\(M=\frac{x\sqrt{x}-1}{x-\sqrt{x}}-\frac{x\sqrt{x}+1}{x+\sqrt{x}}\left(ĐK:x\ge0;x\ne1\right)\)
\(=\frac{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}{\sqrt{x}\left(\sqrt{x}-1\right)}-\frac{\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)}{\sqrt{x}\left(\sqrt{x}+1\right)}\)
\(=\frac{x+\sqrt{x}+1}{\sqrt{x}}-\frac{x-\sqrt{x}+1}{\sqrt{x}}\)
\(=\frac{x+\sqrt{x}+1-x+\sqrt{x}-1}{\sqrt{x}}\)
\(=\frac{2\sqrt{x}}{\sqrt{x}}=2\)
\(A=M:N\left(ĐK:x\ge0;x\ne2\right)\)
\(=2:\frac{x+2}{x-2}\)
\(=\frac{2\left(x-2\right)}{\left(x+2\right)}=\frac{2x-4}{x+2}=\frac{2\left(x+2\right)-6}{x+2}=2-\frac{6}{x+2}\)
Vì: \(x\ge0\)
=> \(x+2\ge2\)
=> \(\frac{6}{x+2}\le\frac{6}{2}=3\)
=> \(-\frac{6}{x+2}\ge-3\)
=> \(2-\frac{6}{x+2}\ge2-3=-1\)
Dấu "=" xảy ra khi x=0(tm)
Vậy..................
