Đề thi cuối kì 2 Toán 10 Chân trời sáng tạo – Đề 07 (có đáp án)

$x^2-\frac{1}{x}$;

$\sqrt{x^2-5x+7}$;

$9x^3-7x^2-12x$;

$5x-4-x^2$.

(f\left( x \right) < 0\) với $x\in \left(0;2\right)$;

$f\left(x\right)>0$;

$f\left(x\right)\ge 0$;

$f\left(x\right)\le 0$.

$S=\left(-\infty ;2\right)\cup \left(2;+\infty \right)$;

$S=\mathbb{R}$;

$$S=\varnothing$$;

$S=\left(-\infty ;-2\right)\cup \left(-2;+\infty \right)$.

$0$;

$1$;

$2$;

$4$.

nhọn;

vuông;

A và B đúng;

A và C sai.

$I\left(-5;-4\right)$;

$I\left(8;4\right)$;

$I\left(-10;-8\right)$;

$I\left(4;2\right)$.

$$\left(-\frac{7}{2};-\frac{25}{2}\right)$$;

$$\left(\frac{11}{4};-\frac{25}{4}\right)$$;

$$\left(\frac{7}{2};\frac{25}{2}\right)$$;

$$\left(\frac{11}{4};\frac{25}{4}\right)$$.

$\{\begin{array}{l}x=1+3t\\ y=-4+2t\end{array}$;

$\{\begin{array}{l}x=3+2t\\ y=1-4t\end{array}$;

$\{\begin{array}{l}x=3+t\\ y=2-4t\end{array}$;

$\{\begin{array}{l}x=1-4t\\ y=3+2t\end{array}$.

$4x-5y-7=0$;

$4x+5y-17=0$;

$4x-5y-17=0$;

$4x+5y+17=0$.

$4x+3y+4=0$;

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

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

$4x-3y-4=0$.

$\left(-2;-2\right)$;

$\left(-\frac{10}{3};-\frac{14}{3}\right)$;

$\left(\frac{2}{5};\frac{14}{5}\right)$;

$\left(-\frac{2}{3};-\frac{10}{3}\right)$.

$$\overrightarrow{n}\left(-2;-1\right)$$;

$$\overrightarrow{n}\left(2;-1\right)$$;

$$\overrightarrow{n}\left(-1;2\right)$$;

$$\overrightarrow{n}\left(1;2\right)$$.

${\left(x+1\right)}^2+{\left(y+4\right)}^2=5$;

${\left(x-1\right)}^2+{\left(y-4\right)}^2=\sqrt{5}$;

${\left(x+1\right)}^2+{\left(y+4\right)}^2=\sqrt{5}$;

${\left(x-1\right)}^2+{\left(y-4\right)}^2=5$.

${\left(x-1\right)}^2+{\left(y+2\right)}^2=9$;

${\left(x+1\right)}^2+{\left(y+2\right)}^2=9$;

${\left(x-1\right)}^2+{\left(y-2\right)}^2=9$;

${\left(x+1\right)}^2+{\left(y-2\right)}^2=9$.

$x^2+y^2+10x+4y+4=0$;

$x^2+y^2+10x+4y-4=0$;

$x^2+y^2+10x-4y-4=0$;

$x^2+y^2+10x-4y+4=0$.

${\left(x-1\right)}^2+{\left(y-1\right)}^2=5$;

${\left(x-1\right)}^2+{\left(y-1\right)}^2=25$;

${\left(x-1\right)}^2+{\left(y-1\right)}^2=1$;

${\left(x-1\right)}^2+{\left(y-1\right)}^2=\frac{1}{5}$.

$\frac{x^2}{9}-\frac{y^2}{16}=1$;

$\frac{x^2}{-9}-\frac{y^2}{16}=1$;

$\frac{x^2}{25}+\frac{{\left(-y\right)}^2}{9}=1$;

$\frac{x^2}{16}+\frac{y^2}{4}=1$.

$$23$$;

$$46$$;

$$92$$;

$\frac{23}{2}$.

$x=9$;

$x+\frac{9}{2}=0$;

$y+9=0$;

$x=-\frac{9}{4}$.

$25$;

$75$;

$100$;

$15$.

$10.10$;

$10!$;

$C_{10}^1$;

$A_{10}^1$.

$4$;

$5$;

$6$;

$7$.

$$24$$;

$$48$$;

$$480$$;

$$60$$.

$C_{24}^2.C_{18}^3$;

$A_{24}^2{A}_{18}^3$;

$A_{42}^5$;

$C_{24}^2{C}_{18}^3.5!$.

$$160$$;

$$80$$;

$$20$$;

$$40$$.

$3^5$;

$2^5$;

$63$;

$1$.

$$x^5-5x^{4}y+10x^3{y}^2-10x^2{y}^3+5x{y}^4-y^5$$;

$$x^5+5x^{4}y+10x^3{y}^2+10x^2{y}^3+5x{y}^4+y^5$$;$d$

$$x^5-5x^{4}y-10x^3{y}^2-10x^2{y}^3-5x{y}^4+y^5$$;

$$x^5+5x^{4}y-10x^3{y}^2+10x^2{y}^3-5x{y}^4+y^5$$.$f\left(x\right)<0$

$P\left(\overline{A}\right)=1-P\left(A\right)$;

$P\left(\overline{A}\right)>1$;

$0\le P\left(\overline{A}\right)\le 1$;

$P\left(A\right)+P\left(\overline{A}\right)=1$.

$A=\left\{\left(4,6\right),\left(6,4\right),\left(3,5\right),\left(5,3\right),\left(5,5\right)\right\}$;

$A=\left\{\left(4,6\right),\left(6,4\right),\left(5,5\right)\right\}$;

$A=\left\{\left(4,6\right),\left(5,5\right)\right\}$;

$A=\left\{\left(4,6\right),\left(5,3\right),\left(5,5\right)\right\}$.

$\overline{A}$: “Trong $3$ quả bóng lấy ra có tối đa một quả bóng đỏ”;

$\overline{A}$: “Trong $3$ quả bóng lấy ra có nhiều nhất một quả bóng đỏ”;

$\overline{A}$: “Trong $3$ quả bóng lấy ra không có bóng đỏ”;

$\overline{A}$: “Trong $3$ quả bóng lấy ra có cả ba quả bóng đỏ”.

$$4$$;

$$8$$;

$$12$$;

$$16$$.

$$\frac{1}{2}$$;

$$\frac{7}{12}$$;

$$\frac{1}{6}$$;

$$\frac{1}{3}$$.

$\frac{2}{7}$;

$\frac{1}{21}$;

$\frac{37}{42}$;

$\frac{5}{42}$.

“ Tổng số chấm trong hai lần gieo là tám”.

“ Lần gieo đầu tiên được mặt hai chấm”.

“ Số chấm trong hai lần gieo là như nhau”.

“ Hai lần gieo đều xuất hiện mặt chấm chẵn”.

$y^2=\frac{5}{2}x$;

$y=5x^2$;

$y^2=10x$;

$y^2=5x$.