Lời giải:

Ta có

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

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

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

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

Lời giải:

Lời giải:

Ta có

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

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

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

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

Lời giải:

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

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

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

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

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

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

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

Lời giải:

Lời giải:

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$B=\frac{\text{2}}{{\text{(}-\text{2)}}^{\text{3}}+\text{1}}=\frac{-\text{2}}{\text{7}}$

Lời giải:

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)

Lời giải:

Ta có

Lời giải:

Lời giải:

Lời giải:

Lời giải:

$\frac{\text{4x}-\text{7}}{{\text{x}}^{\text{2}}-\text{3x}+\text{2}}=\frac{\text{a}}{\text{x}-\text{1}}+\frac{\text{b}}{\text{x}-\text{2}}$

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

$\begin{array}{l}=\frac{\text{ax}-\text{2a}+\text{bx}-\text{b}}{{\text{x}}^{\text{2}}-\text{3x}+\text{2}}\\ =\frac{\text{(a}+\text{b)x}-\text{2a}-\text{b}}{{\text{x}}^{\text{2}}-\text{3x}+\text{2}}\\ =\frac{\text{4x}-\text{7}}{{\text{x}}^{\text{2}}-\text{3x}+\text{2}}\\ \Rightarrow \left\{\begin{array}{l}\text{a}+\text{b}=\text{4}\\ -\text{2a}-\text{b}=-\text{7}\end{array}\right.\\ \iff \left\{\begin{array}{l}\text{a}=\text{3}\\ \text{b}=\text{1}\end{array}\right.\end{array}$

Vậy a = 3, b = 1

Lời giải:

Suy ra a + b = 1 + (-2) = -1

Lời giải:

Ta có

$\begin{array}{l}\frac{\text{1}}{\text{1}-\text{x}}+\frac{\text{1}}{\text{1}+\text{x}}+\frac{\text{2}}{\text{1}+{\text{x}}^{\text{2}}}+\frac{\text{4}}{\text{1}+{\text{x}}^{\text{4}}}+\frac{\text{8}}{\text{1}+{\text{x}}^{\text{8}}}\\ =\frac{\text{1}+\text{x}+\text{1}-\text{x}}{\text{(1}-\text{x)(1}+\text{x)}}+\frac{\text{2}}{\text{1}+{\text{x}}^{\text{2}}}+\frac{\text{4}}{\text{1}+{\text{x}}^{\text{4}}}+\frac{\text{8}}{\text{1}+{\text{x}}^{\text{8}}}\end{array}$

$\begin{array}{l}=\frac{\text{2}}{\text{1}-{\text{x}}^{\text{2}}}+\frac{\text{2}}{\text{1}+{\text{x}}^{\text{2}}}+\frac{\text{4}}{\text{1}+{\text{x}}^{\text{4}}}+\frac{\text{8}}{\text{1}+{\text{x}}^{\text{8}}}\\ =\frac{\text{2(1}+{\text{x}}^{\text{2}}\text{)}+\text{2(1}-{\text{x}}^{\text{2}}\text{)}}{\text{(1}-{\text{x}}^{\text{2}}\text{)(1}+{\text{x}}^{\text{2}}\text{)}}+\frac{\text{4}}{\text{1}+{\text{x}}^{\text{4}}}+\frac{\text{8}}{\text{1}+{\text{x}}^{\text{8}}}\end{array}$

$=\frac{\text{4}}{\text{1}-{\text{x}}^{\text{4}}}+\frac{\text{4}}{\text{1}+{\text{x}}^{\text{4}}}+\frac{\text{8}}{\text{1}+{\text{x}}^{\text{8}}}$

$=\frac{\text{4(1}+{\text{x}}^{\text{4}}\text{)}+\text{4(1}-{\text{x}}^{\text{4}}\text{)}}{\text{(1}-{\text{x}}^{\text{4}}\text{)(1}+{\text{x}}^{\text{4}}\text{)}}+\frac{\text{8}}{\text{1}+{\text{x}}^{\text{8}}}$

=$\frac{\text{8}}{\text{1}-{\text{x}}^{\text{8}}}+\frac{\text{8}}{\text{1}+{\text{x}}^{\text{8}}}$

$=\frac{\text{8(1}+{\text{x}}^{\text{8}}\text{)}+\text{8(1}-{\text{x}}^{\text{8}}\text{)}}{\text{(1}-{\text{x}}^{\text{8}}\text{)(1}+{\text{x}}^{\text{8}}\text{)}}$

$=\frac{\text{16}}{\text{1}-{\text{x}}^{\text{16}}}$

Vậy số cần điền là 16.

Lời giải:

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

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

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

Lời giải:

Lời giải:

Lời giải: