{"id":426,"date":"2026-05-05T04:45:07","date_gmt":"2026-05-05T04:45:07","guid":{"rendered":"https:\/\/79.buffdemo.com\/index.php\/2026\/05\/05\/10-de-giua-hoc-ky-2-toan-11-kntt-nam-2024-2025-co-loi-giai-chi-tiet\/"},"modified":"2026-05-05T07:17:53","modified_gmt":"2026-05-05T07:17:53","slug":"10-de-giua-hoc-ky-2-toan-11-kntt-nam-2024-2025-co-loi-giai-chi-tiet","status":"publish","type":"post","link":"https:\/\/79.buffdemo.com\/index.php\/2026\/05\/05\/10-de-giua-hoc-ky-2-toan-11-kntt-nam-2024-2025-co-loi-giai-chi-tiet\/","title":{"rendered":"10 \u0111\u1ec1 gi\u1eefa h\u1ecdc k\u1ef3 2 To\u00e1n 11 KNTT n\u0103m 2024 \u2013 2025 c\u00f3 l\u1eddi gi\u1ea3i chi ti\u1ebft"},"content":{"rendered":"

10 \u0111\u1ec1 gi\u1eefa h\u1ecdc k\u1ef3 2 To\u00e1n 11 KNTT n\u0103m 2024 \u2013 2025 c\u00f3 l\u1eddi gi\u1ea3i chi ti\u1ebft<\/strong> <\/span>l\u00e0 t\u00e0i li\u1ec7u quan tr\u1ecdng gi\u00fap h\u1ecdc sinh l\u1edbp 11 \u00f4n luy\u1ec7n hi\u1ec7u qu\u1ea3 v\u00e0 \u0111\u00e1nh gi\u00e1 n\u0103ng l\u1ef1c tr\u01b0\u1edbc k\u1ef3 ki\u1ec3m tra gi\u1eefa h\u1ecdc k\u1ef3. B\u1ed9 \u0111\u1ec1 bao g\u1ed3m c\u00e1c d\u1ea1ng b\u00e0i t\u1eadp tr\u1ecdng t\u00e2m nh\u01b0 h\u00e0m s\u1ed1 m\u0169 \u2013 logarit, ph\u01b0\u01a1ng tr\u00ecnh \u2013 b\u1ea5t ph\u01b0\u01a1ng tr\u00ecnh m\u0169 \u2013 loga, quan h\u1ec7 vu\u00f4ng g\u00f3c trong kh\u00f4ng gian, g\u00f3c gi\u1eefa \u0111\u01b0\u1eddng th\u1eb3ng v\u00e0 m\u1eb7t ph\u1eb3ng, hai m\u1eb7t ph\u1eb3ng,\u2026 H\u00e3y c\u00f9ng T\u00e0i Li\u1ec7u \u00d4n Thi<\/strong><\/a> kh\u00e1m ph\u00e1 ngay t\u00e0i li\u1ec7u n\u00e0y \u0111\u1ec3 chu\u1ea9n b\u1ecb th\u1eadt t\u1ed1t cho k\u1ef3 ki\u1ec3m tra gi\u1eefa k\u1ef3 2 nh\u00e9!<\/p>\n

N\u1ed9i dung 10 \u0111\u1ec1 gi\u1eefa h\u1ecdc k\u1ef3 2 To\u00e1n 11 KNTT n\u0103m 2024 \u2013 2025 c\u00f3 l\u1eddi gi\u1ea3i chi ti\u1ebft<\/span><\/strong><\/h2>\n

