a: ĐKXĐ: \(x\notin\left\{0;-1\right\}\)
\(P=\left(\dfrac{1}{x}+\dfrac{x}{x+1}\right):\dfrac{x}{x^2+x}\)
\(=\dfrac{x+1+x^2}{x\left(x+1\right)}\cdot\dfrac{x^2+x}{x}\)
\(=\dfrac{x^2+x+1}{x}\)
b: \(P=\dfrac{x^2+x+1}{x}\)
\(=x+1+\dfrac{1}{x}\)
Ta có: \(x+\dfrac{1}{x}>=2\cdot\sqrt{x\cdot\dfrac{1}{x}}=2\forall x>0\)
=>\(P=x+\dfrac{1}{x}+1>=2\cdot\sqrt{x\cdot\dfrac{1}{x}}+2=3\)
Dấu '=' xảy ra khi \(\left\{{}\begin{matrix}x^2=1\\x>0\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x\in\left\{1;-1\right\}\\x>0\end{matrix}\right.\)
=>x=1
Đúng 1
Bình luận (0)
