{"id":422,"date":"2026-05-05T04:44:55","date_gmt":"2026-05-05T04:44:55","guid":{"rendered":"https:\/\/79.buffdemo.com\/index.php\/2026\/05\/05\/de-chon-hsg-toan-11-nam-2024-2025-truong-thpt-phung-khac-khoan-ha-noi\/"},"modified":"2026-05-05T07:17:41","modified_gmt":"2026-05-05T07:17:41","slug":"de-chon-hsg-toan-11-nam-2024-2025-truong-thpt-phung-khac-khoan-ha-noi","status":"publish","type":"post","link":"https:\/\/79.buffdemo.com\/index.php\/2026\/05\/05\/de-chon-hsg-toan-11-nam-2024-2025-truong-thpt-phung-khac-khoan-ha-noi\/","title":{"rendered":"\u0110\u1ec1 ch\u1ecdn HSG To\u00e1n 11 n\u0103m 2024 \u2013 2025 tr\u01b0\u1eddng THPT Ph\u00f9ng Kh\u1eafc Khoan H\u00e0 N\u1ed9i"},"content":{"rendered":"

\u0110\u1ec1 ch\u1ecdn HSG To\u00e1n 11 n\u0103m 2024 \u2013 2025 tr\u01b0\u1eddng THPT Ph\u00f9ng Kh\u1eafc Khoan H\u00e0 N\u1ed9i<\/strong> l\u00e0 t\u00e0i li\u1ec7u tuy\u1ec3n ch\u1ecdn nh\u1eefng b\u00e0i to\u00e1n n\u00e2ng cao ti\u00eau bi\u1ec3u, bao g\u1ed3m c\u00e1c ch\u1ee7 \u0111\u1ec1 quan tr\u1ecdng nh\u01b0 L\u01b0\u1ee3ng gi\u00e1c, D\u00e3y s\u1ed1, Gi\u1edbi h\u1ea1n, Li\u00ean t\u1ee5c v\u00e0 H\u00ecnh h\u1ecdc kh\u00f4ng gian. \u0110\u1ec1 thi gi\u00fap r\u00e8n luy\u1ec7n t\u01b0 duy logic, k\u1ef9 n\u0103ng l\u1eadp lu\u1eadn v\u00e0 kh\u1ea3 n\u0103ng gi\u1ea3i quy\u1ebft v\u1ea5n \u0111\u1ec1 \u1edf c\u1ea5p \u0111\u1ed9 n\u00e2ng cao. C\u00f9ng T\u00e0i Li\u1ec7u H\u1ecdc T\u1eadp<\/a><\/strong> kh\u00e1m ph\u00e1 ngay nh\u1eefng c\u00e2u h\u1ecfi trong t\u00e0i li\u1ec7u d\u01b0\u1edbi \u0111\u00e2y nh\u00e9.<\/p>\n

N\u1ed9i dung \u0110\u1ec1 ch\u1ecdn HSG To\u00e1n 11 n\u0103m 2024 \u2013 2025 tr\u01b0\u1eddng THPT Ph\u00f9ng Kh\u1eafc Khoan H\u00e0 N\u1ed9i<\/strong><\/h2>\n

\u0110\u1ec1 ch\u1ecdn HSG To\u00e1n 11 n\u0103m 2024 \u2013 2025 tr\u01b0\u1eddng THPT Ph\u00f9ng Kh\u1eafc Khoan H\u00e0 N\u1ed9i<\/strong> g\u1ed3m 6 c\u00e2u t\u1ef1 lu\u1eadn v\u1edbi th\u1eddi gian l\u00e0m b\u00e0i 150 ph\u00fat, bao qu\u00e1t c\u00e1c ch\u1ee7 \u0111\u1ec1 tr\u1ecdng t\u00e2m v\u00e0 n\u00e2ng cao trong ch\u01b0\u01a1ng tr\u00ecnh To\u00e1n 11. M\u1ee5c ti\u00eau c\u1ee7a \u0111\u1ec1 l\u00e0 tuy\u1ec3n ch\u1ecdn nh\u1eefng h\u1ecdc sinh c\u00f3 n\u0103ng l\u1ef1c trong m\u00f4n To\u00e1n \u0111\u1ec3 b\u1ed3i d\u01b0\u1ee1ng v\u00e0 chu\u1ea9n b\u1ecb cho c\u00e1c k\u1ef3 thi h\u1ecdc sinh gi\u1ecfi c\u1ea5p cao h\u01a1n. N\u1ed9i dung \u0111\u1ec1 \u0111\u01b0\u1ee3c chia th\u00e0nh ba m\u1ea3ng ch\u00ednh:<\/p>\n