10 \u0111\u1ec1 gi\u1eefa h\u1ecdc k\u1ef3 2 To\u00e1n 11 KNTT n\u0103m 2024 \u2013 2025 c\u00f3 l\u1eddi gi\u1ea3i chi ti\u1ebft<\/strong> l\u00e0 t\u00e0i li\u1ec7u \u00f4n t\u1eadp quan tr\u1ecdng gi\u00fap h\u1ecdc sinh l\u1edbp 11 h\u1ec7 th\u1ed1ng l\u1ea1i to\u00e0n b\u1ed9 ki\u1ebfn th\u1ee9c tr\u1ecdng t\u00e2m v\u00e0 t\u1ef1 \u0111\u00e1nh gi\u00e1 n\u0103ng l\u1ef1c tr\u01b0\u1edbc k\u1ef3 thi gi\u1eefa h\u1ecdc k\u1ef3 2. B\u1ed9 \u0111\u1ec1 to\u00e1n gi\u1eefa k\u00ec 2 l\u1edbp 11 \u0111\u01b0\u1ee3c x\u00e2y d\u1ef1ng theo h\u01b0\u1edbng ph\u00e1t tri\u1ec3n n\u0103ng l\u1ef1c, t\u1ed5ng h\u1ee3p \u0111\u1ea7y \u0111\u1ee7 c\u00e1c d\u1ea1ng b\u00e0i t\u1eeb nh\u1eadn bi\u1ebft \u0111\u1ebfn v\u1eadn d\u1ee5ng cao, b\u00e1m s\u00e1t ch\u01b0\u01a1ng tr\u00ecnh s\u00e1ch gi\u00e1o khoa v\u00e0 c\u1ea5u tr\u00fac \u0111\u1ec1 thi ch\u00ednh th\u1ee9c.<\/p>\n

Ma tr\u1eadn t\u1ed5ng qu\u00e1t<\/strong><\/h3>\n

CH\u01af\u01a0NG \/ CH\u1ee6 \u0110\u1ec0 1:\u00a0<\/strong>QUAN H\u1ec6 VU\u00d4NG G\u00d3C<\/strong><\/p>\n

10 c\u00e2u \u2013 3,92 \u0111i\u1ec3m<\/p>\n

    \n
  • \u0110\u01af\u1edcNG TH\u1eb2NG VU\u00d4NG G\u00d3C M\u1eb6T PH\u1eb2NG<\/li>\n
  • HAI \u0110\u01af\u1edcNG TH\u1eb2NG VU\u00d4NG G\u00d3C<\/li>\n
  • HAI M\u1eb6T PH\u1eb2NG VU\u00d4NG G\u00d3C<\/li>\n<\/ul>\n

    4 c\u00e2u \u2013 1,12 \u0111i\u1ec3m<\/p>\n

      \n
    • PH\u00c9P CHI\u1ebeU VU\u00d4NG G\u00d3C. G\u00d3C GI\u1eeeA \u0110\u01af\u1edcNG TH\u1eb2NG V\u00c0 M\u1eb6T PH\u1eb2NG<\/li>\n<\/ul>\n

      CH\u01af\u01a0NG \/ CH\u1ee6 \u0110\u1ec0 2: H\u00c0M S\u1ed0 M\u0168 LOGARIT<\/strong><\/p>\n

      7 c\u00e2u \u2013 2,8 \u0111i\u1ec3m<\/p>\n

        \n
      • H\u00c0M S\u1ed0 L\u0168Y TH\u1eeaA<\/li>\n
      • H\u00c0M S\u1ed0 M\u0168 \u2013 H\u00c0M S\u1ed0 L\u00d4GARIT<\/li>\n<\/ul>\n

        7 c\u00e2u \u2013 2,24 \u0111i\u1ec3m<\/p>\n

          \n
        • PH\u01af\u01a0NG TR\u00ccNH , B\u1ea4T PH\u01af\u01a0NG TR\u00ccNH LOGA<\/li>\n
        • PH\u01af\u01a0NG TR\u00ccNH , B\u1ea4T PH\u01af\u01a0NG TR\u00ccNH M\u0168<\/li>\n<\/ul>\n

          N\u1ed9i dung c\u00e2u h\u1ecfi tham kh\u1ea3o\u00a0<\/strong><\/h3>\n

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

          C\u00e2u 1.<\/strong> Nghi\u1ec7m c\u1ee7a ph\u01b0\u01a1ng tr\u00ecnh $2^x = 4$ l\u00e0<\/p>\n

