1eeb khi b\u1ecb Rada ph\u00e1t hi\u1ec7n \u0111\u1ebfn l\u00fac Nh\u1eadt B\u1ea3n ph\u00f3ng t\u00ean l\u1eeda th\u00ec qu\u1ea3 t\u00ean l\u1eeda g\u1eafn \u0111\u1ea7u \u0111\u1ea1n h\u1ea1t nh\u00e2n \u0111i \u0111\u01b0\u1ee3c 1913,4 km.<\/p>\n

c) Sau 15 ph\u00fat ph\u00f3ng t\u00ean l\u1eeda t\u1ea7m trung h\u1ea1 \u0111\u01b0\u1ee3c m\u1ee5c ti\u00eau bi\u1ebft qu\u00e3ng \u0111\u01b0\u1eddng n\u00f3 \u0111i \u0111\u01b0\u1ee3c b\u1eb1ng 1\/2 qu\u00e3ng \u0111\u01b0\u1eddng t\u00ean l\u1eeda Tri\u1ec1u Ti\u00ean \u0111\u00e3 \u0111\u01b0\u1ee3c trong 15 ph\u00fat \u0111\u00f3 khi \u0111\u00f3 gi\u00e1 tr\u1ecb n > 100\n<\/p>\n

d) Gi\u1ea3 s\u1eed h\u00e0m h(t) = ( -5m\/648t\u00b2 + 500m\/9t + a ), a in mathbbR, a > 0 (\u0111\u01a1n v\u1ecb: m\u00e9t) th\u1ec3 hi\u1ec7n \u0111\u1ed9 cao c\u1ee7a qu\u1ea3 t\u00ean l\u1eeda g\u1eafn \u0111\u1ea7u \u0111\u1ea1n h\u1ea1t nh\u00e2n so v\u1edbi m\u1ef1c n\u01b0\u1edbc bi\u1ec3n. Khi qu\u1ea3 t\u00ean l\u1eeda c\u1ee7a Tri\u1ec1u Ti\u00ean \u0111\u1ea1t \u0111\u1ed9 cao l\u1edbn nh\u1ea5t th\u00ec qu\u00e3ng \u0111\u01b0\u1eddng n\u00f3 \u0111i \u0111\u01b0\u1ee3c l\u00e0 483,12 km.<\/p>\n

C\u00e2u 2<\/strong>: Cho h\u00e0m s\u1ed1 f(x) = (x\u00b3)\/(3) – 3x – 6ln(2 – x) + 1\n<\/p>\n

a) \u0110\u1ea1o h\u00e0m c\u1ee7a h\u00e0m s\u1ed1 \u0111\u00e3 cho l\u00e0 f'(x) = (x\u00b2 – 2x\u00b2 – 3x)\/(x – 2)\n<\/p>\n

b) H\u00e0m s\u1ed1 \u0111\u00e3 cho \u0111\u1ed3ng bi\u1ebfn trong kho\u1ea3ng (-\u221e; -1) <\/p>\n

c) T\u1ed5ng c\u00e1c gi\u00e1 tr\u1ecb c\u1ef1c \u0111\u1ea1i v\u00e0 c\u1ef1c ti\u1ec3u c\u1ee7a \u0111\u1ed3 th\u1ecb h\u00e0m s\u1ed1 b\u1eb1ng 14\/3 – 6ln 6\n<\/p>\n

d) H\u00e0m s\u1ed1 g(x) = (\u221ax)\/(x\u00b2 + 2x + 2) c\u00f3 \u0111\u01b0\u1eddng ti\u1ec7m c\u1eadn xi\u00ean c\u00f3 d\u1ea1ng y = ax + b . Khi \u0111\u00f3 a + b = 1\/3\n<\/p>\n

Ph\u1ea7n 3 \u2013 C\u00e2u tr\u1eafc nghi\u1ec7m tr\u1ea3 l\u1eddi ng\u1eafn<\/strong><\/p>\n

T\u01b0\u01a1ng t\u1ef1 ki\u1ebfn th\u1ee9c c\u1ee7a c\u00e1c ph\u1ea7n tr\u00ean, ph\u1ea7n 3 g\u1ed3m 6 c\u00e2u h\u1ecfi v\u1edbi m\u1ee9c \u0111\u1ed9 ph\u00e2n c\u1ea5p t\u1eeb d\u1ec5 \u0111\u1ebfn kh\u00f3. Th\u00ed sinh c\u1ea7n c\u00f3 k\u1ef9 n\u0103ng t\u01b0 duy v\u00e0 gi\u1ea3i \u0111\u1ec1 t\u1ed1t \u1edf ph\u1ea7n n\u00e0y.<\/p>\n

C\u00e2u 1<\/strong>: Cho kh\u1ed1i ch\u00f3p $SABCD$ c\u00f3 \u0111\u00e1y $ABCD$ l\u00e0 h\u00ecnh thoi c\u1ea1nh $2$, $\\angle ABC = 120^\\circ$, $SB = 2$. M\u1eb7t ph\u1eb3ng $(SAD)$ vu\u00f4ng g\u00f3c v\u1edbi \u0111\u00e1y v\u00e0 c\u1ea1nh b\u00ean $SA$ t\u1ea1o v\u1edbi m\u1eb7t ph\u1eb3ng \u0111\u00e1y m\u1ed9t g\u00f3c $60^\\circ$. Th\u1ec3 t\u00edch kh\u1ed1i ch\u00f3p $SABCD$ b\u1eb1ng?<\/p>\n