    \n
  • \u0110\u1ea1i s\u1ed1 v\u00e0 Gi\u1ea3i t\u00edch<\/li>\n
  • T\u01b0 duy logic v\u00e0 k\u1ef9 thu\u1eadt t\u00ednh to\u00e1n<\/li>\n
  • H\u00ecnh h\u1ecdc kh\u00f4ng gian<\/li>\n<\/ul>\n

    D\u01b0\u1edbi \u0111\u00e2y l\u00e0 m\u1ed9t s\u1ed1 c\u00e2u h\u1ecfi tr\u00edch t\u1eeb \u0111\u1ec1<\/strong><\/p>\n

    C\u00e2u 1.<\/strong> (2,0 \u0111i\u1ec3m)<\/p>\n

    a) T\u00ecm t\u1eadp x\u00e1c \u0111\u1ecbnh D c\u1ee7a h\u00e0m s\u1ed1 y = tan(2x – \u03c0\/4) <\/p>\n

    b) X\u00e9t t\u00ednh ch\u1eb5n, l\u1ebb c\u1ee7a h\u00e0m s\u1ed1 y = (sin x – tan x)\/(sin x + cot x) <\/p>\n

    C\u00e2u 2.<\/strong> (2,0 \u0111i\u1ec3m). T\u00ecm nghi\u1ec7m l\u1edbn nh\u1ea5t c\u1ee7a ph\u01b0\u01a1ng tr\u00ecnh 2cos2x – 1 = 0 trong \u0111o\u1ea1n [0; \u03c0] <\/p>\n

    C\u00e2u 3.<\/strong> (4,0 \u0111i\u1ec3m)<\/p>\n

    a) T\u00ecm s\u1ed1 h\u1ea1ng th\u1ee9 10 v\u00e0 x\u00e9t t\u00ednh t\u0103ng, gi\u1ea3m c\u1ee7a d\u00e3y s\u1ed1 u_n = (1)\/(n\u00b2 + n) <\/p>\n

    b) M\u1ed9t qu\u1ea3 b\u00f3ng cao su \u0111\u01b0\u1ee3c th\u1ea3 t\u1eeb \u0111\u1ed9 cao 81 m\u00e9t. M\u1ed7i l\u1ea7n ch\u1ea1m \u0111\u1ea5t qu\u1ea3 b\u00f3ng l\u1ea1i n\u1ea3y l\u00ean hai ph\u1ea7n ba \u0111\u1ed9 cao c\u1ee7a l\u1ea7n r\u01a1i tr\u01b0\u1edbc. T\u00ednh t\u1ed5ng c\u00e1c kho\u1ea3ng c\u00e1ch r\u01a1i v\u00e0 n\u1ea3y c\u1ee7a qu\u1ea3 b\u00f3ng t\u1eeb l\u00fac th\u1ea3 b\u00f3ng cho \u0111\u1ebfn l\u00fac b\u00f3ng kh\u00f4ng n\u1ea3y n\u1eefa.<\/p>\n

