Lý thuyết Phép trừ các phân thức đại số (mới 2023 + Bài Tập) – Toán 8

Đáp án: D

Giải thích: 

Ta có $\frac{\text{x}-\text{2}}{{\text{6x}}^{\text{2}}-\text{6x}}-\frac{\text{1}}{{\text{4x}}^{\text{2}}-\text{4}}$

$\begin{array}{l}=\frac{\text{x}-\text{2}}{\text{6x(x}-\text{1)}}-\frac{\text{1}}{{\text{4(x}}^{\text{2}}-\text{1)}}\\ =\frac{\text{x}-\text{2}}{\text{6x(x}-\text{1)}}-\frac{\text{1}}{\text{4(x}-\text{1)(x}+\text{1)}}\end{array}$

$=\frac{\text{2(x}-\text{2)(x}+\text{1)}}{\text{12x(x}-\text{1)(x}+\text{1)}}-\frac{\text{3x}}{\text{12x(x}-\text{1)(x}+\text{1)}}$

$\begin{array}{l}=\frac{2(x^2-2x+x-2)-3x}{12x(x-1)(x+1)}\\ =\frac{2x^2-5x-4}{12x(x-1)(x+1)}\end{array}$

Đáp án: B

Giải thích:

$C=\frac{\text{1}}{\text{x}-\text{18}}-\frac{\text{1}}{\text{x}+\text{2}}$

$=\frac{\text{x}+\text{2}}{\text{(x}-\text{18)(x}+\text{2)}}-\frac{\text{1(x}-\text{18)}}{\text{(x}-\text{18)(x}+\text{2)}}$

$\begin{array}{l}=\frac{\text{x}+\text{2}-\text{x}+\text{18}}{\text{(x}-\text{18)(x}+\text{2)}}\\ =\frac{\text{20}}{\text{(x}-\text{18)(x}+\text{2)}}\end{array}$

Thay x = 2018 vào

C = $\frac{\text{20}}{\text{(x}-\text{18)(x}+\text{2)}}$ ta được

$\begin{array}{l}C=\frac{\text{20}}{\text{(2018}-\text{18)(2018}+\text{2)}}\\ =\frac{\text{20}}{\text{2000.2020}}=\frac{\text{1}}{\text{202000}}\end{array}$

Đáp án: C

Giải thích:

$\frac{\text{a}}{\text{a}+\text{1}}-\frac{\text{a}}{\text{a}-\text{1}}-\frac{{\text{2a}}^{\text{2}}}{\text{1}-{\text{a}}^{\text{2}}}$

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

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

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

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

Đáp án: C

Giải thích:

*) $\frac{\text{1}}{\text{x}+\text{2}}-\frac{\text{1}}{\text{(x}+\text{2)(4x}+\text{7)}}$ 

$=\frac{\text{1.(4x}+\text{7)}}{\text{(x}+\text{2)(4x}+\text{7)}}-\frac{\text{1}}{\text{(x}+\text{2)(4x}+\text{7)}}$

$=\frac{\text{4x}+\text{7}-\text{1}}{\text{(x}+\text{2)(4x}+\text{7)}}$

$=\frac{\text{4x}+\text{6}}{\text{(x}+\text{2)(4x}+\text{7)}}$nên A sai.

$\begin{array}{l}*)\frac{\text{2}-\text{21x}}{\text{18}}-\frac{\text{4}+\text{x}}{\text{12}}\\ =\frac{\text{2(2-21x)}}{\text{18.2}}-\frac{\text{3(4}+\text{x)}}{\text{12.3}}\\ =\frac{\text{4}-\text{42x}-\text{12}-\text{3x}}{\text{36}}\\ =\frac{-\text{45x}-\text{8}}{\text{36}}\end{array}$

nên B sai.

$\begin{array}{l}*)\frac{\text{2}-\text{21x}}{\text{18}}-\frac{\text{4}+\text{x}}{\text{12}}\\ =\frac{\text{2(2-21x)}}{\text{18.2}}-\frac{\text{3(4}+\text{x)}}{\text{12.3}}\\ =\frac{\text{4}-\text{42x}-\text{12}-\text{3x}}{\text{36}}\\ =\frac{-\text{45x}-\text{8}}{\text{36}}\end{array}$

nên C đúng.

$\begin{array}{l}*)\frac{\text{2}}{\text{x}-\text{5}}+\frac{\text{3x}}{{\text{x}}^{\text{2}}-\text{25}}\\ =\frac{\text{2(x}+\text{5)}}{\text{(x-5)(x}+\text{5)}}+\frac{\text{3x}}{\text{(x}-\text{5)(x}+\text{5)}}\\ =\frac{\text{2x}+\text{10}+\text{3x}}{\text{(x}-\text{5)(x}+\text{5)}}\\ =\frac{\text{5x}+\text{10}}{\text{(x}-\text{5)(x}+\text{5)}}\end{array}$

nên D sai.

Đáp án: D

Giải thích:

Đáp án: A

Giải thích: 

Phân thức đối của phân thức $\frac{\text{3}}{\text{x}+\text{1}}$ là

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

Đáp án: A

Giải thích: 

Đáp án: D

Giải thích:

Ta có $\frac{\text{x}}{\text{x}+\text{1}}-\frac{\text{2}}{\text{x}+\text{1}}=\frac{\text{x}-\text{2}}{\text{x}+\text{1}}$ nên A sai.

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

nên B sai.

*) $\frac{\text{x}}{\text{x}-\text{1}}+\frac{\text{1}}{\text{x}-\text{1}}=\frac{\text{x}+\text{1}}{\text{x}-\text{1}}$ nên C sai.

$\begin{array}{l}*)\frac{\text{x}}{\text{x}+\text{1}}+\frac{\text{1}}{\text{-x}-\text{1}}=\frac{\text{x}}{\text{x}+\text{1}}+\frac{\text{1}}{-\text{(x}+\text{1)}}\\ =\frac{\text{x}}{\text{x}+\text{1}}-\frac{\text{1}}{\text{x}+\text{1}}=\frac{\text{x}-\text{1}}{\text{x}+\text{1}}\end{array}$  

nên D đúng.