{"id":378,"date":"2026-05-05T04:42:37","date_gmt":"2026-05-05T04:42:37","guid":{"rendered":"https:\/\/79.buffdemo.com\/index.php\/2026\/05\/05\/bo-de-kiem-tra-giua-ky-2-toan-10-ket-noi-tri-thuc-nam-hoc-2024-2025\/"},"modified":"2026-05-05T07:15:19","modified_gmt":"2026-05-05T07:15:19","slug":"bo-de-kiem-tra-giua-ky-2-toan-10-ket-noi-tri-thuc-nam-hoc-2024-2025","status":"publish","type":"post","link":"https:\/\/79.buffdemo.com\/index.php\/2026\/05\/05\/bo-de-kiem-tra-giua-ky-2-toan-10-ket-noi-tri-thuc-nam-hoc-2024-2025\/","title":{"rendered":"B\u1ed9 \u0111\u1ec1 ki\u1ec3m tra gi\u1eefa k\u1ef3 2 To\u00e1n 10 K\u1ebft n\u1ed1i tri th\u1ee9c n\u0103m h\u1ecdc 2024 \u2013 2025"},"content":{"rendered":"

B\u1ed9 \u0111\u1ec1 ki\u1ec3m tra gi\u1eefa k\u1ef3 2 To\u00e1n 10 K\u1ebft n\u1ed1i tri th\u1ee9c n\u0103m h\u1ecdc 2024 \u2013 2025<\/strong> l\u00e0 t\u00e0i li\u1ec7u quan tr\u1ecdng gi\u00fap h\u1ecdc sinh l\u1edbp 10 \u00f4n t\u1eadp v\u00e0 \u0111\u00e1nh gi\u00e1 n\u0103ng l\u1ef1c tr\u01b0\u1edbc k\u1ef3 ki\u1ec3m tra ch\u00ednh th\u1ee9c. \u0110\u1ec1 thi bao g\u1ed3m c\u00e1c d\u1ea1ng b\u00e0i t\u1eadp tr\u1ecdng t\u00e2m t\u1eeb h\u00e0m s\u1ed1 b\u1eadc nh\u1ea5t, b\u1eadc hai, ph\u01b0\u01a1ng tr\u00ecnh, b\u1ea5t ph\u01b0\u01a1ng tr\u00ecnh \u0111\u1ebfn h\u00ecnh h\u1ecdc gi\u1ea3i t\u00edch, b\u00e1m s\u00e1t ch\u01b0\u01a1ng tr\u00ecnh s\u00e1ch gi\u00e1o khoa v\u00e0 c\u1ea5u tr\u00fac \u0111\u1ec1 thi gi\u1eefa h\u1ecdc k\u1ef3 2. H\u00e3y c\u00f9ng T\u00e0i Li\u1ec7u \u00d4n Thi<\/a><\/strong> t\u00ecm hi\u1ec3u chi ti\u1ebft v\u1ec1 t\u00e0i li\u1ec7u n\u00e0y \u0111\u1ec3 c\u00f3 s\u1ef1 chu\u1ea9n b\u1ecb t\u1ed1t nh\u1ea5t cho k\u1ef3 ki\u1ec3m tra s\u1eafp t\u1edbi nh\u00e9!<\/p>\n

N\u1ed9i dung B\u1ed9 \u0111\u1ec1 ki\u1ec3m tra gi\u1eefa k\u1ef3 2 To\u00e1n 10 K\u1ebft n\u1ed1i tri th\u1ee9c n\u0103m h\u1ecdc 2024 \u2013 2025<\/strong><\/h2>\n

B\u1ed9 \u0111\u1ec1 ki\u1ec3m tra gi\u1eefa k\u1ef3 2 To\u00e1n 10 K\u1ebft n\u1ed1i tri th\u1ee9c n\u0103m h\u1ecdc 2024 \u2013 2025<\/strong> l\u00e0 t\u00e0i li\u1ec7u \u00f4n t\u1eadp quan tr\u1ecdng gi\u00fap h\u1ecdc sinh l\u1edbp 10 h\u1ec7 th\u1ed1ng l\u1ea1i ki\u1ebfn th\u1ee9c v\u00e0 \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 thi gi\u1eefa k\u00ec 2 to\u00e1n 10 t\u1ed5ng h\u1ee3p \u0111\u1ea7y \u0111\u1ee7 c\u00e1c d\u1ea1ng b\u00e0i t\u1eeb c\u01a1 b\u1ea3n \u0111\u1ebfn n\u00e2ng 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. M\u1ed7i \u0111\u1ec1 thi g\u1ed3m 4 ph\u1ea7n:<\/p>\n

