ta có \((\sqrt{a}-\sqrt{b})^2=a-2\sqrt{ab}+b\)

\(=a-b-2\sqrt{ab}+2b\)

\(=a-b-2\sqrt{b}\left(\sqrt{a}-\sqrt{b}\right)\)

VÌ a>b>0 NÊN \(\sqrt{a}-\sqrt{b}>0\)

suy ra : \(a-b-2\sqrt{b}\left(\sqrt{a}-\sqrt{b}\right)< a-b\)

\(\Leftrightarrow\left(\sqrt{a}-\sqrt{b}\right)^2< \left(\sqrt{a-b}\right)^2\)

VẬY \(\sqrt{a}-\sqrt{b}< \sqrt{a-b}\left(đ.p.c.m\right)\)

dien ak 

nhân cả 2 vế với 2 sau đó đưa về hằng đẳng thức dạng A² + B² + C² =0 suy ra A=B=C=0

Ta có:\(a+b\ge2\sqrt{ab}\Rightarrow a-2\sqrt{ab}+b\ge0\Rightarrow\left(\sqrt{a}\right)^2-2\sqrt{ab}+\left(\sqrt{b}\right)^2\ge0\Rightarrow\left(\sqrt{a}+\sqrt{b}\right)^2\ge0\left(LĐ\right)\)Dấu "=" xảy ra <=> a = b

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

\(P=\left[-\frac{4\sqrt{x}}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\right]\left(-\frac{\sqrt{x}}{2}+\frac{1}{2\sqrt{x}}\right)^2\)

\(P=\left[-\frac{4\sqrt{x}}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\right]\left(\frac{1}{4x}+\frac{1}{4}-\frac{1}{2}\right)\)

\(P=-\frac{4\sqrt{x}.\left(\frac{1}{4x}-\frac{1}{2}+\frac{x}{4}\right)}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\)

\(P=-\frac{4.\frac{x^2-2x+1}{4x}.\sqrt{x}}{\left(\sqrt{x}+1\right)\left(\sqrt{x}+1\right)}\)

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

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

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

thôi ko cần nx đâu,mình làm được rồi,cảm ơn các bạn nha!!!