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

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

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

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

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

\(=\frac{x-\sqrt{x}}{x\sqrt{x}-1}\)

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

\(=\frac{\sqrt{x}}{x+\sqrt{x}+1}\)

\(A=\frac{x+\sqrt{x}+1}{\sqrt{x}+1}ĐKXĐ:x\ge0\)

\(\Rightarrow A=\frac{2+\sqrt{2}+1}{\sqrt{2}+1}=\frac{3+\sqrt{2}}{1+\sqrt{2}}=2\sqrt{2}-1\)

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

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

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

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

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

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

\(C=-A.B=-\frac{x+\sqrt{x}+1}{\sqrt{x}+1}\cdot\frac{-\sqrt{x}}{x+\sqrt{x}+1}=\frac{\sqrt{x}}{\sqrt{x}+1}\)

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

\(ĐểC\in Z\Rightarrow\frac{1}{\sqrt{x}+1}\in Z\)

\(\Rightarrow\sqrt{x}+1\in\left\{1\right\}=\left\{1;-1\right\}\)

\(\Leftrightarrow x\in\left\{0\right\}\)

Kết hợp ĐKXĐ =>...

\(\frac{x}{x+yz}+\frac{y}{y+zx}+\frac{z}{z+xy}=\frac{x}{x\left(x+y+z\right)+yz}+\frac{y}{y\left(x+y+z\right)+zx}+\frac{z}{z\left(x+y+z\right)+xy}\)

\(=\text{Σ}\frac{x}{\left(x+y\right)\left(x+z\right)}=\frac{2\left(xy+yz+xz\right)}{\left(x+y\right)\left(y+z\right)\left(x+z\right)}\)(1)

+) CM bổ đề (cái này khá hữu dụng): \(\left(x+y+z\right)\left(xy+yz+xz\right)\ge3\sqrt[3]{xyz}\cdot3\sqrt[3]{x^2y^2z^2}=9xyz\Leftrightarrow\frac{1}{9}\left(x+y+z\right)\left(xy+yz+xz\right)\ge xyz\)

Có \(\left(x+y\right)\left(y+z\right)\left(x+z\right)=\left(x+y+z\right)\left(xy+yz+xz\right)-xyz\ge\frac{8}{9}\left(x+y+z\right)\left(xy+yz+xz\right)\)

Thay vào (1)-> DPCM

Dấu = xảy ra khi x=y=z=1/3

Thx HD film