          A. $x = -3$. \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 B. $x = 2$.\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 C. $x = -2$.\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 D. $x = 3$.<\/p>\n

          C\u00e2u 2.<\/strong> Trong kh\u00f4ng gian cho \u0111\u01b0\u1eddng th\u1eb3ng $d$, \u0111\u01b0\u1eddng th\u1eb3ng $a$ v\u00e0 m\u1eb7t ph\u1eb3ng $(\\alpha)$. Ch\u1ecdn kh\u1eb3ng \u0111\u1ecbnh \u0111\u00fang.<\/p>\n

          A. G\u00f3c gi\u1eefa \u0111\u01b0\u1eddng th\u1eb3ng $d$ v\u00e0 m\u1eb7t ph\u1eb3ng $(\\alpha)$ l\u00e0 g\u00f3c gi\u1eefa $d$ v\u00e0 h\u00ecnh chi\u1ebfu vu\u00f4ng g\u00f3c c\u1ee7a $d$ tr\u00ean $(\\alpha)$
          \nB. G\u00f3c gi\u1eefa \u0111\u01b0\u1eddng th\u1eb3ng $d$ v\u00e0 m\u1eb7t ph\u1eb3ng $(\\alpha)$ b\u1eb1ng $0^\\circ$ khi v\u00e0 ch\u1ec9 khi $d$ song song v\u1edbi m\u1eb7t ph\u1eb3ng $(\\alpha)$
          \nC. N\u1ebfu $d$ v\u00e0 $a$ c\u00f9ng t\u1ea1o v\u1edbi m\u1eb7t ph\u1eb3ng $(\\alpha)$ m\u1ed9t g\u00f3c $0^\\circ$ th\u00ec $d$ v\u00e0 $a$ \u0111\u1ed3ng ph\u1eb3ng
          \nD. G\u00f3c gi\u1eefa \u0111\u01b0\u1eddng th\u1eb3ng $d$ v\u00e0 m\u1eb7t ph\u1eb3ng $(\\alpha)$ lu\u00f4n l\u00e0 g\u00f3c nh\u1ecdn<\/p>\n

          C\u00e2u 3.<\/strong> T\u1eadp x\u00e1c \u0111\u1ecbnh c\u1ee7a h\u00e0m s\u1ed1 $y = 5^x$ l\u00e0<\/p>\n

          A. $(0; +\\infty)$. \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0B. $\\mathbb{R}$. \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0C. $\\mathbb{R} \\setminus \\{0\\}$. \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0D. $[0; +\\infty)$.<\/p>\n

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

          C\u00e2u 1.<\/strong> Cho h\u00ecnh ch\u00f3p $S.ABCD$ c\u00f3 $SA \\perp (ABCD)$. $ABCD$ l\u00e0 h\u00ecnh vu\u00f4ng v\u00e0 $SA = 2a, AB = a$.<\/p>\n

          a) G\u00f3c gi\u1eefa \u0111\u01b0\u1eddng th\u1eb3ng $SA$ v\u00e0 $(ABCD)$ b\u1eb1ng $90^\\circ$.
          \nb) G\u1ecdi $M$ l\u00e0 \u0111i\u1ec3m thu\u1ed9c $DC$ sao cho $\\frac{\\overline{DM}}{\\overline{DC}} = \\frac{1}{4}$, Tan g\u00f3c gi\u1eefa $SM$ v\u00e0 $(SAB)$ b\u1eb1ng $\\frac{\\sqrt{65}}{4}$.
          \nc) Tan g\u00f3c gi\u1eefa \u0111\u01b0\u1eddng th\u1eb3ng $SC$ v\u00e0 $(ABCD)$ b\u1eb1ng $\\sqrt{2}$.
          \nd) Tan g\u00f3c gi\u1eefa \u0111\u01b0\u1eddng th\u1eb3ng $BC$ v\u00e0 $(SAB)$ b\u1eb1ng $1$.<\/p>\n

