giải pt: x/2(x-3) + x/2x+2 = 2x/(x+1)(x-3)

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1. Đáp án:

    

   Giải thích các bước giải:

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

   $x\left(x+1\right)+x\left(x-3\right)=4x$

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

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

   $=x\left(x-3\right)$

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

   $=\dfrac{x\left(x-3\right)\left(x+1\right)}{x-3}$

   $=x\left(x+1\right)$

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

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

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

   $=2x\cdot \:2$

   $=4x$

   —————————————-

   $x\left(x+1\right)+x\left(x-3\right)=4x$

   $2x^2-2x-4x=4x-4x$

   $2x^2-6x=0$

   `=>x in {0;3}`

   Vậy `x=0` thỏa mãn đề bài.

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2. $\frac{x}{2(x-3)}$+$\frac{x}{2x+2}$ =$\frac{x}{(x+1)(x-3)}$ 

   ⇔$\frac{x}{2(x-3)}$+$\frac{x}{2(x+1)}$ =$\frac{x}{(x+1)(x-3)}$ 

   $MTC: 2(x+1)(x-3)$

   $ĐKXĐ:$ x$\neq$-1 hoặc x$\neq$3

   ⇔$\frac{x(x+1)}{2(x+1)(x-3)}$+$\frac{x(x-3)}{2(x+1)(x-3)}$ =$\frac{2x}{2(x+1)(x-3)}$ 

   $⇒x²+x+x²-3x=4x$

   $⇔x²+x+x²-3x-4x=0$

   $⇔-6x+2x²=0$

   $⇔2x(x-3)=0$

   ⇒$2x=0  $             hoặc        $ x-3=0$

   ⇔$x=0 (nhận)$     hoặc          $ x=3(loại)$

   Vậy S={$0$}

    

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