TOP 40 câu Trắc nghiệm Rút gọn phân thức (có đáp án 2022) - Toán 8

Ta có

$\frac{{\text{(a}+\text{b)}}^{\text{2}}-{\text{c}}^{\text{2}}}{\text{a}+\text{b}+\text{c}}$

$=\frac{\text{(a}+\text{b}+\text{c)(a}+\text{b}-\text{c)}}{\text{(a}+\text{b}+\text{c)}}$

$=\frac{\text{a}+\text{b}-\text{c}}{\text{1}}$ .

Ta có$\frac{{\text{a}}^{\text{2}}-\text{2a}-\text{8}}{{\text{a}}^{\text{2}}+\text{2a}}$

$=\frac{{\text{a}}^{\text{2}}-\text{4a}+\text{2a}-\text{8}}{\text{a(a}+\text{2)}}$

$=\frac{\text{a(a}-\text{4)}+\text{2(a}-\text{4)}}{\text{a(a}+\text{2)}}$

$=\frac{\text{(a}+\text{2)(a}-\text{4)}}{\text{a(a}+\text{2)}}$

$=\frac{\text{a}-\text{4}}{\text{a}}$

Ta có:

$\begin{array}{l}\frac{{\text{x}}^{\text{3}}+{\text{2x}}^{\text{2}}-\text{3x}-\text{6}}{{\text{x}}^{\text{2}}+\text{x}-\text{2}}\\ =\frac{{\text{(x}}^{\text{3}}+{\text{2x}}^{\text{2}}\text{)}-\text{3x}-\text{6}}{{\text{x}}^{\text{2}}-\text{x}+\text{2x}-\text{2}}\\ =\frac{{\text{x}}^{\text{2}}\text{(x}+\text{2)}-\text{3(x}+\text{2)}}{\text{x(x}-\text{1)}+\text{2(x}-\text{1)}}\\ =\frac{\text{(x}+{\text{2)(x}}^{\text{2}}-\text{3)}}{\text{(x}-\text{1)(x}+\text{2)}}\\ =\frac{{\text{x}}^{\text{2}}-\text{3}}{\text{x}-\text{1}}\end{array}$

Ta có:$\frac{\text{54(x}-{\text{3)}}^{\text{3}}}{\text{63(3}-{\text{x)}}^{\text{2}}}$

$=\frac{\text{54(x}-{\text{3)}}^{\text{3}}}{\text{63(x}-{\text{3)}}^{\text{2}}}=\frac{\text{6}}{\text{7}}\text{(x}-\text{3)}$ .

Ta có M =$\frac{{\text{x}}^{\text{2}}+{\text{y}}^{\text{2}}-\text{(1}+\text{2xy)}}{{\text{x}}^{\text{2}}-{\text{y}}^{\text{2}}+\text{1}+\text{2x}}$

$=\frac{{\text{x}}^{\text{2}}+{\text{y}}^{\text{2}}-\text{2xy}-\text{1}}{{\text{x}}^{\text{2}}+\text{2x}+\text{1}-{\text{y}}^{\text{2}}}$

=$\frac{{\text{(x}-\text{y)}}^{\text{2}}-\text{1}}{{\text{(x}+\text{1)}}^{\text{2}}-{\text{y}}^{\text{2}}}$

$=\frac{\text{(x}-\text{y}+\text{1)(x}-\text{y}-\text{1)}}{\text{(x}+\text{1}-\text{y)(x}+\text{1}+\text{y)}}$

$=\frac{\text{x}-\text{y}-\text{1}}{\text{x}+\text{1}+\text{y}}$

Vậy M =$\frac{\text{x}-\text{y}-\text{1}}{\text{x}+\text{1}+\text{y}}$ .

Thay x = 99 và y = 100

vào M = $\frac{\text{x}-\text{y}-\text{1}}{\text{x}+\text{1}+\text{y}}$ ta được

M =$\frac{\text{99}-\text{100}-\text{1}}{\text{99}+\text{1}+\text{100}}=\frac{-\text{2}}{\text{200}}=\frac{-\text{1}}{\text{100}}$ .

Ta có

$\frac{{\text{x}}^{\text{2}}-\text{xy}-\text{x}+\text{y}}{{\text{x}}^{\text{2}}+\text{xy}-\text{x}-\text{y}}$

$=\frac{\text{x(x}-\text{y)}-\text{(x}-\text{y)}}{\text{x(x}+\text{y)}-\text{(x}+\text{y)}}$

$=\frac{\text{(x}-\text{1)(x}-\text{y)}}{\text{(x}-\text{1)(x}+\text{y)}}$

$=\frac{\text{x}-\text{y}}{\text{x}+\text{y}}$.

Vậy mẫu thức của phân thức đã rút gọn là x + y.

Ta có $\frac{{\text{6x}}^{\text{2}}{\text{y}}^{\text{3}}\text{(x}+\text{3y)}}{{\text{18x}}^{\text{2}}\text{y(x}+{\text{3y)}}^{\text{2}}}$

$=\frac{{\text{6x}}^{\text{2}}\text{y.(x}+{\text{3y).y}}^{\text{2}}}{{\text{6x}}^{\text{2}}\text{y(x}+\text{3y).3(x}+\text{3y)}}$

$=\frac{{\text{y}}^{\text{2}}}{\text{3(x}+\text{3y)}}$.