C\u00e2u 2<\/strong>: M\u1ed9t nh\u00f3m c\u00e1c k\u1ef9 s\u01b0 mu\u1ed1n x\u00e2y d\u1ef1ng m\u1ed9t c\u00e2y c\u1ea7u v\u00f2m d\u00e0n th\u00e9p v\u1edbi gi\u00e1 \u0111\u1ee1 d\u01b0\u1edbi b\u1eb1ng th\u00e9p cao c\u1ea5p c\u00f3 h\u00ecnh d\u00e1ng l\u00e0 m\u1ed9t \u0111\u01b0\u1eddng cong Parabol n\u1ed1i t\u1eeb 2 c\u1ed9t tr\u1ee5 $A$ v\u00e0 $B$ n\u1eb1m b\u00ean d\u01b0\u1edbi c\u00e2y c\u1ea7u, bi\u1ebft hai c\u1ed9t tr\u1ee5 c\u00e1ch nhau $400\\, \\text{m}$, kho\u1ea3ng c\u00e1ch t\u1eeb tr\u1ee5 $A$ \u0111\u1ebfn c\u00e2y c\u1ea7u l\u00e0 $50\\, \\text{m}$ v\u00e0 $AB$ song song v\u1edbi m\u1eb7t \u0111\u01b0\u1eddng. G\u1eafn h\u1ec7 tr\u1ee5c t\u1ecda \u0111\u1ed9 $Oxy$ v\u00e0o c\u00e2y c\u1ea7u v\u1edbi \u0111\u01a1n v\u1ecb tr\u1ee5c t\u1ecda \u0111\u1ed9 l\u00e0 $10\\, \\text{m}$. Gi\u00e1 \u0111\u1ee1 d\u01b0\u1edbi b\u1eb1ng th\u00e9p l\u00e0 \u0111\u01b0\u1eddng cong Parabol t\u1ea1o v\u1edbi 2 tr\u1ee5c t\u1ecda \u0111\u1ed9 c\u00e1c h\u00ecnh ph\u1eb3ng c\u00f3 di\u1ec7n t\u00edch $S_1, S_2$ nh\u01b0 h\u00ecnh v\u1ebd b\u00ean, bi\u1ebft r\u1eb1ng: S\u2081 – 2S\u2082 = 2200\/21.\n<\/p>\n

\"\"<\/p>\n

\u0110i\u1ec3m cao nh\u1ea5t c\u1ee7a gi\u00e1 \u0111\u1ee1 d\u01b0\u1edbi b\u1eb1ng th\u00e9p cao c\u1ea5p c\u00e1ch m\u1eb7t \u0111\u01b0\u1eddng c\u00e2y c\u1ea7u bao nhi\u00eau m\u00e9t (l\u00e0m tr\u00f2n \u0111\u1ebfn h\u00e0ng ph\u1ea7n m\u01b0\u1eddi)?<\/p>\n

T\u1ea3i \u0110\u1ec1 thi th\u1eed t\u1ed1t nghi\u1ec7p THPT m\u00f4n To\u00e1n 2024 \u2013 2025 tr\u01b0\u1eddng THPT L\u00ea Th\u00e1nh T\u00f4ng TP HCM l\u1ea7n 2<\/strong><\/h2>\n

C\u00e1c b\u1ea1n h\u1ecdc sinh \u0111\u1eebng b\u1ecf l\u1ee1 t\u00e0i li\u1ec7u h\u1eefu \u00edch n\u00e0y, b\u1edfi ch\u00fang s\u1ebd h\u1ed7 tr\u1ee3 c\u00e1c b\u1ea1n trong qu\u00e1 tr\u00ecnh \u00f4n t\u1eadp v\u00e0 gi\u1ea3i \u0111\u1ec1 thi. H\u00e3y t\u1ea3i \u0110\u1ec1 thi th\u1eed t\u1ed1t nghi\u1ec7p THPT m\u00f4n To\u00e1n 2024 \u2013 2025 tr\u01b0\u1eddng THPT L\u00ea Th\u00e1nh T\u00f4ng TP HCM l\u1ea7n 2<\/strong> qua \u0111\u01b0\u1eddng link d\u01b0\u1edbi \u0111\u00e2y v\u00e0 luy\u1ec7n t\u1eadp ngay h\u00f4m nay. T\u00e0i li\u1ec7u n\u00e0y c\u0169ng s\u1ebd l\u00e0 ch\u00eca kh\u00f3a gi\u00fap b\u1ea1n b\u1ee9c ph\u00e1 v\u00e0 \u0111\u1ea1t \u0111\u01b0\u1ee3c k\u1ebft qu\u1ea3 nh\u01b0 mong \u0111\u1ee3i trong k\u00ec thi ch\u00ednh th\u1ee9c \u0111\u1ea5y.<\/p>\n