Question

Câu 2 Cho biểu thức :P=\(\dfrac{1}{\sqrt{x}-1}+\dfrac{1}{\sqrt{x}+1}+\dfrac{2x}{x-1}\)với \(x\ge0\)và\(x\ne1\)

a Rút gọn biểu thức P

b Tìm x để \(P>-\dfrac{1}{2}\)

Answer 1

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

Answer 2

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

b: Để P>-1/2 thì P+1/2>0

\(\Leftrightarrow\dfrac{2\sqrt{x}}{\sqrt{x}-1}+\dfrac{1}{2}>0\)

=>\(\dfrac{5\sqrt{x}-1}{2\left(\sqrt{x}-1\right)}>0\)

=>1/5<căn x<1

=>1/25<x<1