Question

[shunnokeshi]

P=\(\left(\frac{2x\sqrt{x}+x-\sqrt{x}}{x\sqrt{x}-1}-\frac{x+\sqrt{x}}{x-1}\right)\)). \(\frac{x-1}{2x+\sqrt{x}-1}\)+\(\frac{\sqrt{x}}{2\sqrt{x}-1}\)

a) rút gọn P

b)tìm gtnn của P

Answer 1

[Nguyễn Minh Đăng]

a) đk: \(x\ge0;x\ne\left\{\frac{1}{4};1\right\}\)

\(P=\left(\frac{2x\sqrt{x}+x-\sqrt{x}}{x\sqrt{x}-1}-\frac{x+\sqrt{x}}{x-1}\right)\cdot\frac{x-1}{2x+\sqrt{x}-1}+\frac{\sqrt{x}}{2\sqrt{x}-1}\)

\(P=\left[\frac{\left(2x+\sqrt{x}-1\right)\sqrt{x}}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}-\frac{\left(\sqrt{x}+1\right)\sqrt{x}}{x-1}\right]\cdot\frac{x-1}{2x+\sqrt{x}-1}+\frac{\sqrt{x}}{2\sqrt{x}-1}\)

\(P=\frac{\left(x-1\right)\sqrt{x}}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}-\frac{\left(\sqrt{x}+1\right)\sqrt{x}}{2x+\sqrt{x}-1}+\frac{\sqrt{x}}{2\sqrt{x}-1}\)

\(P=\frac{x+\sqrt{x}}{x+\sqrt{x}+1}-\frac{\sqrt{x}}{2\sqrt{x}-1}+\frac{\sqrt{x}}{2\sqrt{x}-1}\)

\(P=\frac{x+\sqrt{x}}{x+\sqrt{x}+1}\)

Answer 2

[Nguyễn Minh Đăng]

b) Ta có: 

\(P=\frac{x+\sqrt{x}}{x+\sqrt{x}+1}=\frac{\left(x+\sqrt{x}+1\right)-1}{x+\sqrt{x}+1}=1-\frac{1}{x+\sqrt{x}+1}\)

Mà \(x+\sqrt{x}\ge0\left(\forall x\right)\)

\(\Leftrightarrow x+\sqrt{x}+1\ge1\left(\forall x\right)\)

\(\Leftrightarrow\frac{1}{x+\sqrt{x}+1}\le1\left(\forall x\right)\)

\(\Leftrightarrow P=1-\frac{1}{x+\sqrt{x}+1}\ge0\left(\forall x\right)\)

Dấu "=" xảy ra khi: \(x+\sqrt{x}=0\Leftrightarrow x=0\)

Vậy Min(P) = 0 khi x = 0