          C\u00e2u 2.<\/strong> Cho ph\u01b0\u01a1ng tr\u00ecnh $4^x \u2013 m.2^{x+1} + 2m \u2013 1 = 0$ $(*)$ v\u1edbi $m$ l\u00e0 tham s\u1ed1.<\/p>\n

          a) Khi $m = 0$ ph\u01b0\u01a1ng tr\u00ecnh c\u00f3 hai nghi\u1ec7m ph\u00e2n bi\u1ec7t.
          \nb) Khi $m = 1$ ph\u01b0\u01a1ng tr\u00ecnh c\u00f3 t\u1eadp nghi\u1ec7m l\u00e0 $S = \\{0\\}$.
          \nc) S\u1ed1 gi\u00e1 tr\u1ecb nguy\u00ean d\u01b0\u01a1ng c\u1ee7a $m$ \u0111\u1ec3 ph\u01b0\u01a1ng tr\u00ecnh c\u00f3 nghi\u1ec7m duy nh\u1ea5t l\u00e0 1.
          \nd) G\u1ecdi $m_0$ l\u00e0 gi\u00e1 tr\u1ecb th\u1ef1c \u0111\u1ec3 ph\u01b0\u01a1ng tr\u00ecnh c\u00f3 2 nghi\u1ec7m $x_1, x_2$ sao cho $x_1 + x_2 = 6$. Khi \u0111\u00f3, $m_0 > 32$.<\/p>\n

          PH\u1ea6N III: C\u00e2u tr\u1eafc nghi\u1ec7m tr\u1ea3 l\u1eddi ng\u1eafn<\/strong><\/p>\n

          C\u00e2u 1.<\/strong> Cho h\u00ecnh ch\u00f3p $S.ABC$ c\u00f3 \u0111\u00e1y l\u00e0 tam gi\u00e1c \u0111\u1ec1u c\u1ea1nh $a$, $SB \\perp (ABC)$ v\u00e0 $SB = 4a$. T\u00ednh g\u00f3c gi\u1eefa \u0111\u01b0\u1eddng th\u1eb3ng $SC$ v\u00e0 m\u1eb7t ph\u1eb3ng $(SAB)$?<\/p>\n

          C\u00e2u 2.<\/strong> M\u1ed9t ng\u01b0\u1eddi g\u1eedi ti\u1ebft ki\u1ec7m 100 tri\u1ec7u \u0111\u1ed3ng v\u00e0o ng\u00e2n h\u00e0ng theo th\u1ec3 th\u1ee9c l\u00e3i k\u00e9p k\u00ec h\u1ea1n 6 th\u00e1ng v\u1edbi l\u00e3i su\u1ea5t 8% m\u1ed9t n\u0103m. Gi\u1ea3 s\u1eed l\u00e3i su\u1ea5t kh\u00f4ng thay \u0111\u1ed5i. H\u1ecfi sau bao nhi\u00eau th\u00e1ng ng\u01b0\u1eddi \u0111\u00f3 nh\u1eadn \u0111\u01b0\u1ee3c \u00edt nh\u1ea5t 120 tri\u1ec7u \u0111\u1ed3ng?<\/p>\n

          PH\u1ea6N IV: C\u00e2u h\u1ecfi t\u1ef1 lu\u1eadn<\/strong><\/p>\n

          C\u00e2u 1.<\/strong> Theo th\u1ed1ng k\u00ea, trong n\u0103m 2021 di\u1ec7n t\u00edch nu\u00f4i ngao b\u00e3i tri\u1ec1u c\u1ee7a t\u1ec9nh Th\u00e1i B\u00ecnh l\u00e0 840 $(ha)$. Bi\u1ebft r\u1eb1ng di\u1ec7n t\u00edch nu\u00f4i ngao b\u00e3i tri\u1ec1u m\u1ed7i n\u0103m t\u0103ng 6% so v\u1edbi di\u1ec7n t\u00edch c\u1ee7a n\u0103m li\u1ec1n tr\u01b0\u1edbc. K\u1ec3 t\u1eeb sau n\u0103m 2021, n\u0103m n\u00e0o l\u00e0 n\u0103m \u0111\u1ea7u ti\u00ean t\u1ec9nh Th\u00e1i B\u00ecnh c\u00f3 di\u1ec7n t\u00edch nu\u00f4i ngao b\u00e3i tri\u1ec1u \u0111\u1ea1t tr\u00ean 1400 $(ha)$?<\/p>\n

