{"id":532,"date":"2026-05-05T04:50:45","date_gmt":"2026-05-05T04:50:45","guid":{"rendered":"https:\/\/79.buffdemo.com\/index.php\/2026\/05\/05\/de-thi-thu-tot-nghiep-thpt-mon-toan-2024-2025-truong-thpt-tran-nguyen-han-vinh-phuc\/"},"modified":"2026-05-05T07:24:08","modified_gmt":"2026-05-05T07:24:08","slug":"de-thi-thu-tot-nghiep-thpt-mon-toan-2024-2025-truong-thpt-tran-nguyen-han-vinh-phuc","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-tran-nguyen-han-vinh-phuc\/","title":{"rendered":"\u0110\u1ec1 thi th\u1eed t\u1ed1t nghi\u1ec7p THPT m\u00f4n To\u00e1n 2024 \u2013 2025 tr\u01b0\u1eddng THPT Tr\u1ea7n Nguy\u1ec5n H\u00e3n V\u0129nh Ph\u00fac"},"content":{"rendered":"

\u0110\u1ec1 thi th\u1eed t\u1ed1t nghi\u1ec7p THPT m\u00f4n To\u00e1n 2024 \u2013 2025 tr\u01b0\u1eddng THPT Tr\u1ea7n Nguy\u1ec5n H\u00e3n V\u0129nh Ph\u00fac<\/strong> c\u00f3 th\u1eddi gian l\u00e0m b\u00e0i 90 ph\u00fat, g\u1ed3m 3 ph\u1ea7n v\u1edbi nhi\u1ec1u d\u1ea1ng c\u00e2u h\u1ecfi kh\u00e1c nhau. \u0110\u1ec1 thi k\u00e8m theo ph\u1ea7n \u0111\u00e1p \u00e1n chi ti\u1ebft v\u00e0 h\u01b0\u1edbng d\u1eabn gi\u1ea3i cho t\u1ea5t c\u1ea3 c\u00e1c c\u00e2u h\u1ecfi, gi\u00fap h\u1ecdc sinh c\u00f3 th\u1ec3 t\u1ef1 \u0111\u00e1nh gi\u00e1 v\u00e0 h\u1ecdc t\u1eadp. T\u00e0i Li\u1ec7u \u00d4n Thi<\/a><\/strong> s\u1ebd c\u00f9ng c\u00e1c b\u1ea1n kh\u00e1m ph\u00e1 \u0111\u1ec1 thi chi ti\u1ebft n\u00e0y h\u00f4m nay.<\/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 Tr\u1ea7n Nguy\u1ec5n H\u00e3n V\u0129nh Ph\u00fac<\/span><\/strong><\/h2>\n

\u0110\u1ec1 thi th\u1eed t\u1ed1t nghi\u1ec7p THPT m\u00f4n To\u00e1n 2024 \u2013 2025 tr\u01b0\u1eddng THPT Tr\u1ea7n Nguy\u1ec5n H\u00e3n V\u0129nh Ph\u00fac<\/strong> n\u00e0y \u0111\u01b0\u1ee3c bi\u00ean so\u1ea1n v\u1edbi nhi\u1ec1u d\u1ea1ng b\u00e0i t\u1eadp \u0111a d\u1ea1ng, t\u1eeb c\u01a1 b\u1ea3n \u0111\u1ebfn n\u00e2ng cao nh\u01b0: d\u00e3y s\u1ed1 v\u00e0 c\u1ea5p s\u1ed1, h\u00e0m s\u1ed1 v\u00e0 \u0111\u1ea1o h\u00e0m, b\u1ea5t ph\u01b0\u01a1ng tr\u00ecnh logarit, h\u00ecnh h\u1ecdc kh\u00f4ng gian, v\u00e0 x\u00e1c su\u1ea5t th\u1ed1ng k\u00ea \u2013 gi\u00fap h\u1ecdc sinh r\u00e8n luy\u1ec7n k\u1ef9 n\u0103ng gi\u1ea3i quy\u1ebft nhi\u1ec1u d\u1ea1ng b\u00e0i kh\u00e1c nhau trong ch\u01b0\u01a1ng tr\u00ecnh.<\/p>\n

Ph\u1ea7n I c\u00f3 12 c\u00e2u tr\u1eafc nghi\u1ec7m nhi\u1ec1u ph\u01b0\u01a1ng \u00e1n l\u1ef1a ch\u1ecdn, m\u1ed7i c\u00e2u ch\u1ec9 c\u00f3 m\u1ed9t \u0111\u00e1p \u00e1n \u0111\u00fang.<\/strong> <\/span><\/p>\n

