Question

Cho bt \(B=\frac{x^5+x^2}{x^3-x^2+x}\)      \(vs\)\(x\ne0\)

a,Rút gọn B

b,Tìm x dể B=0

Tìm Gtnn của bt B

Answer 1

a,\(B=\frac{x^5+x^2}{x^3-x^2+x}\left(ĐKXĐ:x\ne0\right)\)

\(\Rightarrow B=\frac{x^2\left(x^3+1\right)}{x\left(x^2-x+1\right)}=\frac{x\left(x+1\right)\left(x^2-x+1\right)}{x^2-x+1}=x^2+x\)

b,Để \(B=0\Rightarrow x^2+x=0\Leftrightarrow x\left(x+1\right)=0\Rightarrow\orbr{\begin{cases}x=0\\x=-1\end{cases}}\)

c,\(B=x^2+x=x^2+2.\frac{1}{2}x+\frac{1}{4}-\frac{1}{4}=\left(x+\frac{1}{2}\right)^2+\left(-\frac{1}{4}\right)\ge-\frac{1}{4}\)

Vậy MIn = -1/4 <=> x = -1/2