ĐK: \(x>0,x\ne1\).

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

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

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

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

\(x-\sqrt{x}+1=x-\sqrt{x}+\frac{1}{4}+\frac{3}{4}=\left(\sqrt{x}-\frac{1}{2}\right)^2+\frac{3}{4}\ge\frac{3}{4}\)

Dấu \(=\)khi \(x=\frac{1}{4}\).

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

\(\Leftrightarrow Qx-\sqrt{x}\left(Q+2\right)+Q=0\)

Với \(Q=0\Rightarrow x=0\)không thỏa mãn. 

Với \(Q\ne0\): 

Đặt \(\sqrt{x}=t>0\).

\(Qt^2-t\left(Q+2\right)+Q=0\)

\(\Delta=\left(Q+2\right)^2-4Q^2=-3Q^2+4Q+4\)

Phương trình có nghiệm suy ra \(-3Q^2+4Q+4\ge0\Leftrightarrow-\frac{2}{3}\le Q\le2\)

mà \(Q\inℤ\)\(\Rightarrow Q\in\left\{0,1,2\right\}\).

Với từng giá trị \(Q\)ta thế trực tiếp tìm giá trị của \(x\).

\(a,\frac{x+2\sqrt{x}}{x-4}\)

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

\(\frac{\sqrt{x}}{\sqrt{x}-2}\)

\(b,\frac{x+10\sqrt{x}+25}{3x+15\sqrt{x}}\)

\(\frac{\left(\sqrt{x}+5\right)^2}{3\sqrt{x}\left(\sqrt{x}+5\right)}\)

\(\frac{\sqrt{x}+5}{3\sqrt{x}}\)

\(c,\frac{\sqrt{x}-3x}{9x-6\sqrt{x}+1}\)

\(\frac{-\sqrt{x}\left(3\sqrt{x}-1\right)}{\left(3\sqrt{x}-1\right)^2}\)

\(\frac{-\sqrt{x}}{3\sqrt{x}-1}\)

hình 9 hả bn

\(\sqrt{6+\sqrt{32}}-\sqrt{11-\sqrt{72}}\)

\(=\sqrt{6+2\sqrt{8}}-\sqrt{11-2.3\sqrt{2}}\)

\(=\sqrt{\left(2+\sqrt{2}\right)^2}-\sqrt{\left(3-\sqrt{2}\right)^2}\)

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

\(\hept{\begin{cases}x^2+y^2=2x\left(1\right)\\\left(x-1\right)^3+y^3=1\left(2\right)\end{cases}}\)

Pt (1): \(x^2+y^2=2x\)

\(\Leftrightarrow x^2-2x+1=1-y^2\)

\(\Leftrightarrow\left(x-1\right)^2=1-y^2\)

\(\Leftrightarrow\left(x-1\right)^6=\left(1-y^2\right)^3\)(*)

PT(2): \(\left(x-1\right)^3+y^3=1\)

\(\Leftrightarrow\left(x-1\right)^3=1-y^3\)

\(\Leftrightarrow\left(x-1\right)^6=\left(1-y^3\right)^2\)(**)

Từ (*) và (**) \(\Rightarrow\left(1-y^2\right)^3=\left(1-y^3\right)^2\)

\(\Leftrightarrow\left(1-y\right)^3\left(1+y\right)^3-\left(1-y\right)^2\left(1+y+y^2\right)^2=0\)

\(\Leftrightarrow\left(1-y\right)^2\left[\left(1-y\right)\left(1+y\right)^3-\left(1+y+y^2\right)^2\right]=0\)

\(\Leftrightarrow\left(1-y\right)^2\left(-2y^4-4y^3-3y^2\right)=0\)

\(\Leftrightarrow-y^2\left(1-y\right)^2\left(2y^2+4y+3\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}y^2=0\\\left(1-y\right)^2=0\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}y=0\\y=1\end{cases}}\)

Thay y = 0 vào pt (2) ta đc

\(\left(x-1\right)^3=1\)

\(\Leftrightarrow x=2\)

Thay y = 1 vào pt(2) ta đc

\(\left(x-1\right)^3=0\)

\(\Leftrightarrow x=1\)

Vậy hpt có nghiệm (x,y) là (2,0);(1,1)