PH\u1ea6N 1 \u2013 C\u00c2U H\u1eceI TR\u1eaeC NGHI\u1ec6M (3 \u0111i\u1ec3m)<\/strong><\/p>\n

C\u00f3 12 c\u00e2u tr\u1eafc nghi\u1ec7m theo nhi\u1ec1u m\u1ee9c \u0111\u1ed9 t\u1eeb nh\u1eadn bi\u1ebft, th\u00f4ng hi\u1ec3u \u0111\u1ebfn v\u1eadn d\u1ee5ng bao g\u1ed3m c\u00e1c ki\u1ebfn th\u1ee9c v\u1ec1 h\u00e0m s\u1ed1, ph\u01b0\u01a1ng tr\u00ecnh, b\u1ea5t ph\u01b0\u01a1ng tr\u00ecnh, h\u00ecnh h\u1ecdc gi\u1ea3i t\u00edch, h\u1ec7 ph\u01b0\u01a1ng tr\u00ecnh v\u00e0 \u1ee9ng d\u1ee5ng th\u1ef1c t\u1ebf. M\u1ed9t s\u1ed1 c\u00e2u h\u1ecfi n\u1ed5i b\u1eadt:<\/p>\n

C\u00e2u 1. <\/strong>T\u00ecm kh\u1eb3ng \u0111\u1ecbnh \u0111\u00fang trong c\u00e1c kh\u1eb3ng \u0111\u1ecbnh sau?<\/p>\n

A. f(x) = 3x\u00b2 + 2x – 5 \u00a0l\u00e0 tam th\u1ee9c b\u1eadc hai.
\nB. f(x) = 2x – 4 \u00a0l\u00e0 tam th\u1ee9c b\u1eadc hai.
\nC. f(x) = 3x\u00b3 + 2x – 1 l\u00e0 tam th\u1ee9c b\u1eadc hai.
\nD. f(x) = x\u2074 – x\u00b2 + 1 \u00a0l\u00e0 tam th\u1ee9c b\u1eadc hai.<\/p>\n

C\u00e2u 2. <\/strong>Cho h\u00e0m s\u1ed1 y = f(x) = 2x\u00b2 – 1 . T\u00ednh f(2) <\/p>\n

A. f(2) = 2
B. f(2) = 3
C. f(2) = 7
D. f(2) = 5 <\/p>\n

C\u00e2u 3.<\/strong> H\u00e0m s\u1ed1 n\u00e0o d\u01b0\u1edbi \u0111\u00e2y l\u00e0 h\u00e0m s\u1ed1 b\u1eadc nh\u1ea5t?<\/p>\n

A. y = 2x + 1\/x
B. y = 2
C. y = x+1\/x-2
D. y = 2x + \u221a2 <\/p>\n

PH\u1ea6N 2 \u2013 C\u00c2U H\u1eceI \u0110\u00daNG\/SAI (2 \u0111i\u1ec3m)<\/strong><\/p>\n

C\u00e2u h\u1ecfi \u0111\u00fang\/sai ki\u1ec3m tra kh\u1ea3 n\u0103ng nh\u1eadn di\u1ec7n ki\u1ebfn th\u1ee9c v\u1ec1 h\u00e0m s\u1ed1, ph\u01b0\u01a1ng tr\u00ecnh, b\u1ea5t ph\u01b0\u01a1ng tr\u00ecnh, h\u00ecnh h\u1ecdc gi\u1ea3i t\u00edch, vect\u01a1, v\u00e0 \u1ee9ng d\u1ee5ng th\u1ef1c t\u1ebf. M\u1ed9t s\u1ed1 c\u00e2u h\u1ecfi trong b\u1ed9 \u0111\u1ec1:<\/p>\n

