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 $x=1$