Đề cương ôn tập Giữa kì 2 Toán 10 trường THPT Số 2 Nguyễn Bỉnh Khiêm (Gia Lai) năm 2025-2026

$M(3;5)$

$N(2;6)$

$P(0;-3)$

$Q(0;1)$

$y=-5x^3-5x^2-3$

$y=\dfrac{1-x}{3-5x}$

$y=-3x+\sqrt{-5x-5}$

$y=-x^2-5x-3$

$I\left(-\dfrac{1}{2}; -\dfrac{21}{4}\right)$

$I\left(\dfrac{1}{2}; -\dfrac{1}{4}\right)$

$I(-1;-4)$

$I(1;6)$

$f(x)\ge 0, \forall x \in \mathbb{R}$

$f(x)\le 0, \forall x \in \mathbb{R}$

$f(x)<0, \forall x \in \mathbb{R}$

$f(x)>0, \forall x \in \mathbb{R}$

$(-\infty;1]\cup[4; +\infty)$

$(-\infty;1)\cup(4;+\infty)$

$(-\infty;1]$

$[1; +\infty)$

$\vec{n}=(1;-2)$

$\vec{n}=(2;1)$

$\vec{n}=(-2;3)$

$\vec{n}=(1;3)$

$2x - y - 4 = 0$

$2x - y = 0$

$x + 2y + 5 = 0$

$x + 2y - 5 = 0$

$30^0$

$45^0$

$150^0$

$60^0$

$x^{2} + (y - 5)^{2} = 2$

$x^{2} + (y - 5)^{2} = 16$

$x^{2} + (y + 5)^{2} = 16$

$(x - 5)^{2} + y^{2} = 4$

$-4x + 3y - 7 = 0$

$4x + 3y + 1 = 0$

$3x + 4y - 1 = 0$

$3x - 4y + 7 = 0$

$\dfrac{x^{2}}{9} + \dfrac{y^{2}}{36} = 1$

$\dfrac{x^{2}}{7} - \dfrac{y^{2}}{9} = 0$

$\dfrac{x^{2}}{7} - \dfrac{y^{2}}{9} = 1$

$\dfrac{x^{2}}{9} + \dfrac{y^{2}}{36} = 2$