{"id":418,"date":"2026-05-05T04:44:42","date_gmt":"2026-05-05T04:44:42","guid":{"rendered":"https:\/\/79.buffdemo.com\/index.php\/2026\/05\/05\/10-de-on-tap-giua-hoc-ki-2-toan-11-nam-2024-2025-co-loi-giai-chi-tiet\/"},"modified":"2026-05-05T07:17:28","modified_gmt":"2026-05-05T07:17:28","slug":"10-de-on-tap-giua-hoc-ki-2-toan-11-nam-2024-2025-co-loi-giai-chi-tiet","status":"publish","type":"post","link":"https:\/\/79.buffdemo.com\/index.php\/2026\/05\/05\/10-de-on-tap-giua-hoc-ki-2-toan-11-nam-2024-2025-co-loi-giai-chi-tiet\/","title":{"rendered":"10 \u0111\u1ec1 \u00f4n t\u1eadp gi\u1eefa h\u1ecdc k\u00ec 2 To\u00e1n 11 n\u0103m 2024 \u2013 2025 c\u00f3 l\u1eddi gi\u1ea3i chi ti\u1ebft"},"content":{"rendered":"

10 \u0111\u1ec1 \u00f4n t\u1eadp gi\u1eefa h\u1ecdc k\u00ec 2 To\u00e1n 11 n\u0103m 2024 \u2013 2025 c\u00f3 l\u1eddi gi\u1ea3i chi ti\u1ebft<\/strong> l\u00e0 ngu\u1ed3n t\u00e0i li\u1ec7u h\u1eefu \u00edch gi\u00fap b\u1ea1n c\u1ee7ng c\u1ed1 ki\u1ebfn th\u1ee9c v\u00e0 r\u00e8n luy\u1ec7n k\u1ef9 n\u0103ng tr\u01b0\u1edbc k\u1ef3 thi quan tr\u1ecdng. V\u1edbi h\u1ec7 th\u1ed1ng c\u00e2u h\u1ecfi \u0111a d\u1ea1ng, b\u00e1m s\u00e1t ch\u01b0\u01a1ng tr\u00ecnh h\u1ecdc, c\u00f9ng l\u1eddi gi\u1ea3i chi ti\u1ebft, b\u1ed9 \u0111\u1ec1 n\u00e0y s\u1ebd gi\u00fap b\u1ea1n t\u1ef1 tin chinh ph\u1ee5c \u0111i\u1ec3m s\u1ed1 cao. \u0110\u1eebng b\u1ecf l\u1ee1 c\u01a1 h\u1ed9i \u00f4n t\u1eadp hi\u1ec7u qu\u1ea3, c\u00f9ng T\u00e0i Li\u1ec7u Mi\u1ec5n Ph\u00ed<\/strong><\/a> kh\u00e1m ph\u00e1 ngay nh\u00e9!<\/p>\n

N\u1ed9i dung 10 \u0111\u1ec1 \u00f4n t\u1eadp gi\u1eefa h\u1ecdc k\u00ec 2 To\u00e1n 11 n\u0103m 2024 \u2013 2025 c\u00f3 l\u1eddi gi\u1ea3i chi ti\u1ebft<\/span><\/strong><\/h2>\n

10 \u0111\u1ec1 \u00f4n t\u1eadp gi\u1eefa h\u1ecdc k\u00ec 2 To\u00e1n 11 n\u0103m 2024 \u2013 2025 c\u00f3 l\u1eddi gi\u1ea3i chi ti\u1ebft<\/strong> <\/span>bao g\u1ed3m 10 \u0111\u1ec1 thi \u00f4n t\u1eadp, t\u1ed5ng h\u1ee3p \u0111a d\u1ea1ng c\u00e1c c\u00e2u h\u1ecfi To\u00e1n 11 t\u1eeb tr\u1eafc nghi\u1ec7m \u0111\u1ebfn t\u1ef1 lu\u1eadn, \u0111\u01b0\u1ee3c h\u1ec7 th\u1ed1ng theo 4 ph\u1ea7n. N\u1ed9i dung b\u00e1m s\u00e1t ch\u01b0\u01a1ng tr\u00ecnh v\u1edbi c\u00e1c ch\u1ee7 \u0111\u1ec1 quan tr\u1ecdng nh\u01b0 h\u00e0m s\u1ed1 m\u0169, h\u00e0m s\u1ed1 logarit v\u00e0 quan h\u1ec7 vu\u00f4ng g\u00f3c trong kh\u00f4ng gian. M\u1ed7i \u0111\u1ec1 thi c\u00f3 th\u1eddi gian l\u00e0m b\u00e0i 90 ph\u00fat, g\u1ed3m nhi\u1ec1u d\u1ea1ng b\u00e0i t\u1eadp gi\u00fap h\u1ecdc sinh c\u1ee7ng c\u1ed1 ki\u1ebfn th\u1ee9c v\u00e0 r\u00e8n luy\u1ec7n k\u1ef9 n\u0103ng gi\u1ea3i to\u00e1n.<\/p>\n

