Đề thi Giữa kì 2 Toán 10 năm 2023-2024 trường THPT chuyên Quốc học Huế

$x^2 + y^2 + 3x - 2y + 1 = 0.$

$x^2 + y^2 - 2xy - 2y + 4 = 0.$

$x^2 + y^2 + x + y + 2 = 0.$

$x^2 + y^2 - 4x + 7 = 0.$

$m > -\dfrac{5}{4}$.

$m > -5.$

$m < \dfrac{5}{4}$.

$m < 5.$

$2x - y + 6 = 0.$

$x + 2y - 3 = 0.$

$x + 2y + 3 = 0.$

$2x - y - 6 = 0.$

$M\left(-\dfrac{2}{5} ; \dfrac{3}{5}\right).$

$M\left(-\dfrac{3}{5} ; -\dfrac{2}{5}\right).$

$M\left(\dfrac{3}{5} ; \dfrac{2}{5}\right).$

$M\left(-\dfrac{2}{5} ; -\dfrac{3}{5}\right).$

$d(A, \Delta) = \dfrac{20}{3\sqrt{2}}$.

$d(A, \Delta) = \dfrac{21}{5}$.

$d(A, A) = 4$.

$d(A, A) = \dfrac{20}{\sqrt{7}}$

$x = 8$.

$x = 4$.

$x = 3.$

$x = 6.$

$(x - 4)^2 + (y - 1)^2 = 4.$

$(x + 4)^2 + (y + 1)^2 = 4.$

$(x - 4)^2 + (y - 1)^2 = 2.$

$(x + 4)^2 + (y + 1)^2 = 2.$

$H(-2 ; -8)$.

$G(5 ; -1)$.

$F(1 ; 1)$.

$E(2 ; 0)$

$m \leq \dfrac{1}{4}$

$m > \dfrac{1}{4}$

$m \geq \dfrac{1}{4}$

$m < \dfrac{1}{4}$

$x = 2$.

$x = -2$.

$x = -1$.

$x = 1$.

S = \left(-\infty ; -\dfrac{1}{2}\right) \cup \\[2 ; +\infty\right)

S = \\[ -2 ; \dfrac{1}{2}\right]

S = \\[ -\dfrac{1}{2} ; 2 \right]

S = (-\infty ; -2] \cup \\[ \dfrac{1}{2} ; +\infty)

A. $f(2) = 8$.

B. $f(2) = 7$.

C. $f(2) = 6$.

D. $f(2) = 5$.

A. $I(-2; -1), R = 3$.

B. $I(2; 1), R = 3$.

C. $I(-2; -1), R = \sqrt{3}$.

D. $I(2; 1), R = \sqrt{3}$.

A. $\Delta' : 3x + y - 6 = 0$.

B. $\Delta' : 3x + y - 2 = 0$.

C. $\Delta' : 3x + y - 1 = 0$.

D. $\Delta' : 3x + y + 8 = 0$.

A. $y = \dfrac{x^2}{x + 1}$.

B. $y = 2x + 2$.

C. $y = \sqrt{x^2 + 4}$.

D. $y = 1 - x^2$.

A. $x - 2y + 2 = 0$.

B. $x + 2y + 2 = 0$.

C. $x - 2y - 2 = 0$.

D. $x + 2y - 2 = 0$.

A. $\left(-\infty ; -\dfrac{1}{4}\right)$.

B. $\left[-\dfrac{1}{4} ; +\infty \right)$.

C. $\left(-\dfrac{1}{2} ; +\infty \right)$.

D. $\left(-\infty ; -\dfrac{1}{2} \right)$.

A. $\Delta : x - y + 2 = 0$ và $\Delta' : -2x + 2y - 4 = 0$.

B. $\Delta : y + 1 = 0$ và $\Delta' : x - 1 = 0$.

C. $\Delta : x + 2y + 6 = 0$ và $\Delta' : 2x + y - 4 = 0$.

D. $\Delta : x + y - 1 = 0$ và $\Delta' : 2x + 2y + 3 = 0$.

A. $x^2 + 3x - 1 = 0$.

B. $x^2 - 5x - 1 = 0$.

C. $x^2 - x - 1 = 0$.

D. $x^2 - 3x - 1 = 0$.

A. $\cos \varphi = \dfrac{1}{\sqrt{26}}$.

B. $\cos \varphi = \dfrac{1}{\sqrt{10}}$.

C. $\cos \varphi = \dfrac{5}{\sqrt{26}}$.

D. $\cos \varphi = \dfrac{3}{\sqrt{10}}$.

A. $\vec{u} = (-3; 2)$.

B. $\vec{u} = (-2; 3)$.

C. $\vec{u} = (2; -3)$.

D. $\vec{u} = (4; -6)$.

A. $D = (-\infty ; 2]$.

B. $D = (-\infty ; -2]$.

C. $D = [2; +\infty)$.

D. $D = [-2; +\infty)$.

$3x - 2y - 10 = 0$

$3x - 2y - 14 = 0$

$2x + 3y - 11 = 0$

$2x + 3y - 5 = 0$

$y = 2x + 7$

$x = 4y - 1$

$\begin{cases} x = 1 \\ y = 2 - t \end{cases}$

$3x - y + 1 = 0$

$\left[-\dfrac{2}{3}; + \infty \right)$

$\left[-\dfrac{1}{3}; + \infty \right)$

$(-\infty ; -\dfrac{2}{3}]$

$(-\infty ; -\dfrac{1}{3}]$

$\vec{n} = (4 ; -1)$

$\vec{n} = (1 ; 4)$

$\vec{n} = (0 ; 2)$

$\vec{n} = (1 ; 0)$

$\begin{cases} x = -2 - 5t \\ y = 7 + 2t \end{cases}$

$\begin{cases} x = 5 + 7t \\ y = 2 + 2t \end{cases}$

$\begin{cases} x = 5 + 2t \\ y = 2 + 7t \end{cases}$

$\begin{cases} x = 5 - 2t \\ y = 2 + 7t \end{cases}$

$\{1 ; 0 ; -4 ; 8 \}$

$\{-2 ; 0 ; 2 ; 3 ; 5 \}$

$\{1 ; 0 ; -4 ; -1 ; 8\}$

$\{0 ; 2 ; 5\}$

$\left(-\dfrac{b}{a} ; -\dfrac{\Delta}{4a} \right)$

$\left(-\dfrac{b}{a} ; \dfrac{\Delta}{4a} \right)$

$\left(-\dfrac{b}{2a} ; -\dfrac{\Delta}{4a} \right)$

$\left(-\dfrac{b}{2a} ; \dfrac{\Delta}{4a} \right)$

$I(2 ; -6)$

$I(1 ; -3)$

$I(-1 ; 3)$

$I(-2 ; 6)$