C\u00e2u 13. <\/strong>Trong m\u1eb7t ph\u1eb3ng v\u1edbi h\u1ec7 t\u1ecda \u0111\u1ed9 Oxy , cho tam gi\u00e1c ABC c\u00f3 M(2;1) l\u00e0 trung \u0111i\u1ec3m c\u1ea1nh AC , \u0111i\u1ec3m H(0;-3) l\u00e0 ch\u00e2n \u0111\u01b0\u1eddng cao k\u1ebb t\u1eeb A . \u0110i\u1ec3m E(23;-2) thu\u1ed9c \u0111\u01b0\u1eddng th\u1eb3ng ch\u1ee9a trung tuy\u1ebfn k\u1ebb t\u1eeb C . Bi\u1ebft \u0111i\u1ec3m A thu\u1ed9c \u0111\u01b0\u1eddng th\u1eb3ng d: 2x + 3y – 5 = 0 v\u00e0 \u0111i\u1ec3m C c\u00f3 ho\u00e0nh \u0111\u1ed9 d\u01b0\u01a1ng.<\/p>\n\n\n\n\n\n\n\n
M\u1ec7nh \u0111\u1ec1\u00a0<\/strong><\/td>\n\u0110\u00fang<\/strong><\/td>\nSai<\/strong><\/td>\n<\/tr>\n
(a) Ph\u01b0\u01a1ng tr\u00ecnh \u0111\u01b0\u1eddng th\u1eb3ng BC l\u00e0 x + 3y – 9 = 0 .<\/td>\n<\/td>\n<\/td>\n<\/tr>\n
(b) \u0110\u01b0\u1eddng th\u1eb3ng CE c\u00f3 ph\u01b0\u01a1ng tr\u00ecnh l\u00e0 x + 17y + 11 = 0 .<\/td>\n<\/td>\n<\/td>\n<\/tr>\n
(c) Trung \u0111i\u1ec3m c\u1ee7a \u0111o\u1ea1n th\u1eb3ng AB c\u00f3 t\u1ecda \u0111\u1ed9 l\u00e0 (-5\/2, -1\/2 ) .<\/td>\n<\/td>\n<\/td>\n<\/tr>\n
(d) \u0110o\u1ea1n th\u1eb3ng BC c\u00f3 \u0111\u1ed9 d\u00e0i b\u1eb1ng 27.<\/td>\n<\/td>\n<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

C\u00e2u 14. <\/strong>\u0110\u1ec3 x\u00e2y d\u1ef1ng ph\u01b0\u01a1ng \u00e1n kinh doanh cho m\u1ed9t lo\u1ea1i s\u1ea3n ph\u1ea9m, doanh nghi\u1ec7p t\u00ednh to\u00e1n l\u1ee3i nhu\u1eadn y (\u0111\u1ed3ng) theo c\u00f4ng th\u1ee9c sau: y = -86x^2 + 86000x \u2013 18146000 trong \u0111\u00f3 x l\u00e0 s\u1ed1 s\u1ea3n ph\u1ea9m \u0111\u01b0\u1ee3c b\u00e1n ra.<\/p>\n

\n\n\n\n\n\n\n\n\n
M\u1ec7nh \u0111\u1ec1<\/strong><\/th>\n\u0110\u00fang<\/strong><\/th>\nSai<\/strong><\/th>\n<\/tr>\n<\/thead>\n
(a) Doanh nghi\u1ec7p b\u1ecb l\u1ed7 khi b\u00e1n t\u1eeb 303 \u0111\u1ebfn 698 s\u1ea3n ph\u1ea9m.<\/td>\n<\/td>\n<\/td>\n<\/tr>\n
(b) Doanh nghi\u1ec7p c\u00f3 l\u00e3i khi b\u00e1n t\u1ed1i \u0111a 302 s\u1ea3n ph\u1ea9m ho\u1eb7c b\u00e1n t\u1ed1i thi\u1ec3u 697 s\u1ea3n ph\u1ea9m.<\/td>\n<\/td>\n<\/td>\n<\/tr>\n
(c) Doanh nghi\u1ec7p c\u00f3 l\u00e3i khi b\u00e1n t\u1eeb 303 \u0111\u1ebfn 697 s\u1ea3n ph\u1ea9m.<\/td>\n<\/td>\n<\/td>\n<\/tr>\n
(d) Doanh nghi\u1ec7p b\u1ecb l\u1ed7 khi b\u00e1n t\u1ed1i \u0111a 302 s\u1ea3n ph\u1ea9m ho\u1eb7c b\u00e1n t\u1ed1i thi\u1ec3u 698 s\u1ea3n ph\u1ea9m.<\/td>\n<\/td>\n<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<\/div>\n

PH\u1ea6N 3 \u2013 C\u00c2U H\u1eceI TR\u1ea2 L\u1edcI NG\u1eaeN (2 \u0111i\u1ec3m)<\/strong><\/p>\n