N\u1ed9i dung ki\u1ebfn th\u1ee9c \u0111\u01b0\u1ee3c \u00f4n t\u1eadp g\u1ed3m<\/strong><\/h3>\n

\u0110\u1ea1i s\u1ed1: <\/strong><\/p>\n

    \n
  • Ch\u01b0\u01a1ng 4: H\u00e0m s\u1ed1 m\u0169 v\u00e0 h\u00e0m s\u1ed1 logarit<\/li>\n<\/ul>\n

    H\u00ecnh h\u1ecdc: <\/strong>Ch\u01b0\u01a1ng 8: Quan h\u1ec7 vu\u00f4ng g\u00f3c trong kh\u00f4ng gian<\/p>\n

      \n
    • B\u00e0i 1: Hai \u0111\u01b0\u1eddng th\u1eb3ng vu\u00f4ng g\u00f3c<\/li>\n
    • B\u00e0i 2: \u0110\u01b0\u1eddng th\u1eb3ng vu\u00f4ng g\u00f3c v\u1edbi m\u1eb7t ph\u1eb3ng<\/li>\n<\/ul>\n

      M\u1ed9t s\u1ed1 c\u00e2u h\u1ecfi minh h\u1ecda n\u1ed5i b\u1eadt\u00a0<\/strong><\/h3>\n

      PH\u1ea6N 1: C\u00e2u tr\u1eafc nghi\u1ec7m nhi\u1ec1u ph\u01b0\u01a1ng \u00e1n l\u1ef1a ch\u1ecdn \u2013 12 c\u00e2u<\/strong><\/p>\n

      C\u00e2u 1.<\/strong> Cho $x, y > 0$ v\u00e0 $\\alpha, \\beta \\in \\mathbb{R}$. Kh\u1eb3ng \u0111\u1ecbnh n\u00e0o sau \u0111\u00e2y sai?<\/p>\n

      A. $(xy)^\\alpha = x^\\alpha \\cdot y^\\alpha$ \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0B. $x^\\alpha + y^\\alpha = (x+y)^\\alpha$\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0C. $(x^\\alpha)^\\beta = x^{\\alpha\\beta}$\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0D. $x^\\alpha \\cdot x^\\beta = x^{\\alpha+\\beta}$<\/p>\n

      C\u00e2u 2.<\/strong> V\u1edbi $a$ l\u00e0 s\u1ed1 th\u1ef1c d\u01b0\u01a1ng t\u00f9y \u00fd, $\\log_5 a^3$ b\u1eb1ng<\/p>\n

      A. $\\frac{1}{3} \\log_5 a$ \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0B. $\\frac{1}{3} + \\log_5 a$ \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 C. $3 + \\log_5 a$ \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 D. 3$\\log_5 a$<\/p>\n

      PH\u1ea6N 2: C\u00e2u tr\u1eafc nghi\u1ec7m \u0111\u00fang sai \u2013 2 c\u00e2u<\/strong><\/p>\n

      C\u00e2u 1.<\/strong> Cho h\u00e0m s\u1ed1 $y = 4^x$.<\/p>\n

      a) H\u00e0m s\u1ed1 \u0111\u00e3 cho c\u00f3 t\u1eadp x\u00e1c \u0111\u1ecbnh $D = \\mathbb{R}$.
      \nb) H\u00e0m s\u1ed1 \u0111\u00e3 cho c\u00f3 t\u1eadp gi\u00e1 tr\u1ecb l\u00e0 $T = (-\\infty; +\\infty)$.
      \nc) H\u00e0m s\u1ed1 \u0111\u00e3 cho ngh\u1ecbch bi\u1ebfn tr\u00ean kho\u1ea3ng $(-\\infty; +\\infty)$.
      \nd) \u0110\u1ed3 th\u1ecb h\u00e0m s\u1ed1 \u0111\u00e3 cho c\u00f3 h\u00ecnh sau b\u00ean:<\/p>\n

