{"id":380,"date":"2026-05-05T04:42:43","date_gmt":"2026-05-05T04:42:43","guid":{"rendered":"https:\/\/79.buffdemo.com\/index.php\/2026\/05\/05\/de-hoc-sinh-gioi-cap-truong-toan-10-nam-2024-2025-truong-thpt-bo-ha-bac-giang\/"},"modified":"2026-05-05T07:15:25","modified_gmt":"2026-05-05T07:15:25","slug":"de-hoc-sinh-gioi-cap-truong-toan-10-nam-2024-2025-truong-thpt-bo-ha-bac-giang","status":"publish","type":"post","link":"https:\/\/79.buffdemo.com\/index.php\/2026\/05\/05\/de-hoc-sinh-gioi-cap-truong-toan-10-nam-2024-2025-truong-thpt-bo-ha-bac-giang\/","title":{"rendered":"\u0110\u1ec1 h\u1ecdc sinh gi\u1ecfi c\u1ea5p tr\u01b0\u1eddng To\u00e1n 10 n\u0103m 2024 \u2013 2025 tr\u01b0\u1eddng THPT B\u1ed1 H\u1ea1 B\u1eafc Giang"},"content":{"rendered":"

\u0110\u1ec1 h\u1ecdc sinh gi\u1ecfi c\u1ea5p tr\u01b0\u1eddng To\u00e1n 10 n\u0103m 2024 \u2013 2025 tr\u01b0\u1eddng THPT B\u1ed1 H\u1ea1 B\u1eafc Giang<\/strong> \u0111\u01b0\u1ee3c bi\u00ean so\u1ea1n nh\u1eb1m \u0111\u00e1nh gi\u00e1 n\u0103ng l\u1ef1c v\u00e0 b\u1ed3i d\u01b0\u1ee1ng t\u01b0 duy To\u00e1n h\u1ecdc n\u00e2ng cao. V\u1edbi c\u1ea5u tr\u00fac b\u00e1m s\u00e1t ch\u01b0\u01a1ng tr\u00ecnh m\u1edbi nh\u1ea5t n\u0103m 2024 \u2013 2025, \u0111\u1ec1 thi gi\u00fap h\u1ecdc sinh r\u00e8n luy\u1ec7n k\u1ef9 n\u0103ng gi\u1ea3i b\u00e0i t\u1eadp chuy\u00ean s\u00e2u, ph\u00e1t tri\u1ec3n t\u01b0 duy logic v\u00e0 chu\u1ea9n b\u1ecb t\u1ed1t nh\u1ea5t cho c\u00e1c k\u1ef3 thi l\u1edbn. \u0110\u1ec3 t\u00ecm hi\u1ec3u chi ti\u1ebft n\u1ed9i dung v\u00e0 ph\u01b0\u01a1ng ph\u00e1p l\u00e0m b\u00e0i hi\u1ec7u qu\u1ea3, h\u00e3y c\u00f9ng T\u00e0i Li\u1ec7u Mi\u1ec5n Ph\u00ed<\/strong><\/a> tham kh\u1ea3o b\u00e0i vi\u1ebft sau!<\/p>\n

N\u1ed9i dung \u0110\u1ec1 h\u1ecdc sinh gi\u1ecfi c\u1ea5p tr\u01b0\u1eddng To\u00e1n 10 n\u0103m 2024 \u2013 2025 tr\u01b0\u1eddng THPT B\u1ed1 H\u1ea1 B\u1eafc Giang<\/strong><\/h2>\n

\u0110\u1ec1 h\u1ecdc sinh gi\u1ecfi c\u1ea5p tr\u01b0\u1eddng To\u00e1n 10 n\u0103m 2024 \u2013 2025 tr\u01b0\u1eddng THPT B\u1ed1 H\u1ea1 B\u1eafc Giang<\/strong> c\u00f3 2 m\u00e3 \u0111\u1ec1 (M\u00e3 \u0111\u1ec1 101 v\u00e0 M\u00e3 \u0111\u1ec1 103). M\u1ed7i \u0111\u1ec1 thi g\u1ed3m 5 trang v\u00e0 c\u00f3 c\u1ea5u tr\u00fac g\u1ed3m hai ph\u1ea7n ch\u00ednh:<\/p>\n

