Đáp án:
Giải thích các bước giải:
$\dfrac{x}{2\left(x-3\right)}\cdot \:2\left(x-3\right)\left(x+1\right)+\dfrac{x}{2x+2}\cdot \:2\left(x-3\right)\left(x+1\right)=\dfrac{2x}{\left(x+1\right)\left(x-3\right)}\cdot \:2\left(x-3\right)\left(x+1\right)$
$x\left(x+1\right)+x\left(x-3\right)=4x$
*$\dfrac{x}{2x+2}\cdot \:2\left(x-3\right)\left(x+1\right)$
$=\dfrac{x\cdot \:2\left(x-3\right)\left(x+1\right)}{2x+2}$
$=x\left(x-3\right)$
*$=\dfrac{x\cdot \:2\left(x-3\right)\left(x+1\right)}{2\left(x-3\right)}$
$=\dfrac{x\left(x-3\right)\left(x+1\right)}{x-3}$
$=x\left(x+1\right)$
*$\dfrac{2x}{\left(x+1\right)\left(x-3\right)}\cdot \:2\left(x-3\right)\left(x+1\right)$
$=\dfrac{2x\cdot \:2\left(x-3\right)\left(x+1\right)}{\left(x+1\right)\left(x-3\right)}$
$=\dfrac{2x\cdot \:2\left(x+1\right)}{x+1}$
$=2x\cdot \:2$
$=4x$
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$x\left(x+1\right)+x\left(x-3\right)=4x$
$2x^2-2x-4x=4x-4x$
$2x^2-6x=0$
`=>x in {0;3}`
Vậy `x=0` thỏa mãn đề bài.