      \"\"<\/p>\n

      C\u00e2u 2.<\/strong> Cho h\u00ecnh ch\u00f3p $S.ABCD$ c\u00f3 \u0111\u00e1y $ABCD$ l\u00e0 h\u00ecnh thoi t\u00e2m $O$. Bi\u1ebft $SA = SC$ v\u00e0 $SB = SD$.<\/p>\n

      a) $SO \\perp (ABCD)$
      \nb) $CD \\perp (SBD)$
      \nc) $OA$ l\u00e0 h\u00ecnh chi\u1ebfu c\u1ee7a $SA$ tr\u00ean m\u1eb7t ph\u1eb3ng $(SBD)$.
      \nd) Tam gi\u00e1c $CAB$ l\u00e0 h\u00ecnh chi\u1ebfu c\u1ee7a tam gi\u00e1c $SAB$ tr\u00ean m\u1eb7t ph\u1eb3ng $(ABCD)$.<\/p>\n

      PH\u1ea6N 3: C\u00e2u tr\u1eafc nghi\u1ec7m tr\u1ea3 l\u1eddi ng\u1eafn \u2013 4 c\u00e2u\u00a0<\/strong><\/p>\n

      C\u00e2u 1.<\/strong> Cho h\u00e0m s\u1ed1 $f(x) = \\frac{1}{2} \\log_2 \\left( \\frac{2x}{1 \u2013 x} \\right)$.<\/p>\n

      T\u00ednh gi\u00e1 tr\u1ecb bi\u1ec3u th\u1ee9c $M = f \\left( \\frac{1}{2025} \\right) + f \\left( \\frac{2}{2025} \\right) + \\dots + f \\left( \\frac{2023}{2025} \\right) + f \\left( \\frac{2024}{2025} \\right)$.<\/p>\n

      Tr\u1ea3 l\u1eddi:\u00a0 \u2026\u2026\u2026\u2026\u2026\u2026.<\/p>\n

      C\u00e2u 2.<\/strong> COVID19 l\u00e0 m\u1ed9t lo\u1ea1i b\u1ec7nh vi\u00eam \u0111\u01b0\u1eddng h\u00f4 h\u1ea5p c\u1ea5p do ch\u1ee7ng m\u1edbi c\u1ee7a virus corona (nCoV) b\u1eaft ngu\u1ed3n t\u1eeb Trung Qu\u1ed1c (\u0111\u1ea7u th\u00e1ng 12\/2019) g\u00e2y ra v\u1edbi t\u1ed1c \u0111\u1ed9 truy\u1ec1n b\u1ec7nh r\u1ea5t nhanh (t\u00ednh \u0111\u1ebfn 7\/4\/2020 \u0111\u00e3 c\u00f3 1 360 039 ng\u01b0\u1eddi nhi\u1ec5m b\u1ec7nh). Gi\u1ea3 s\u1eed ban \u0111\u1ea7u c\u00f3 1 ng\u01b0\u1eddi b\u1ecb nhi\u1ec5m b\u1ec7nh v\u00e0 c\u1ee9 sau 1 ng\u00e0y s\u1ebd l\u00e2y sang 4 ng\u01b0\u1eddi kh\u00e1c. T\u1ea5t c\u1ea3 nh\u1eefng ng\u01b0\u1eddi nhi\u1ec5m b\u1ec7nh l\u1ea1i ti\u1ebfp t\u1ee5c l\u00e2y sang nh\u1eefng ng\u01b0\u1eddi kh\u00e1c v\u1edbi t\u1ed1c \u0111\u1ed9 nh\u01b0 tr\u00ean (1 ng\u01b0\u1eddi l\u00e2y 4 ng\u01b0\u1eddi). H\u1ecfi sau 7 ng\u00e0y s\u1ebd c\u00f3 t\u1ed5ng c\u1ed9ng bao nhi\u00eau ng\u01b0\u1eddi nhi\u1ec5m b\u1ec7nh? (Bi\u1ebft r\u1eb1ng nh\u1eefng ng\u01b0\u1eddi nhi\u1ec5m b\u1ec7nh kh\u00f4ng ph\u00e1t hi\u1ec7n b\u1ea3n th\u00e2n b\u1ecb b\u1ec7nh v\u00e0 kh\u00f4ng ph\u00f2ng tr\u00e1nh c\u00e1ch li, do trong th\u1eddi gian \u1ee7 b\u1ec7nh v\u1eabn l\u00e2y b\u1ec7nh sang ng\u01b0\u1eddi kh\u00e1c).<\/p>\n

