26 câu Trắc nghiệm Toán 8: Ôn tập chương 2 có đáp án (Nhận biết)

ĐK: 3x² + 6x ≠ 0 $\iff$ 3x(x + 2) ≠ 0  

$\iff \left\{\begin{array}{l}x \ne 0\\ x \ne -2\end{array}\right.$

Đáp án cần chọn là: C

$$\begin{gathered} \frac{x^2 + 2x + 4}{3x} \\ = \frac{(x-2)(x^2 + 2x + 4)}{3x(x-2)} \\ =\frac{x^3 - 8}{3x(x-2)} \end{gathered}$$

Vậy đa thức cần tìm là 3x(x – 2)

Đáp án cần chọn là: A

Ta có:

 $$\begin{gathered} \frac{5{(y-x)}^2}{5x^2-5xy} = \frac{5{(x-y)}^2}{5x(x-y)} = \frac{x-y}{x} \\ \Rightarrow \frac{x-y}{x} = \frac{x-y}{P} \\ \Rightarrow P = x \end{gathered}$$

=> P = x

Đáp án cần chọn là: D

$$\begin{gathered} \frac{3x-1}{2xy} - \frac{5x-2}{2xy} \\ = \frac{3x-1-5x+2}{2xy} \\ =\frac{-2x+1}{2xy} \end{gathered}$$

Đáp án cần chọn là: C

Ta có:

$$\begin{gathered} \frac{x^3}{x^2+1}+\frac{x}{x^2+1} \\ =\frac{x^3+x}{x^2+1} \\ =\frac{x(x^2+1)}{x^2+1} = x \end{gathered}$$

Đáp án cần chọn là D

$$\begin{gathered} \frac{2x+5}{5x^2{y}^2} + \frac{8}{5x{y}^2} + \frac{2x-1}{x^2{y}^2} \\ = \frac{2x+5+8x+10x-5}{5x^2{y}^2} \\ =\frac{20x}{5x^2{y}^2} = \frac{4}{x{y}^2} \end{gathered}$$

Đáp án D

Gọi phân thức cần điền là P, khi đó

$$\begin{gathered} P= \frac{2x-6}{x+3} - \frac{x+1}{2} \\ =\frac{2(2x-6)-(x+3)(x+1)}{2(x+3)} \\ = \frac{4x-12-x^2-x-3x-3}{2(x+3)} \\ =\frac{-x^2-15}{2(x+3)} \end{gathered}$$

Đáp án cần chọn là: C

$$\begin{gathered} \frac{1}{x} + \frac{1}{x(x+1)} +...+\frac{1}{(x+9)(x+10)} \\ = \frac{1}{x} + \frac{1}{x} +0 + ... + 0 - \frac{1}{x + 10} \\ = \frac{2}{x} - \frac{1}{x+10} \\ =\frac{2x+20-x}{x(x+10)} = \frac{x+20}{x(x+10)} \end{gathered}$$

Đáp án cần chọn là đáp án A

Điều kiện: $x\ne -1, x\ne -2, x\ne \frac{-1}{2}$

$$\begin{gathered} \frac{1}{x+2}+\frac{1}{(x+1)(x+2)}+\frac{1}{(x+1)(2x+1)} \\ = \frac{2x^2+x+2x+1+2x+x+1+2}{(x+1)(x+2)(2x+1)} \\ =\frac{2(x^2+3x+2)}{(x+1)(x+2)(2x+1)} \\ = \frac{2\left[x\left(x+1\right)+2\left(x+1\right)\right]}{(x+1)(x+2)(2x+1)} \\ = \frac{2(x+1)(x+2)}{(x+1)(x+2)(2x+1)} \\ =\frac{2}{2x+1} \end{gathered}$$

Đáp án cần chọn là C

$$\begin{gathered} ĐK: x\ne 1 \\ \frac{4x^2-3x+5}{x^3-1} - \frac{1-2x}{x^2+x+1} - \frac{6}{x-1} \\ = \frac{4x^2-3x+5-(1-2x)(x-1)-6(x^2+x+1)}{(x-1)(x^2+x+1)} \\ =\frac{-12}{x^3-1} \end{gathered}$$

Đáp án A

$$\begin{gathered} ĐK: x\ne \left\{-2,2,3\right\} \\ P=\frac{3}{x-3}-\frac{x^2}{4-x^2}-\frac{4x-12}{x^3-3x^2-4x+12} \\ P=\frac{3(x^2-4)}{(x-3)(x^2-4)}+\frac{x^2(x-3)}{(x-3)(x^2-4)}-\frac{4x-12}{(x-3)(x^2-4)} \\ P= \frac{x^3-4x}{(x-3)(x-2)(x+2)} \\ P=\frac{x}{x-3} \\ Đáp án B \end{gathered}$$

$$\begin{gathered} P = \frac{3x+15}{x^2-4}.\frac{x-2}{x+5} \\ = \frac{3(x+5)}{(x-2)(x+2)}.\frac{x-2}{x+5} \\ =\frac{3}{x+2} \end{gathered}$$

$$\begin{gathered} P = \frac{x^2+4x+5}{5(x^3+1)}.\frac{2x}{x^2+4}.\frac{3(x^3+1)}{x^4+4x^2+5} \\ =\frac{6x}{5(x^2+4)} \end{gathered}$$

