`color{#D0D0D0}{O}color{#C8C8C8}{m}color{#C0C0C0}{i}color{#B8B8B8}{s}color{#B0B0B0}{e}`

`1)` `\frac{1}{y(x-y)} - \frac{1}{x(x-y)}`

Điều kiện xác định: `x \ne y`

`\frac{1}{y(x-y)} - \frac{1}{x(x-y)}`

`= \frac{x}{xy(x-y)} - \frac{y}{xy(x-y)}`

`= \frac{x-y}{xy(x-y)}`

`= \frac{1}{xy}`

`2)` `\frac{x+6}{2x+6} + \frac{2x+3}{x(x+3)}`

Điều kiện xác định: $x \neq 0$ và $x \neq -3$

`\frac{x+6}{2x+6} + \frac{2x+3}{x(x+3)}`

`= \frac{x+6}{2(x+3)} + \frac{2x+3}{x(x+3)}`

`= \frac{x(x+6)}{2x(x+3)} + \frac{2(2x+3)}{2x(x+3)}`

`= \frac{x^2+6x + 4x+6}{2x(x+3)}`

`= \frac{x^2+10x+6}{2x(x+3)}`

`3)` `\frac{x}{5x+5} - \frac{x}{10x-10}`

Điều kiện xác định: $x \neq 1$ và $x \neq -1$

`\frac{x}{5x+5} - \frac{x}{10x-10}`

`= \frac{x}{5(x+1)} - \frac{x}{10(x-1)}`

`= \frac{2x(x-1)}{10(x+1)(x-1)} - \frac{x(x+1)}{10(x+1)(x-1)}`

`= \frac{2x^2-2x - (x^2+x)}{10(x+1)(x-1)}`

`= \frac{2x^2-2x - x^2-x}{10(x+1)(x-1)}`

`= \frac{x^2-3x}{10(x+1)(x-1)}`

`= \frac{x(x-3)}{10(x+1)(x-1)}`

`4)` `\frac{1}{xy-x^2} - \frac{1}{y^2-xy}`

Điều kiện xác định: $x \neq 0$, $y \neq 0$, $x \neq y$

`\frac{1}{xy-x^2} - \frac{1}{y^2-xy}`

`= \frac{1}{x(y-x)} - \frac{1}{y(y-x)}`

`= \frac{y}{xy(y-x)} - \frac{x}{xy(y-x)}`

`= \frac{y-x}{xy(y-x)}`

`= \frac{1}{xy}$`

`5)` `\frac{1}{y^2-xy} + \frac{1}{x^2-xy}`

Điều kiện xác định: $x \neq 0$, $y \neq 0$, $x \neq y$

`\frac{1}{y^2-xy} + \frac{1}{x^2-xy}`

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

`= \frac{1}{y(y-x)} - \frac{1}{x(y-x)}`

`= \frac{x}{xy(y-x)} - \frac{y}{xy(y-x)}`

`= \frac{x-y}{xy(y-x)}`

`= \frac{-1}{xy}`

`6)` `\frac{x}{y^2-xy} - \frac{y}{xy-x^2}`

Điều kiện xác định: $x \neq 0$, $y \neq 0$, $x \neq y$

`\frac{x}{y^2-xy} - \frac{y}{xy-x^2}`

`= \frac{x}{y(y-x)} - \frac{y}{x(y-x)}`

`= \frac{x}{y(y-x)} + \frac{y}{x(x-y)}`

`= \frac{x}{y(y-x)} + \frac{y}{-x(y-x)}`

`= \frac{x}{y(y-x)} - \frac{y}{x(y-x)}`

`= \frac{x^2}{xy(y-x)} - \frac{y^2}{xy(y-x)}`

`= \frac{x^2-y^2}{xy(y-x)}`

`= \frac{(x-y)(x+y)}{-xy(x-y)}`

`= \frac{-(x+y)}{xy}`

`= \frac{-x-y}{xy}`