C\u00e1c c\u00e2u h\u1ecfi bao g\u1ed3m c\u00e1c ch\u1ee7 \u0111\u1ec1 v\u1ec1 d\u00e3y s\u1ed1, \u0111\u1ea1o h\u00e0m, b\u1ea5t ph\u01b0\u01a1ng tr\u00ecnh, x\u00e1c su\u1ea5t, \u0111\u1ed3 th\u1ecb h\u00e0m s\u1ed1, c\u1ef1c tr\u1ecb v\u00e0 \u0111\u1ea1o h\u00e0m, h\u00ecnh h\u1ecdc kh\u00f4ng gian v\u00e0 vector.<\/em><\/p>\n

C\u00e2u 1.<\/strong> Cho d\u00e3y s\u1ed1 (u_n) v\u1edbi u_n = 2n – 5 . Kh\u1eb3ng \u0111\u1ecbnh n\u00e0o sau \u0111\u00e2y l\u00e0 \u0111\u00fang?<\/p>\n

A. D\u00e3y s\u1ed1 (u_n) l\u00e0 c\u1ea5p s\u1ed1 c\u1ed9ng v\u1edbi c\u00f4ng sai d = 2
\nB. D\u00e3y s\u1ed1 (u_n) l\u00e0 c\u1ea5p s\u1ed1 nh\u00e2n v\u1edbi c\u00f4ng b\u1ed9i q = 2
\nC. D\u00e3y s\u1ed1 (u_n) l\u00e0 c\u1ea5p s\u1ed1 nh\u00e2n v\u1edbi c\u00f4ng sai q = -5
\nD. D\u00e3y s\u1ed1 (u_n) l\u00e0 c\u1ea5p s\u1ed1 c\u1ed9ng v\u1edbi c\u00f4ng sai d = -5 <\/p>\n

C\u00e2u 2.<\/strong> \u0110\u1ea1o h\u00e0m c\u1ee7a h\u00e0m s\u1ed1 y = 3^x l\u00e0<\/p>\n

A. y\u2019 = 3^x \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0B. y\u2019 = x \u00b7 3^x-1 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0C. y\u2019 = 3^x ln 3 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0D. y\u2019 = (3^x)\/(ln 3) <\/p>\n

Ph\u1ea7n II c\u00f3 4 c\u00e2u tr\u1eafc nghi\u1ec7m \u0111\u00fang\/sai v\u1edbi nhi\u1ec1u \u00fd, y\u00eau c\u1ea7u th\u00ed sinh ph\u00e2n t\u00edch v\u00e0 ch\u1ecdn \u0111\u00fang ho\u1eb7c sai cho t\u1eebng \u00fd.<\/strong> <\/span><\/p>\n

Ph\u1ea7n n\u00e0y t\u1eadp trung v\u00e0o h\u00e0m s\u1ed1, h\u00ecnh h\u1ecdc kh\u00f4ng gian, v\u1eadt l\u00fd \u1ee9ng d\u1ee5ng (chuy\u1ec3n \u0111\u1ed9ng) v\u00e0 x\u00e1c su\u1ea5t.<\/em><\/p>\n

C\u00e2u 1.<\/strong> Cho h\u00e0m s\u1ed1 y = 2x\u00b3 + 2(m+1)x\u00b2 + 6x + 4 + 2m . Khi \u0111\u00f3:<\/p>\n

a) Khi m = -1 th\u00ec h\u00e0m s\u1ed1 \u0111\u1ed3ng bi\u1ebfn tr\u00ean (-\u221e; +\u221e) <\/p>\n

b) H\u00e0m s\u1ed1 kh\u00f4ng c\u00f3 c\u1ef1c tr\u1ecb khi m = 1 <\/p>\n

c) C\u00f3 3 gi\u00e1 tr\u1ecb nguy\u00ean d\u01b0\u01a1ng c\u1ee7a tham s\u1ed1 m \u0111\u1ec3 h\u00e0m s\u1ed1 \u0111\u1ed3ng bi\u1ebfn tr\u00ean m in (2;5) <\/p>\n

d) H\u00e0m s\u1ed1 \u0111\u1ea1t c\u1ef1c ti\u1ec3u t\u1ea1i x = 2 th\u00ec m in (2;5) <\/p>\n

C\u00e2u 2.<\/strong> Trong kh\u00f4ng gian Oxyz cho h\u00ecnh ch\u00f3p S.ABCD c\u00f3 S(0;0;3,5) , ABCD l\u00e0 h\u00ecnh ch\u1eef nh\u1eadt v\u1edbi A(0;0;0), B(4;0;0), D(0;10;0) (H\u00ecnh v\u1ebd).<\/p>\n

\"\"<\/p>\n

a) T\u1ecda \u0111\u1ed9 \u0111i\u1ec3m C(4;10;0) .<\/p>\n

b) T\u1ecda \u0111\u1ed9 c\u1ee7a vect\u01a1 SC\u20d7 l\u00e0 (4;10;-3,5) .<\/p>\n

c) Ph\u01b0\u01a1ng tr\u00ecnh m\u1eb7t ph\u1eb3ng (SBD) l\u00e0 x\/4 + y\/10 + z\/3,5 = 1 .<\/p>\n

d) G\u00f3c gi\u1eefa \u0111\u01b0\u1eddng th\u1eb3ng SC v\u00e0 m\u1eb7t ph\u1eb3ng (SBD) (l\u00e0m tr\u00f2n \u0111\u1ebfn h\u00e0ng \u0111\u01a1n v\u1ecb c\u1ee7a \u0111\u1ed9) l\u00e0 20^\u00b0 .<\/p>\n

Ph\u1ea7n III c\u00f3 6 c\u00e2u tr\u1eafc nghi\u1ec7m tr\u1ea3 l\u1eddi ng\u1eafn.<\/strong><\/span><\/p>\n

\u0110\u1ec1 thi \u0111\u00f2i h\u1ecfi th\u00ed sinh ph\u1ea3i gi\u1ea3i c\u00e1c b\u00e0i to\u00e1n ph\u1ee9c t\u1ea1p h\u01a1n v\u1ec1 h\u00ecnh h\u1ecdc kh\u00f4ng gian, c\u1ef1c ti\u1ec3u h\u00e0m s\u1ed1, th\u1ec3 t\u00edch h\u00ecnh h\u1ecdc v\u00e0 x\u00e1c su\u1ea5t.<\/em><\/p>\n

C\u00e2u 1.<\/strong> Kim t\u1ef1 th\u00e1p \u1edf Ai C\u1eadp c\u00f3 h\u00ecnh d\u1ea1ng l\u00e0 m\u1ed9t h\u00ecnh ch\u00f3p t\u1ee9 gi\u00e1c \u0111\u1ec1u c\u00f3 c\u1ea1nh \u0111\u00e1y l\u00e0 262 m v\u00e0 c\u1ea1nh b\u00ean l\u00e0 230 m . Gi\u1ea3 s\u1eed, t\u1eeb m\u1ed9t m\u1eb7t b\u1ec1n c\u1ee7a kim t\u1ef1 th\u00e1p ta c\u1ea7n \u0111\u00e0o m\u1ed9t con \u0111\u01b0\u1eddng ng\u1eafn nh\u1ea5t \u0111\u1ec3 \u0111i \u0111\u1ebfn t\u00e2m c\u1ee7a \u0111\u00e1y kim t\u1ef1 th\u00e1p, khi \u0111\u00f3 qu\u00e3ng \u0111\u01b0\u1eddng ng\u1eafn nh\u1ea5t c\u00f3 \u0111\u1ed9 d\u00e0i kho\u1ea3ng bao nhi\u00eau m\u00e9t (l\u00e0m tr\u00f2n \u0111\u1ebfn h\u00e0ng \u0111\u01a1n v\u1ecb)?<\/p>\n

\"\"<\/p>\n

C\u00e2u 2.<\/strong> M\u1ed9t c\u00e1i ao h\u00ecnh ABCDE , \u1edf gi\u1eefa ao c\u00f3 m\u1ed9t m\u1ea3nh v\u01b0\u1eddn h\u00ecnh tr\u00f2n c\u00f3 b\u00e1n k\u00ednh 10 (m) . Ng\u01b0\u1eddi ta mu\u1ed1n b\u1eafc m\u1ed9t c\u1ea7u c\u1ea7u t\u1eeb B c\u1ee7a ao \u0111\u1ebfn v\u01b0\u1eddn. T\u00ednh g\u1ea7n \u0111\u00fang \u0111\u1ed9 d\u00e0i t\u1ed1i thi\u1ec3u l c\u1ee7a c\u00e2y c\u1ea7u bi\u1ebft: Hai b\u1edd AE v\u00e0 BC n\u1eb1m tr\u00ean hai \u0111\u01b0\u1eddng th\u1eb3ng vu\u00f4ng g\u00f3c v\u1edbi nhau, hai \u0111\u01b0\u1eddng th\u1eb3ng n\u00e0y c\u1eaft nhau t\u1ea1i \u0111i\u1ec3m O .<\/p>\n