Ta có:$\frac{{\text{5x}}^{\text{2}}-\text{10xy}+{\text{5y}}^{\text{2}}}{{\text{x}}^{\text{2}}-{\text{y}}^{\text{2}}}$

$=\frac{{\text{5(x}}^{\text{2}}-\text{2xy}+{\text{y}}^{\text{2}}\text{)}}{\text{(x}-\text{y)(x}+\text{y)}}$

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

$=\frac{\text{5(x}-\text{y)}}{\text{x}+\text{y}}$ 

Ta có:

T =$\frac{{\text{3a}}^{\text{2}}+\text{6ab}+{\text{3b}}^{\text{2}}}{\text{a}+\text{b}}$

$=\frac{{\text{3(a}}^{\text{2}}+\text{2ab}+{\text{b}}^{\text{2}}\text{)}}{\text{a}+\text{b}}$

$=\frac{\text{3(a}+{\text{b)}}^{\text{2}}}{\text{a}+\text{b}}$

$=\text{3(a}+\text{b)}$

Mà a + b = 3

=> T = 3.3 = 9.     

Ta có:

$\frac{{\text{2x}}^{\text{3}}-{\text{7x}}^{\text{2}}-\text{12x}+\text{45}}{{\text{3x}}^{\text{3}}-{\text{19x}}^{\text{2}}+\text{33x}-\text{9}}$

$=\frac{{\text{2x}}^{\text{3}}+{\text{5x}}^{\text{2}}-{\text{12x}}^{\text{2}}-\text{30x}+\text{18x}+\text{45}}{{\text{3x}}^{\text{3}}-{\text{x}}^{\text{2}}-{\text{18x}}^{\text{2}}+\text{6x}+\text{27x}-\text{9}}$

=$\frac{{\text{x}}^{\text{2}}\text{(2x}+\text{5)}-\text{6x(2x}+\text{5)}+\text{9(2x}+\text{5)}}{{\text{x}}^{\text{2}}\text{(3x}-\text{1)}-\text{6x(3x}-\text{1)}+\text{9(3x}-\text{1)}}$

=$\frac{\text{(2x}+{\text{5)(x}}^{\text{2}}-\text{6x}+\text{9)}}{\text{(3x}-{\text{1)(x}}^{\text{2}}-\text{6x}+\text{9)}}$

$=\frac{\text{2x}+\text{5}}{\text{3x}-\text{1}}$

Ta có:

B =$\frac{{\text{x}}^{\text{3}}-{\text{x}}^{\text{2}}\text{y}+{\text{xy}}^{\text{2}}}{{\text{x}}^{\text{3}}+{\text{y}}^{\text{3}}}$

$=\frac{{\text{x(x}}^{\text{2}}-\text{xy}+{\text{y}}^{\text{2}}\text{)}}{\text{(x}+{\text{y)(x}}^{\text{2}}-\text{xy}+{\text{y}}^{\text{2}}\text{)}}$

$=\frac{\text{x}}{\text{x}+\text{y}}$ .

Thay x = -5; y = 10 vào biểu thức B

ta được:

B = $\frac{-\text{5}}{-\text{5}+\text{10}}=\frac{-\text{5}}{\text{5}}$ = -1.

Ta có:

Q = $\frac{\text{10}}{{\text{x}}^{\text{2}}-\text{4x}+\text{9}}$

$=\frac{\text{10}}{{\text{x}}^{\text{2}}-\text{4x}+\text{4}+\text{5}}$

$=\frac{\text{10}}{{\text{(x}-\text{2)}}^{\text{2}}+\text{5}}$.

Ta có:

$\begin{array}{l}\frac{{\text{y}}^{\text{2}}-{\text{x}}^{\text{2}}}{{\text{x}}^{\text{2}}-\text{3xy}+{\text{2y}}^{\text{2}}}\\ =\frac{\text{(y}-\text{x)(y}+\text{x)}}{{\text{x}}^{\text{2}}-\text{xy}-\text{2xy}+{\text{2y}}^{\text{2}}}\\ =\frac{\text{(x}+\text{y)(y}-\text{x)}}{\text{x(x}-\text{y)}-\text{2y(x}-\text{y)}}\\ =\frac{-\text{(x}+\text{y)(x}-\text{y)}}{\text{(x}-\text{y)(x}-\text{2y)}}\\ =\frac{-\text{(x}+\text{y)}}{\text{x}-\text{2y}}\end{array}$

Vậy mẫu thức của phân thức đã rút gọn là x - 2y.

