Question

Cho biểu thức: \(P=\left(\dfrac{2\sqrt{x}}{x\sqrt{x}+\sqrt{x}-x-1}-\dfrac{1}{\sqrt{x}-1}\right):\left(1+\dfrac{\sqrt{x}}{x+1}\right)\)

a) Rút gọn P?

b) Tính x để P\(\le\)0?

Answer 1

a: \(P=\left(\dfrac{2\sqrt{x}}{\left(x+1\right)\left(\sqrt{x}-1\right)}-\dfrac{1}{\sqrt{x}-1}\right):\dfrac{x+\sqrt{x}+1}{x+1}\)

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

\(=\dfrac{-\left(\sqrt{x}-1\right)^2}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}=\dfrac{-\sqrt{x}+1}{x+\sqrt{x}+1}\)

b: Để P<=0 thì \(-\sqrt{x}+1< =0\)

\(\Leftrightarrow\sqrt{x}-1>=0\)

hay x>=1

Kết hợp ĐKXĐ, ta được: x>1