Ph\u1ea7n I: Tr\u1eafc nghi\u1ec7m (14 \u0111i\u1ec3m)<\/strong><\/span><\/p>\n

    \n
  • 20 c\u00e2u h\u1ecfi tr\u1eafc nghi\u1ec7m kh\u00e1ch quan v\u1edbi 4 l\u1ef1a ch\u1ecdn.<\/li>\n
  • 4 c\u00e2u h\u1ecfi tr\u1eafc nghi\u1ec7m \u0111\u00fang\/sai.<\/li>\n
  • 6 c\u00e2u h\u1ecfi tr\u1eafc nghi\u1ec7m y\u00eau c\u1ea7u tr\u1ea3 l\u1eddi ng\u1eafn.<\/li>\n<\/ul>\n

    1.1 C\u00e2u h\u1ecfi tr\u1eafc nghi\u1ec7m nhi\u1ec1u l\u1ef1a ch\u1ecdn (20 c\u00e2u)<\/strong><\/p>\n

    C\u00e1c c\u00e2u h\u1ecfi tr\u1eafc nghi\u1ec7m bao g\u1ed3m nhi\u1ec1u ch\u1ee7 \u0111\u1ec1 kh\u00e1c nhau, tr\u1ea3i d\u00e0i t\u1eeb h\u00ecnh h\u1ecdc, \u0111\u1ea1i s\u1ed1 \u0111\u1ebfn \u1ee9ng d\u1ee5ng th\u1ef1c t\u1ebf.<\/p>\n

    C\u00e2u 1:<\/strong> Trong m\u1eb7t ph\u1eb3ng t\u1ecda \u0111\u1ed9 Oxy , cho \u0111\u01b0\u1eddng tr\u00f2n (C) : (x+1)^2 + (y-2)^2 = 9 v\u00e0 \u0111i\u1ec3m M(2,-2) . T\u1eeb \u0111i\u1ec3m M k\u1ebb c\u00e1c ti\u1ebfp tuy\u1ebfn MA, MB c\u1ee7a \u0111\u01b0\u1eddng tr\u00f2n v\u1edbi hai ti\u1ebfp \u0111i\u1ec3m l\u00e0 A, B . G\u1ecdi E(a,b) l\u00e0 t\u00e2m \u0111\u01b0\u1eddng tr\u00f2n n\u1ed9i ti\u1ebfp tam gi\u00e1c MAB . Gi\u00e1 tr\u1ecb c\u1ee7a bi\u1ec3u th\u1ee9c 5(a + b) b\u1eb1ng<\/p>\n

    A. 5
    \nB. 4
    \nC. 7
    \nD. 2<\/p>\n

    C\u00e2u 2:<\/strong> Cho ba l\u1ef1c F\u2081\u20d7 = MA\u20d7, F\u2082\u20d7 = MB\u20d7, F\u2083\u20d7 = MC\u20d7 c\u00f9ng t\u00e1c \u0111\u1ed9ng v\u00e0o m\u1ed9t v\u1eadt t\u1ea1i \u0111i\u1ec3m M v\u00e0 v\u1eadt \u0111\u1ee9ng y\u00ean. Cho bi\u1ebft c\u01b0\u1eddng \u0111\u1ed9 c\u1ee7a F\u2081\u20d7, F\u2082\u20d7 \u0111\u1ec1u b\u1eb1ng 50N v\u00e0 g\u00f3c angle AMB = 60^\u00b0 .<\/p>\n

    Khi \u0111\u00f3 c\u01b0\u1eddng \u0111\u1ed9 l\u1ef1c c\u1ee7a F_3 l\u00e0:<\/p>\n

    A. 50\u221a3 N
    \nB. 50\u221a2 N
    \nC. 25\u221a3 N
    \nD. 100\u221a3 N <\/p>\n

    1.2 C\u00e2u h\u1ecfi tr\u1eafc nghi\u1ec7m \u0111\u00fang\/sai (4 c\u00e2u)<\/strong><\/p>\n

    D\u1ea1ng b\u00e0i n\u00e0y y\u00eau c\u1ea7u h\u1ecdc sinh ki\u1ec3m tra t\u00ednh \u0111\u00fang\/sai c\u1ee7a c\u00e1c m\u1ec7nh \u0111\u1ec1 to\u00e1n h\u1ecdc.<\/p>\n

