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

Đáp án: C

Giải thích:

Ta có

$\frac{\text{2}-\text{3x}}{{\text{6x}}^{\text{2}}\text{y}}+\frac{\text{2x}+\text{1}}{{\text{6x}}^{\text{2}}\text{y}}+\frac{\text{2x}-\text{3}}{{\text{6x}}^{\text{2}}\text{y}}$

$=\frac{\text{2}-\text{3x}+\text{2x}+\text{1}+\text{2x}-\text{3}}{{\text{6x}}^{\text{2}}\text{y}}$

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

= $\frac{\text{x}}{{\text{6x}}^{\text{2}}\text{y}}=\frac{\text{1}}{\text{6xy}}$.

Đáp án: D

Giải thích:

Đáp án: B

Giải thích:

Ta có

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

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

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

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

Đáp án: D

Giải thích:

* $\frac{\text{3x}-\text{4}}{{\text{4x}}^{\text{2}}{\text{y}}^{\text{5}}}+\frac{\text{9x}+\text{4}}{{\text{4x}}^{\text{2}}{\text{y}}^{\text{5}}}$

$=\frac{\text{3x-4}+\text{9x}+\text{4}}{{\text{4x}}^{\text{2}}{\text{y}}^{\text{5}}}$ 

$=\frac{\text{12x}}{{\text{4x}}^{\text{2}}{\text{y}}^{\text{5}}}=\frac{\text{3}}{{\text{xy}}^{\text{5}}}$nên A sai.

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

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

$=\frac{\text{3x}+\text{3}}{\text{3}}=\text{x}+\text{1}$nên B sai.

$*\frac{\text{x}+\text{8}}{\text{x}-\text{1}}+\frac{\text{2x}-\text{1}}{\text{x}-\text{1}}+\frac{\text{6x}+\text{2}}{\text{x}-\text{1}}$

$\begin{array}{l}=\frac{\text{x}+\text{8+(2x}-\text{1)+(6x}+\text{2)}}{\text{x}-\text{1}}\\ =\frac{\text{x}+\text{8+2x-1+6x+2}}{\text{x}-\text{1}}\\ =\frac{\text{9x}+9}{\text{x}-\text{1}}=\frac{\text{9(x+1)}}{\text{x}-\text{1}}\end{array}$

nên C sai.

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

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

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

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

Đáp án: A

Giải thích:

Đáp án: C

Giải thích:

Ta có

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

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

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

Thay x = -2 vào $B=\frac{\text{2}}{{\text{x}}^{\text{3}}+\text{1}}$  ta được$B=\frac{\text{2}}{{\text{(}-\text{2)}}^{\text{3}}+\text{1}}=\frac{-\text{2}}{\text{7}}$

Đáp án: B

Giải thích:

Quy tắc: Muốn cộng hai phân thức cùng mẫu thức ta cộng các tử thức với nhau và giữ nguyên mẫu thức.

$\frac{\text{A}}{\text{B}}+\frac{\text{C}}{\text{B}}=\frac{\text{A}+\text{C}}{\text{B}}$ (B ≠ 0)

Đáp án: D

Giải thích:

Ta có:

Đáp án: B

Giải thích:

Đáp án: D

Giải thích: