a)Ta có:
\(x^5+x^2=x^5-x^4+x^3+x^4-x^3+x^2\)
\(=x^2\left(x^3-x^2+x\right)+x\left(x^3-x^2+x\right)\)
\(=\left(x^2+x\right)\left(x^3-x^2+x\right)\)
Thay vào A ta có:\(A=\frac{x^5+x^2}{x^3-x^2+x}=\frac{\left(x^2+x\right)\left(x^3-x^2+x\right)}{x^3-x^2+x}=x^2+x\)
b)\(A-\left|A\right|=0\Leftrightarrow x^2+x-\left|x^2+x\right|=0\)
\(\left|x^2+x\right|=x^2+x\)\(\Leftrightarrow\orbr{\begin{cases}x^2+x=x^2+x\\x^2+x=-x^2-x\end{cases}}\)
giải tiếp chắc dễ
c)\(A=x^2+x\)\(=x^2-x+\frac{1}{4}-\frac{1}{4}\)
\(=\left(x+\frac{1}{2}\right)^2-\frac{1}{4}\ge-\frac{1}{4}\)
Dấu = khi \(\left(x+\frac{1}{2}\right)^2=0\Leftrightarrow x+\frac{1}{2}=0\Leftrightarrow x=-\frac{1}{2}\)
Vậy MinA=\(-\frac{1}{4}\Leftrightarrow x=-\frac{1}{2}\)