    C\u00e2u 4.<\/strong> (1,5 \u0111i\u1ec3m)<\/p>\n

    M\u1ed9t ng\u01b0\u1eddi g\u1eedi v\u00e0o ng\u00e2n h\u00e0ng s\u1ed1 ti\u1ec1n ti\u1ebft ki\u1ec7m l\u00e0 73 tri\u1ec7u \u0111\u1ed3ng theo h\u00ecnh th\u1ee9c l\u00e3i k\u00e9p, nh\u1eb1m m\u1ee5c \u0111\u00edch sau 5 n\u0103m thu \u0111\u01b0\u1ee3c s\u1ed1 ti\u1ec1n l\u00e0 100 tri\u1ec7u \u0111\u1ed3ng. Tuy nhi\u00ean v\u00ec k\u1ebf ho\u1ea1ch t\u00e0i ch\u00ednh thay \u0111\u1ed5i n\u00ean ng\u01b0\u1eddi \u0111\u00f3 kh\u00f4ng r\u00fat ti\u1ec1n ra m\u00e0 \u0111\u1ec3 sau 10 n\u0103m m\u1edbi r\u00fat to\u00e0n b\u1ed9 g\u1ed1c v\u00e0 l\u00e3i. Gi\u1ea3 s\u1eed trong su\u1ed1t qu\u00e1 tr\u00ecnh g\u1eedi 10 n\u0103m, l\u00e3i su\u1ea5t c\u1ee7a ng\u00e2n h\u00e0ng kh\u00f4ng thay \u0111\u1ed5i. T\u00ednh s\u1ed1 ti\u1ec1n c\u1ea3 g\u1ed1c v\u00e0 l\u00e3i m\u00e0 ng\u01b0\u1eddi \u0111\u00f3 nh\u1eadn v\u1ec1 sau 10 n\u0103m g\u1eedi? (K\u1ebft qu\u1ea3 l\u00e0m tr\u00f2n \u0111\u1ebfn h\u00e0ng \u0111\u01a1n v\u1ecb v\u00e0 t\u00ednh theo \u0111\u01a1n v\u1ecb tri\u1ec7u \u0111\u1ed3ng).<\/p>\n

    C\u00e2u 5.<\/strong> (4,0 \u0111i\u1ec3m)<\/p>\n

    a. T\u00ednh $\\lim u_n$ v\u1edbi $u_n = \\frac{1}{1.3} + \\frac{1}{3.5} + \\frac{1}{5.7} + \\dots + \\frac{1}{(2n-1)(2n+1)}$<\/p>\n

    b. T\u00ecm $m$ \u0111\u1ec3 h\u00e0m s\u1ed1 f(x) = \\begin{cases}
    \n\\frac{\\sqrt{1+x} \u2013 3\\sqrt{1+x}}{x}, & \\text{khi } x < 0, \\\\
    \n\\frac{m + x^3 \u2013 3x + 1}{x+2}, & \\text{khi } x \\geq 0
    \n\\end{cases} li\u00ean t\u1ee5c t\u1ea1i \u0111i\u1ec3m $x_0 = 0$.<\/p>\n

    T\u1ea3i \u0110\u1ec1 ch\u1ecdn HSG To\u00e1n 11 n\u0103m 2024 \u2013 2025 tr\u01b0\u1eddng THPT Ph\u00f9ng Kh\u1eafc Khoan H\u00e0 N\u1ed9i<\/strong><\/h2>\n

    \u0110\u1ec1 ch\u1ecdn HSG To\u00e1n 11 n\u0103m 2024 \u2013 2025 tr\u01b0\u1eddng THPT Ph\u00f9ng Kh\u1eafc Khoan H\u00e0 N\u1ed9i<\/strong> l\u00e0 ngu\u1ed3n luy\u1ec7n t\u1eadp ch\u1ea5t l\u01b0\u1ee3ng gi\u00fap h\u1ecdc sinh n\u1eafm ki\u1ebfn th\u1ee9c v\u00e0 r\u00e8n luy\u1ec7n k\u1ef9 n\u0103ng gi\u1ea3i to\u00e1n n\u00e2ng cao. V\u1edbi h\u1ec7 th\u1ed1ng c\u00e2u h\u1ecfi c\u00f3 t\u00ednh ph\u00e2n lo\u1ea1i, t\u00e0i li\u1ec7u ph\u00f9 h\u1ee3p cho nh\u1eefng b\u1ea1n \u0111ang \u00f4n luy\u1ec7n \u0111\u1ec3 chinh ph\u1ee5c c\u00e1c k\u1ef3 thi h\u1ecdc sinh gi\u1ecfi. H\u00e3y click v\u00e0o \u0111\u01b0\u1eddng d\u1eabn d\u01b0\u1edbi \u0111\u00e2y \u0111\u1ec3 t\u1ea3i to\u00e0n b\u1ed9 t\u00e0i li\u1ec7u to\u00e1n l\u1edbp 11 nh\u00e9!<\/p>\n