    C\u00e2u 1: <\/strong>Cho tam gi\u00e1c ABC c\u00f3 tr\u1ecdng t\u00e2m G , M l\u00e0 trung \u0111i\u1ec3m c\u1ee7a \u0111o\u1ea1n BC , AD l\u00e0 \u0111\u01b0\u1eddng ph\u00e2n gi\u00e1c trong c\u1ee7a g\u00f3c A . Bi\u1ebft:\u00a0<\/span>\\[
    \nAB = 5, \\quad AC = 7, \\quad \\widehat{BAC} = 120^{\\circ}.
    \n\\] C\u00e1c m\u1ec7nh \u0111\u1ec1 sau \u0111\u00fang hay sai?<\/span><\/p>\n

    A. AB\u20d7 \u00b7 AC\u20d7 = 35\/2 .<\/span>
    \nB. AD\u20d7 = 5\/12 AB\u20d7 + 7\/12 AC\u20d7 .<\/span>
    \nC. MB\u20d7 + MC\u20d7 = 0\u20d7 .<\/span>
    \nD. AG\u20d7 = 1\/3 AB\u20d7 + 1\/3 AC\u20d7 .<\/span><\/p>\n

    C\u00e2u 2:<\/strong> Trong m\u1eb7t ph\u1eb3ng v\u1edbi h\u1ec7 t\u1ecda \u0111\u1ed9 Oxy , cho \u0111\u01b0\u1eddng th\u1eb3ng: d: x + 2y + 8 = 0 v\u00e0 c\u00e1c \u0111i\u1ec3m: A(1,2), B(-2,1), C(-5,3). \u0110i\u1ec3m M(a,b) thu\u1ed9c \u0111\u01b0\u1eddng th\u1eb3ng d sao cho: <\/span>P = MA^2 + MB^2 + MC^2 \u0111\u1ea1t gi\u00e1 tr\u1ecb nh\u1ecf nh\u1ea5t. C\u00e1c m\u1ec7nh \u0111\u1ec1 sau \u0111\u00fang hay sai?<\/span><\/p>\n

    A. BC = \u221a13 .
    \nB. \u0110\u01b0\u1eddng tr\u00f2n t\u00e2m A v\u00e0 ti\u1ebfp x\u00fac v\u1edbi \u0111\u01b0\u1eddng th\u1eb3ng d c\u00f3 b\u00e1n k\u00ednh l\u00e0 R = (13)\/(\u221a5) .
    \nC. G\u1ecdi E = BC cap d . Khi \u0111\u00f3 E(20,-14) .
    \nD. a + b = -6 .<\/span><\/p>\n

    1.3 D\u1ea1ng c\u00e2u h\u1ecfi ng\u1eafn<\/strong><\/p>\n

    C\u00e2u 1:\u00a0<\/strong>Trong m\u1eb7t ph\u1eb3ng v\u1edbi h\u1ec7 t\u1ecda \u0111\u1ed9 Oxy , cho \u0111\u01b0\u1eddng tr\u00f2n (C) c\u00f3 ph\u01b0\u01a1ng tr\u00ecnh: \\[
    \n(x-3)^2 + (y+1)^2 = 18
    \n\\]\nv\u00e0 \u0111\u01b0\u1eddng th\u1eb3ng:
    \n\\[
    \nd: x + y \u2013 2 = 0.
    \n\\] \u0110i\u1ec3m M(a; b) ( a > 0 ) thu\u1ed9c \u0111\u01b0\u1eddng th\u1eb3ng d sao cho qua M k\u1ebb \u0111\u01b0\u1ee3c 2 ti\u1ebfp tuy\u1ebfn MA, MB \u0111\u1ebfn (C) ( A, B l\u00e0 hai ti\u1ebfp \u0111i\u1ec3m) v\u00e0 tam gi\u00e1c ABC \u0111\u1ec1u. Khi \u0111\u00f3 a\u00b2 + b\u00b2 b\u1eb1ng bao nhi\u00eau?<\/p>\n

    C\u00e2u 2:\u00a0<\/strong>M\u1ed9t chi\u1ebfc c\u1ed5ng h\u00ecnh parabol bao g\u1ed3m m\u1ed9t c\u1eeda ch\u00ednh h\u00ecnh ch\u1eef nh\u1eadt \u1edf gi\u1eefa v\u00e0 hai c\u00e1nh c\u1eeda ph\u1ee5 hai b\u00ean nh\u01b0 h\u00ecnh v\u1ebd.<\/p>\n