C\u00e2u h\u1ecfi tr\u1ea3 l\u1eddi ng\u1eafn g\u1ed3m 4 c\u00e2u h\u1ecfi li\u00ean quan \u0111\u1ebfn h\u00e0m s\u1ed1, ph\u01b0\u01a1ng tr\u00ecnh, b\u1ea5t ph\u01b0\u01a1ng tr\u00ecnh, h\u00ecnh h\u1ecdc gi\u1ea3i t\u00edch, h\u1ec7 ph\u01b0\u01a1ng tr\u00ecnh, v\u00e0 b\u00e0i to\u00e1n th\u1ef1c t\u1ebf. Sau \u0111\u00e2y l\u00e0 m\u1ed9t s\u1ed1 c\u00e2u h\u1ecfi tham kh\u1ea3o:<\/p>\n

C\u00e2u 15.\u00a0<\/strong>Cho h\u00e0m s\u1ed1 <\/span>\\[
\nf(x) =
\n\\begin{cases}
\nx \u2013 \\sqrt{x^2 + m^2} & \\text{khi } x < 1, \\quad \\text{v\u1edbi } m \\text{ l\u00e0 tham s\u1ed1,} \\\\
\n2x \u2013 1 & \\text{khi } x \\geq 1.
\n\\end{cases}
\n\\] Bi\u1ebft r\u1eb1ng \u0111\u1ed3 th\u1ecb h\u00e0m s\u1ed1 c\u1eaft tr\u1ee5c tung t\u1ea1i \u0111i\u1ec3m c\u00f3 tung \u0111\u1ed9 b\u1eb1ng 3. H\u00e3y t\u00ednh P = f(-4) + f(1) . Vi\u1ebft k\u1ebft qu\u1ea3 d\u01b0\u1edbi d\u1ea1ng s\u1ed1 th\u1eadp ph\u00e2n.<\/p>\n

C\u00e2u 16.<\/strong> M\u1ed9t c\u00f4ng ty du l\u1ecbch b\u00e1o gi\u00e1 ti\u1ec1n tham quan c\u1ee7a m\u1ed9t nh\u00f3m kh\u00e1ch du l\u1ecbch nh\u01b0 sau:<\/p>\n

    \n
  • 50 kh\u00e1ch \u0111\u1ea7u ti\u00ean c\u00f3 gi\u00e1 v\u00e9 300.000 \u0111\u1ed3ng\/ng\u01b0\u1eddi.<\/li>\n
  • N\u1ebfu s\u1ed1 l\u01b0\u1ee3ng kh\u00e1ch l\u1edbn h\u01a1n 50, m\u1ed7i ng\u01b0\u1eddi t\u0103ng th\u00eam s\u1ebd l\u00e0m gi\u00e1 v\u00e9 gi\u1ea3m 5.000 \u0111\u1ed3ng\/ng\u01b0\u1eddi cho to\u00e0n b\u1ed9 h\u00e0nh kh\u00e1ch.<\/li>\n<\/ul>\n

    G\u1ecdi x \u00a0<\/span><\/span>l\u00e0 s\u1ed1 l\u01b0\u1ee3ng kh\u00e1ch v\u01b0\u1ee3t qu\u00e1 50 ng\u01b0\u1eddi c\u1ee7a nh\u00f3m. Bi\u1ebft chi ph\u00ed th\u1ef1c t\u1ebf c\u1ee7a chuy\u1ebfn du l\u1ecbch l\u00e0 15.080.000 \u0111\u1ed3ng. H\u00e3y x\u00e1c \u0111\u1ecbnh s\u1ed1 nguy\u00ean l\u1edbn nh\u1ea5t c\u1ee7a x \u0111\u1ec3 c\u00f4ng ty kh\u00f4ng b\u1ecb l\u1ed7.<\/p>\n

    PH\u1ea6N 4 \u2013 C\u00c2U H\u1eceI T\u1ef0 LU\u1eacN (3 \u0111i\u1ec3m)<\/strong><\/p>\n

    C\u00e2u h\u1ecfi t\u1ef1 lu\u1eadn y\u00eau c\u1ea7u tr\u00ecnh b\u00e0y l\u1eddi gi\u1ea3i chi ti\u1ebft, ki\u1ec3m tra t\u01b0 duy l\u1eadp lu\u1eadn v\u00e0 v\u1eadn d\u1ee5ng ki\u1ebfn th\u1ee9c v\u1ec1 h\u00e0m s\u1ed1, ph\u01b0\u01a1ng tr\u00ecnh, b\u1ea5t ph\u01b0\u01a1ng tr\u00ecnh, h\u00ecnh h\u1ecdc gi\u1ea3i t\u00edch, h\u1ec7 ph\u01b0\u01a1ng tr\u00ecnh v\u00e0 b\u00e0i to\u00e1n th\u1ef1c t\u1ebf. M\u1ed9t s\u1ed1 c\u00e2u h\u1ecfi trong b\u1ed9 \u0111\u1ec1:<\/p>\n

