Giúpp toii các bạn oii,c.onn

Giải thích các bước giải:

Bài 2:

a.Ta có:

$\dfrac{x^2-36}{2x+10}\cdot \dfrac3{6-x}$ 

$=\dfrac{(x-6)(x+6)}{2(x+5)}\cdot \dfrac3{-(x-6)}$

$=\dfrac{3(x+6)}{-2(x+5)}$

b.Ta có:

$\dfrac{5x-15}{4x+4}:\dfrac{x^2-9}{x^2+2x+1}$

$=\dfrac{5(x-3)}{4(x+1)}:\dfrac{(x-3)(x+3)}{(x+1)^2}$

$=\dfrac{5(x-3)}{4(x+1)}\cdot\dfrac{(x+1)^2}{(x-3)(x+3)}$

$=\dfrac{5(x+1)}{4(x+3)}$

c.Ta có:

$\dfrac3{2x^2+2x}+\dfrac{2x-1}{x^2-1}-\dfrac2x$

$=\dfrac{3\left(x-1\right)}{2x\left(x+1\right)\left(x-1\right)}+\dfrac{\left(2x-1\right)\cdot \:2x}{2x\left(x+1\right)\left(x-1\right)}-\dfrac{4\left(x+1\right)\left(x-1\right)}{2x\left(x+1\right)\left(x-1\right)}$

$=\dfrac{3\left(x-1\right)+\left(2x-1\right)\cdot \:2x-4\left(x+1\right)\left(x-1\right)}{2x\left(x+1\right)\left(x-1\right)}$

$=\dfrac{x+1}{2x\left(x+1\right)\left(x-1\right)}$

$=\dfrac{1}{2x\left(x-1\right)}$

Bài 1:

a.Ta có:

$\dfrac{3x}{5x+5y}-\dfrac{x}{10x-10y}$

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

$=\dfrac{6x\left(x-y\right)}{10\left(x+y\right)\left(x-y\right)}-\dfrac{x\left(x+y\right)}{10\left(x+y\right)\left(x-y\right)}$

$=\dfrac{6x\left(x-y\right)-x\left(x+y\right)}{10\left(x+y\right)\left(x-y\right)}$

$=\dfrac{5x^2-7xy}{10\left(x+y\right)\left(x-y\right)}$

b.Ta có:

$\dfrac{xy}{2x-y}-\dfrac{x^2-1}{y-2x}$

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

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