Đáp án B

$$\begin{gathered} P = \frac{x-1}{2-x}.\frac{x+2}{x-1}.\frac{-(2-x)}{(x+2)(2-x)} \\ = \frac{1}{x-2} \end{gathered}$$

Đáp án D

$$\begin{gathered} Q=\frac{x}{x^2-1}:\frac{5x}{x^2+2x+1} \\ = \frac{x}{(x-1)(x+1)}.\frac{{(x+1)}^2}{5x} \\ =\frac{x+1}{5(x-1)} \end{gathered}$$

Đáp án D

$$\begin{gathered} ĐK: x\ne \left\{0, - 1, -2, -3, -4, -5, -6\right\} \\ P = \frac{1}{x+6} = 1 \\ x = -5 \left(KTM\right) \end{gathered}$$

Đáp án D.

$$\begin{gathered} P=\frac{x+6}{x^2+1}.\frac{3x^2-3x+3+3x}{(x-6)(x+6)} \\ =\frac{x-6}{x^2+1}.\frac{3(x^2+1)}{(x-6)(x+6)} \\ =\frac{3}{x+6} \end{gathered}$$

Đáp án D

$$\begin{gathered} M = \frac{5x^2+10xy}{x^2+2xy+4y^2}.\frac{x^3-8y^3}{x+2y} \\ = 5x(x-2y) \end{gathered}$$

Đáp án B

$$\begin{gathered} \frac{-x-1}{3x+1}:\frac{x^2-1}{9x^2-1} \\ = \frac{-(x+1)}{3x+1}.\frac{(3x+1)(3x-1)}{(x+1)(x-1)} \\ =\frac{1-3x}{x-1} \end{gathered}$$

Đáp án A

$$\begin{gathered} C=\frac{2(x+1)(x-3)}{(x-3)(x^2-1)(x-2)} \\ =\frac{2}{(x-1)(x-2)} \end{gathered}$$

Đáp án D

$$\begin{gathered} ĐK: x\ne \left\{\pm 3\right\} \\ Q=\frac{x^2+3x+3x+9}{(x^2+9)(x+3)}:\frac{x^2+9-6x}{(x-3)(x^2+9)} \\ =\frac{x+3}{x-3} \end{gathered}$$

Đáp án D

$$\begin{gathered} ĐK: \left\{\begin{array}{l}{x}^2+3x-4\ne 0\\ x+4\ne 0\\ 1-x\ne 0\end{array}\Rightarrow \right.\left\{\begin{array}{l}x\ne 1\\ x\ne -4\end{array}\right. \\ P=\frac{10x-(2x-3)(x-1)-(x+1)(x+4)}{(x-1)(x+4)} \\ =\frac{-3x^2+10x-7}{(x-1)(x+\_4)} \\ =\frac{-3x+7}{x+4} \end{gathered}$$

Đáp án C

$$\begin{gathered} P=\frac{-3x+7}{x+4} \\ Khi x=-1 \\ \to P = \frac{10}{3} \end{gathered}$$

Đáp án C

$$\begin{gathered} P=\frac{-3x+7}{x+4}, \left\{\begin{array}{l}x\ne 1\\ x\ne -4\end{array}\right. \\ P+1=-2+\frac{19}{x+4} \\ x\in Z, P+1\in Z \Rightarrow x+4\in Ư\left(19\right)=\left\{\pm 1, \pm 19\right\} \\ \Rightarrow x\in \left\{-23, -5, -3, 15\right\} \end{gathered}$$

Đáp án B

$$\begin{gathered} x+y+z=0\Rightarrow x+y=-z \\ \Rightarrow x^2+2xy+y^2=z^2 \\ \Rightarrow x^2+y^2-z^2=-2xy \\ \Rightarrow \left\{\begin{array}{l}{y}^2+z^2-x^2=-2yz\\ z^2+x^2-y^2=-2xz\end{array}\right. \\ \Rightarrow A=\frac{xy}{-2xy}+\frac{yz}{-2yz}+\frac{zx}{-2zx}=\frac{-3}{2} \end{gathered}$$

Suy ra A < -1

Đáp án D

$$\begin{gathered} \frac{5}{x^2-6x+9+1}=\frac{5}{{(x-3)}^2+1} \\ {(x-3)}^2\ge 0 \Rightarrow x-3{)}^2 + 1\ge 1 \\ \Rightarrow \frac{5}{{(x-3)}^2+1} \le 5 \\ GTLN = 5 \\ "=" xảy ra khi x = 3 \end{gathered}$$

Đáp án A