Question

cho bt

A=\(\dfrac{x\sqrt{x}-1}{x-\sqrt{x}}\)-\(\dfrac{x\sqrt{x}+1}{x+\sqrt{x}}\)+\(\dfrac{x+1}{\sqrt{x}}\)

a, rút gọn P

b, tìm x để A=\(\dfrac{1}{2}\)

c, chứng minh rằng A< 0

Answer 1

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

\(=\dfrac{x+\sqrt{x}+1-x+\sqrt{x}-1+x+1}{\sqrt{x}}\)

\(=\dfrac{x+2\sqrt{x}+1}{\sqrt{x}}\)

b: Để A=1/2 thì \(\dfrac{\left(\sqrt{x}+1\right)^2}{\sqrt{x}}=\dfrac{1}{2}\)

\(\Leftrightarrow2x+4\sqrt{x}+2-\sqrt{x}=0\)

\(\Leftrightarrow2x+3\sqrt{x}+2=0\)(1)

Đặt \(\sqrt{x}=a\)(a>=0)

(1) trở thành \(2a^2+3a+2=0\)

\(\Delta=3^2-4\cdot2\cdot2=9-16=-7< 0\)

Do đó: (1) vô nghiệm