x^2 + 2x x – 5 50 – x A= ———————- + —————- + ————————- 2x

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1. a, $A=\dfrac{x^2+2x}{2x+10}+\dfrac{x-5}{x}+\dfrac{50-x}{2x(x+5)}$     ĐK: $x\neq-5;x\neq0$

       $A=\dfrac{x^2+2x}{2(x+5)}+\dfrac{x-5}{x}+\dfrac{50-x}{2x(x+5)}$

       $A=\dfrac{x(x^2+2x)+2(x+5)(x-5)+50-x}{2x(x+5)}$

       $A=\dfrac{x^3+2x^2+2x^2-50+50-x}{2x(x+5)}$

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

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

       $A=\dfrac{x^2+4x-1}{2(x+5)}$

   Vậy $A=\dfrac{x^2+4x-1}{2(x+5)}$ với $x\neq-5;x\neq0$

   b) Ta có: $x^2-x=0$

   $⇔x(x-1)=0$

   \(⇔\left[ \begin{array}{l}x=0\\x-1=0\end{array} \right.\)

   \(⇔\left[ \begin{array}{l}x=0(ktm)\\x=1(tm)\end{array} \right.\)

   Thay $x=1$ vào A ta được:

    $A=\dfrac{1^2+4.1-1}{2(1+5)}=\dfrac{1}{3}$

   Vậy $A=\dfrac{1}{3}$ khi $x=1$

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