    \"\"<\/p>\n

    Bi\u1ebft chi\u1ec1u cao c\u1ed5ng parabol l\u00e0 4m , c\u1eeda ch\u00ednh (\u1edf gi\u1eefa parabol) cao 3m v\u00e0 r\u1ed9ng 4m . T\u00ednh kho\u1ea3ng c\u00e1ch gi\u1eefa hai ch\u00e2n c\u1ed5ng parabol \u1ea5y (\u0111o\u1ea1n AB tr\u00ean h\u00ecnh v\u1ebd).<\/p>\n

    Ph\u1ea7n II: T\u1ef1 lu\u1eadn (6 \u0111i\u1ec3m)<\/strong><\/span><\/p>\n

      \n
    • 4 b\u00e0i to\u00e1n t\u1ef1 lu\u1eadn, y\u00eau c\u1ea7u gi\u1ea3i chi ti\u1ebft.<\/li>\n<\/ul>\n

      G\u1ed3m 4 b\u00e0i to\u00e1n y\u00eau c\u1ea7u tr\u00ecnh b\u00e0y chi ti\u1ebft v\u00e0 l\u1eadp lu\u1eadn ch\u1eb7t ch\u1ebd.<\/p>\n

      C\u00e2u 1:<\/strong> Gi\u1ea3i ph\u01b0\u01a1ng tr\u00ecnh <\/span>\\[
      \n\\sqrt{3x + 1} + \\sqrt{4x \u2013 3} = \\sqrt{5x + 4}.
      \n\\]\n<\/p>\n

      C\u00e2u 2: Cho h\u00e0m s\u1ed1 y = x\u00b2 + 2x + 2 c\u00f3 \u0111\u1ed3 th\u1ecb \\( (P).<\/span><\/strong> T\u00ecm gi\u00e1 tr\u1ecb c\u1ee7a tham s\u1ed1 m \u0111\u1ec3 \u0111\u01b0\u1eddng th\u1eb3ng d: y = (2 \u2013 2m)x + m v\u00e0 \u0111\u1ed3 th\u1ecb (P) c\u00f3 \u0111i\u1ec3m chung A(x\u2081, y\u2081) , B(x\u2082, y\u2082) sao cho bi\u1ec3u th\u1ee9c Q = 3(x_1 + x_2) \u2013 (x_1^2 + x_2^2) \u0111\u1ea1t gi\u00e1 tr\u1ecb l\u1edbn nh\u1ea5t.<\/span><\/strong><\/p>\n

      T\u1ea3i \u0110\u1ec1 h\u1ecdc sinh gi\u1ecfi c\u1ea5p tr\u01b0\u1eddng To\u00e1n 10 n\u0103m 2024 \u2013 2025 tr\u01b0\u1eddng THPT B\u1ed1 H\u1ea1 B\u1eafc Giang<\/strong><\/h2>\n

      B\u1ea1n \u0111\u00e3 s\u1eb5n s\u00e0ng th\u1eed s\u1ee9c v\u1edbi \u0111\u1ec1 thi hsg to\u00e1n 10 ch\u01b0a? \u0110\u00e2y ch\u00ednh l\u00e0 c\u01a1 h\u1ed9i \u0111\u1ec3 b\u1ea1n ki\u1ec3m tra n\u0103ng l\u1ef1c c\u1ee7a m\u00ecnh v\u00e0 chinh ph\u1ee5c nh\u1eefng c\u00e2u h\u1ecfi h\u00f3c b\u00faa nh\u1ea5t! T\u1ea3i ngay \u0110\u1ec1 h\u1ecdc sinh gi\u1ecfi c\u1ea5p tr\u01b0\u1eddng To\u00e1n 10 n\u0103m 2024 \u2013 2025 tr\u01b0\u1eddng THPT B\u1ed1 H\u1ea1 B\u1eafc Giang<\/strong>\u00a0v\u00e0 b\u1eaft \u0111\u1ea7u \u00f4n luy\u1ec7n \u0111\u1ec3 \u0111\u1ea1t \u0111i\u1ec3m cao nh\u1ea5t trong k\u1ef3 thi n\u00e0y nh\u00e9!<\/p>\n