+$\frac{{\text{x}}^{\text{2}}-\text{3xy}}{{\text{21y}}^{\text{2}}-\text{7xy}}$

$=\frac{\text{x(x -3y)}}{\text{7y(3y-x)}}$

$=\frac{\text{x(x}-\text{3y)}}{-\text{7y(x}-\text{3y)}}=\frac{-\text{x}}{\text{7y}}$

nên A sai

+$\frac{\text{2x}+\text{4}}{{\text{x}}^{\text{2}}-\text{x}-\text{6}}$

$=\frac{\text{2(x}+\text{2)}}{{\text{x}}^{\text{2}}-\text{3x}+\text{2x}-\text{6}}$

$=\frac{\text{2(x}+\text{2)}}{\text{x(x}-\text{3)}+\text{2(x}-\text{3)}}$

$=\frac{\text{2(x}+\text{2)}}{\text{(x}+\text{2)(x}-\text{3)}}$

$=\frac{\text{2}}{\text{x}-\text{3}}$  nên B sai.

+ $\frac{\text{2x}-\text{6y}}{{\text{x}}^{\text{2}}-{\text{9y}}^{\text{2}}}$

$=\frac{\text{2(x-3y)}}{\text{(x-3y)(x}+\text{3y)}}$

$=\frac{\text{2}}{\text{x}+\text{3y}}$ nên C sai.

+  nên D đúng.

Ta có A = $\frac{{\text{2x}}^{\text{2}}-\text{4x}+\text{2}}{{\text{(x}-\text{1)}}^{\text{2}}}$

$=\frac{{\text{2(x}}^{\text{2}}-\text{2x}+\text{1)}}{{\text{(x}-\text{1)}}^{\text{2}}}$

$=\frac{\text{2(x}-{\text{1)}}^{\text{2}}}{{\text{(x}-\text{1)}}^{\text{2}}}=\text{2}$.

Ta có $\frac{\text{2xy}-{\text{x}}^{\text{2}}}{{\text{2y}}^{\text{2}}-\text{xy}}$

$=\frac{\text{x(2y-x)}}{\text{y(2y-x)}}=\frac{\text{x}}{\text{y}}$ nên A đúng.

+) $\frac{\text{(x}-\text{2)(x}+\text{4)}}{{\text{x}}^{\text{2}}+\text{7x}+\text{12}}$

$=\frac{\text{(x}-\text{2)(x}+\text{4)}}{{\text{x}}^{\text{2}}+\text{3x}+\text{4x}+\text{12}}$

$=\frac{\text{(x}-\text{2)(x}+\text{4)}}{\text{(x}+\text{3)(x}+\text{4)}}$

$=\frac{\text{x}-\text{2}}{\text{x}+\text{3}}$ nên B đúng.

+) $\frac{\text{(2x}-\text{4)(x}-\text{3)}}{{\text{(x}}^{\text{3}}-\text{27)(x}-\text{2)}}$

$=\frac{\text{2(x-2)(x-3)}}{{\text{(x-3)(x}}^{\text{2}}+\text{3x}+\text{9)(x}-\text{2)}}$

$=\frac{\text{2}}{{\text{x}}^{\text{2}}+\text{3x}+\text{9}}$ nên C sai.

+) $\frac{{\text{25xy}}^{\text{2}}}{{\text{40x}}^{\text{3}}{\text{y}}^{\text{2}}}$

$=\frac{{\text{5xy}}^{\text{2}}\text{.5}}{{\text{5xy}}^{\text{2}}{\text{.8x}}^{\text{2}}}=\frac{\text{5}}{{\text{8x}}^{\text{2}}}$ nên D đúng.

Ta có: $\frac{{\text{x}}^{\text{3}}-\text{1}}{{\text{x}}^{\text{2}}-\text{2x}+\text{1}}$

$=\frac{\text{(x}-{\text{1)(x}}^{\text{2}}+\text{x}+\text{1)}}{{\text{(x}-\text{1)}}^{\text{2}}}$

$=\frac{{\text{x}}^{\text{2}}+\text{x}+\text{1}}{\text{x}-\text{1}}$ nên A đúng.

+ $\frac{\text{5(x}-\text{y)}-\text{3(y}-\text{x)}}{\text{10(x}-\text{y)}}$

$=\frac{\text{5(x}-\text{y)}+\text{3(x}-\text{y)}}{\text{10(x-y)}}$

$=\frac{\text{8(x}-\text{y)}}{\text{10(x}-\text{y)}}=\frac{\text{4}}{\text{5}}$ nên B đúng.

+ $\frac{{\text{x}}^{\text{2}}+\text{4x}+\text{3}}{\text{2x}+\text{6}}$

$=\frac{{\text{x}}^{\text{2}}+\text{x}+\text{3x}+\text{3}}{\text{2(x}+\text{3)}}$

$=\frac{\text{x(x}+\text{1)}+\text{3(x}+\text{1)}}{\text{2(x}+\text{3)}}$

$=\frac{\text{(x}+\text{1)(x}+\text{3)}}{\text{2(x}+\text{3)}}=\frac{\text{x}+\text{1}}{\text{2}}$

 nên C đúng.

+ $\frac{{\text{16x}}^{\text{2}}\text{y(x}+{\text{y)}}^{\text{2}}}{\text{12xy(x}+\text{y)}}$

$=\frac{\text{16x.xy(x}+{\text{y)}}^{\text{2}}}{\text{12xy(x}+\text{y)}}$

$=\frac{\text{4x(x}+\text{y)}}{\text{3}}$ nên D sai.