ĐK:  \(x\ge0\)

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

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

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

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

Nhận thấy \(P\ge0\)nên P lớn nhất khi 1/P nhỏ nhất

=>  \(\frac{x-\sqrt{x}+1}{\sqrt{x}}\)nhỏ nhất

Ta có:  \(\frac{x-\sqrt{x}+1}{\sqrt{x}}=\sqrt{x}-1+\frac{1}{\sqrt{x}}\ge2\sqrt{\sqrt{x}.\frac{1}{\sqrt{x}}}-1=1\)

=> MIN 1/P = 1  khi x = 1

Vậy MAX P = 1 khi x = 1

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

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

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

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

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

\(=\frac{\sqrt{x}}{x-\sqrt{x}+1}=\frac{\sqrt{x}}{\left(\sqrt{x}-1^2\right)+\sqrt{x}}\le1\)

\(Pmax=1\Leftrightarrow\left(\sqrt{x}-1\right)^2=0\Leftrightarrow\sqrt{x}-1=0\Leftrightarrow x=1\)

kb nha.

\(x\ge16\)

\(\Rightarrow\)\(\sqrt{x}\ge4\)

\(\Rightarrow\)\(\sqrt{x}-2\ge2\)

\(\Rightarrow\)\(\frac{1}{\sqrt{x}-2}\le\frac{1}{2}\)

\(\Rightarrow\)\(\frac{3}{\sqrt{x}-2}\le\frac{3}{2}\)

Dấu "=" xảy ra khi x = 16

Vậy MAX \(\frac{3}{\sqrt{x}-2}=\frac{3}{2}\)khi x = 16

\(A=\frac{2\sqrt{x}-9}{x-5\sqrt{x}+6}-\frac{\sqrt{x}+3}{\sqrt{x}-2}-\frac{2\sqrt{x}+1}{3-\sqrt{x}}\)

\(=\frac{2\sqrt{x}-9}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}-\frac{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}+\frac{2\sqrt{x}\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}-2\right)}\)

\(=\frac{2\sqrt{x}-9-\left(x-9\right)+2x-4\sqrt{x}}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}\)

\(=\frac{x-2\sqrt{x}}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}\)

\(=\frac{\sqrt{x}}{\sqrt{x-3}}\)

sai r bạn ơi