a: \(P=\frac{\sqrt{x}}{\sqrt{x}-1}+\frac{3}{\sqrt{x}+1}-\frac{6\sqrt{x}-4}{x-1}\)

\(=\frac{\sqrt{x}\left(\sqrt{x}+1\right)+3\left(\sqrt{x}-1\right)-6\sqrt{x}+4}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)

\(=\frac{x+\sqrt{x}+3\sqrt{x}-3-6\sqrt{x}+4}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}=\frac{x-2\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)

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

b: Thay x=9 vào P, ta được:

\(P=\frac{\sqrt9-1}{\sqrt9+1}=\frac{3-1}{3+1}=\frac24=\frac12\)

cvfbhm,

Xin lỗi em ko biết làm , em vẫn chưa lên lớp 9

\(x^2-3x+1=0\)

\(a=1,b=-3,c=1\)

\(\Delta=b^2-4ac\)\(=\left(-3\right)^2-4\times1\times1\)\(=5>0\)

\(\Rightarrow pt\)CÓ 2 NO PHÂN BIỆT

\(x_1=\frac{-b-\sqrt{\Delta}}{2a}=\frac{3-\sqrt{5}}{2}\)\(;x_2=\frac{-b+\sqrt{\Delta}}{2a}=\frac{3+\sqrt{5}}{2}\)

VẬY....

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

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

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