          C\u00e2u 2.<\/strong> M\u1ed9t ng\u01b0\u1eddi g\u1eedi ti\u1ebft ki\u1ec7m 20.000.000 \u0111\u1ed3ng lo\u1ea1i k\u1ef3 h\u1ea1n m\u1ed9t n\u0103m v\u00e0o ng\u00e2n h\u00e0ng v\u1edbi l\u00e3i su\u1ea5t 6,5% m\u1ed9t n\u0103m. T\u00ednh s\u1ed1 ti\u1ec1n sau 5 n\u0103m 2 th\u00e1ng ng\u01b0\u1eddi \u0111\u00f3 nh\u1eadn \u0111\u01b0\u1ee3c bao nhi\u00eau ti\u1ec1n c\u1ea3 g\u1ed1c l\u1eabn l\u00e3i. Bi\u1ebft n\u1ebfu r\u00fat tr\u01b0\u1edbc k\u1ef3 h\u1ea1n th\u00ec ng\u00e2n h\u00e0ng tr\u1ea3 theo l\u00e3i su\u1ea5t kh\u00f4ng k\u1ef3 h\u1ea1n l\u00e0 0,01% m\u1ed9t ng\u00e0y (1 th\u00e1ng t\u00ednh 30 ng\u00e0y).<\/p>\n

          T\u1ea3i 10 \u0111\u1ec1 gi\u1eefa h\u1ecdc k\u1ef3 2 To\u00e1n 11 KNTT n\u0103m 2024 \u2013 2025 c\u00f3 l\u1eddi gi\u1ea3i chi ti\u1ebft<\/span><\/strong><\/h2>\n

          10 \u0111\u1ec1 gi\u1eefa h\u1ecdc k\u1ef3 2 To\u00e1n 11 KNTT n\u0103m 2024 \u2013 2025 c\u00f3 l\u1eddi gi\u1ea3i chi ti\u1ebft <\/strong>s\u1ebd gi\u00fap h\u1ecdc sinh c\u1ee7ng c\u1ed1 ki\u1ebfn th\u1ee9c, r\u00e8n luy\u1ec7n k\u1ef9 n\u0103ng gi\u1ea3i b\u00e0i t\u1eadp v\u00e0 l\u00e0m quen v\u1edbi c\u1ea5u tr\u00fac \u0111\u1ec1 thi ch\u00ednh th\u1ee9c. V\u1edbi h\u1ec7 th\u1ed1ng b\u00e0i t\u1eadp \u0111a d\u1ea1ng, b\u00e1m s\u00e1t ch\u01b0\u01a1ng tr\u00ecnh h\u1ecdc, t\u00e0i li\u1ec7u n\u00e0y l\u00e0 tr\u1ee3 th\u1ee7 \u0111\u1eafc l\u1ef1c gi\u00fap c\u00e1c em t\u1ef1 tin \u0111\u1ea1t \u0111i\u1ec3m cao trong k\u1ef3 ki\u1ec3m tra s\u1eafp t\u1edbi. \u0110\u1eebng b\u1ecf l\u1ee1 c\u01a1 h\u1ed9i \u00f4n t\u1eadp hi\u1ec7u qu\u1ea3 \u2013 c\u00e1c b\u1ea1n c\u00f3 th\u1ec3 click v\u00e0o \u0111\u01b0\u1eddng link d\u01b0\u1edbi \u0111\u00e2y \u0111\u1ec3 t\u1ea3i to\u00e0n b\u1ed9 t\u00e0i li\u1ec7u!<\/p>\n