    C\u00e2u 19.<\/strong> M\u1ed9t ng\u01b0\u1eddi n\u00f4ng d\u00e2n th\u1ea3 1000 con c\u00e1 gi\u1ed1ng v\u00e0o h\u1ed3 nu\u00f4i v\u1eeba m\u1edbi \u0111\u00e0o. Bi\u1ebft r\u1eb1ng sau m\u1ed7i n\u0103m th\u00ec s\u1ed1 l\u01b0\u1ee3ng c\u00e1 trong h\u1ed3 t\u0103ng th\u00eam x l\u1ea7n s\u1ed1 l\u01b0\u1ee3ng c\u00e1 ban \u0111\u1ea7u v\u00e0 x kh\u00f4ng \u0111\u1ed5i. B\u1eb1ng c\u00e1ch thay \u0111\u1ed5i k\u0129 thu\u1eadt nu\u00f4i v\u00e0 th\u1ee9c \u0103n cho c\u00e1, h\u1ecfi sau hai n\u0103m \u0111\u1ec3 s\u1ed1 c\u00e1 trong h\u1ed3 l\u00e0 36000 con th\u00ec t\u1ed1c \u0111\u1ed9 t\u0103ng s\u1ed1 l\u01b0\u1ee3ng c\u00e1 trong h\u1ed3 l\u00e0 bao nhi\u00eau? Bi\u1ebft t\u1ed1c \u0111\u1ed9 t\u0103ng m\u1ed7i n\u0103m l\u00e0 kh\u00f4ng \u0111\u1ed5i.<\/p>\n

    C\u00e2u 20.<\/strong> Trong m\u1eb7t ph\u1eb3ng Oxy cho tam gi\u00e1c ABC c\u00f3 t\u1ecda \u0111\u1ed9 c\u00e1c \u0111\u1ec9nh A(1;1) , B(-2;5) . \u0110\u1ec9nh C thu\u1ed9c \u0111\u01b0\u1eddng th\u1eb3ng d: x – 4 = 0 , tr\u1ecdng t\u00e2m G c\u1ee7a tam gi\u00e1c ABC thu\u1ed9c \u0111\u01b0\u1eddng th\u1eb3ng d\u2019: 2x – 3y + 6 = 0 . T\u00ednh di\u1ec7n t\u00edch tam gi\u00e1c ABC .<\/p>\n

    T\u1ea3i B\u1ed9 \u0111\u1ec1 ki\u1ec3m tra gi\u1eefa k\u1ef3 2 To\u00e1n 10 K\u1ebft n\u1ed1i tri th\u1ee9c n\u0103m h\u1ecdc 2024 \u2013 2025<\/span><\/strong><\/h2>\n

    \u0110\u1ec3 b\u1eaft \u0111\u1ea7u l\u00e0m quen v\u00e0 chu\u1ea9n b\u1ecb t\u1ed1t cho k\u1ef3 thi gi\u1eefa h\u1ecdc k\u1ef3 2 m\u00f4n To\u00e1n l\u1edbp 10, B\u1ed9 \u0111\u1ec1 ki\u1ec3m tra gi\u1eefa k\u1ef3 2 To\u00e1n 10 K\u1ebft n\u1ed1i tri th\u1ee9c n\u0103m h\u1ecdc 2024 \u2013 2025<\/strong> s\u1ebd l\u00e0 t\u00e0i li\u1ec7u h\u1eefu \u00edch gi\u00fap b\u1ea1n \u00f4n t\u1eadp hi\u1ec7u qu\u1ea3 v\u00e0 \u0111\u1ea1t \u0111i\u1ec3m s\u1ed1 cao trong k\u1ef3 thi s\u1eafp t\u1edbi. B\u1ed9 \u0111\u1ec1 \u0111\u01b0\u1ee3c bi\u00ean so\u1ea1n b\u00e1m s\u00e1t ch\u01b0\u01a1ng tr\u00ecnh s\u00e1ch gi\u00e1o khoa K\u1ebft n\u1ed1i tri th\u1ee9c, gi\u00fap b\u1ea1n h\u1ec7 th\u1ed1ng l\u1ea1i ki\u1ebfn th\u1ee9c quan tr\u1ecdng, 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. T\u1ea3i ngay b\u1ed9 \u0111\u1ec1 theo \u0111\u01b0\u1eddng link d\u01b0\u1edbi \u0111\u00e2y v\u00e0 luy\u1ec7n t\u1eadp h\u1eb1ng ng\u00e0y nh\u00e9!<\/p>\n