      Tr\u1ea3 l\u1eddi:\u00a0 \u2026\u2026\u2026\u2026\u2026\u2026.<\/p>\n

      PH\u1ea6N 4: C\u00e2u t\u1ef1 lu\u1eadn \u2013 3 c\u00e2u\u00a0<\/strong><\/p>\n

      C\u00e2u 1.<\/strong> T\u00ecm t\u1ea5t c\u1ea3 c\u00e1c gi\u00e1 tr\u1ecb c\u1ee7a tham s\u1ed1 a \u0111\u1ec3 h\u00e0m s\u1ed1 $y = \\log_{a^2 \u2013 2a + 1} x$ ngh\u1ecbch bi\u1ebfn tr\u00ean kho\u1ea3ng (0; +\u221e) .<\/p>\n

      C\u00e2u 2.<\/strong> Gi\u1ea3i ph\u01b0\u01a1ng tr\u00ecnh v\u00e0 b\u1ea5t ph\u01b0\u01a1ng tr\u00ecnh sau:<\/p>\n

      a) (\u221a2 + 1)^x+1 – (\u221a2 – 1)^x\/x-1 = 0 .\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0b) ln(x+3) \u2265 ln(2x-8) .<\/p>\n

      T\u1ea3i 10 \u0111\u1ec1 \u00f4n t\u1eadp gi\u1eefa h\u1ecdc k\u00ec 2 To\u00e1n 11 n\u0103m 2024 \u2013 2025 c\u00f3 l\u1eddi gi\u1ea3i chi ti\u1ebft<\/span><\/strong><\/h2>\n

      10 \u0111\u1ec1 \u00f4n t\u1eadp gi\u1eefa h\u1ecdc k\u00ec 2 To\u00e1n 11 n\u0103m 2024 \u2013 2025 c\u00f3 l\u1eddi gi\u1ea3i chi ti\u1ebft<\/strong> l\u00e0 t\u00e0i li\u1ec7u kh\u00f4ng th\u1ec3 b\u1ecf qua \u0111\u1ec3 h\u1ecdc sinh l\u1edbp 11 h\u1ec7 th\u1ed1ng l\u1ea1i ki\u1ebfn th\u1ee9c v\u00e0 n\u00e2ng cao k\u1ef9 n\u0103ng l\u00e0m b\u00e0i. B\u1ed9 \u0111\u1ec1 b\u00e1m s\u00e1t ch\u01b0\u01a1ng tr\u00ecnh, gi\u00fap b\u1ea1n r\u00e8n luy\u1ec7n t\u01b0 duy to\u00e1n h\u1ecdc v\u00e0 l\u00e0m quen v\u1edbi c\u00e1c d\u1ea1ng b\u00e0i th\u01b0\u1eddng g\u1eb7p trong \u0111\u1ec1 ki\u1ec3m tra. H\u00e3y t\u1eadn d\u1ee5ng ngay ngu\u1ed3n t\u00e0i li\u1ec7u ch\u1ea5t l\u01b0\u1ee3ng n\u00e0y \u0111\u1ec3 chu\u1ea9n b\u1ecb th\u1eadt t\u1ed1t cho k\u1ef3 thi! T\u1ea3i \u0111\u1ea7y \u0111\u1ee7 b\u1ed9 \u0111\u1ec1 to\u00e1n gi\u1eefa k\u00ec 2 l\u1edbp 11 qua \u0111\u01b0\u1eddng link d\u01b0\u1edbi \u0111\u00e2y!<\/p>\n