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a) đk: \(\hept{\begin{cases}x\ge0\\x\ne1\end{cases}}\)
Ta có:
\(P=\left(\frac{3x-\sqrt{9x}-3}{x+\sqrt{x}-2}+\frac{1}{\sqrt{x}-1}+\frac{1}{\sqrt{x}+2}\right)\div\frac{1}{x-1}\)
\(P=\frac{3x-3\sqrt{x}-3+\sqrt{x}+2+\sqrt{x}-1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}\cdot\left(x-1\right)\)
\(P=\frac{3x-\sqrt{x}-2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}\cdot\left(\sqrt{x}-1\right)\cdot\left(\sqrt{x}+1\right)\)
\(P=\frac{\left(\sqrt{x}-1\right)\left(3\sqrt{x}+2\right)\left(\sqrt{x}+1\right)}{\sqrt{x}+2}\)
\(P=\frac{\left(3\sqrt{x}+2\right)\left(x-1\right)}{\sqrt{x}+2}\)
a: \(A=\dfrac{x+\sqrt{x}+1+\sqrt{x}+2+\sqrt{x}-1}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}\cdot\dfrac{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}{1}\)
\(=\dfrac{x+3\sqrt{x}+2}{\sqrt{x}+2}\cdot\dfrac{\sqrt{x}+1}{1}=\left(\sqrt{x}+1\right)^2\)
b: Để 1/A là số tự nhiên thì \(\sqrt{x}+1\) là số tự nhiên
hay \(x=k^2\left(k\in N;k\ne1\right)\)
a: \(A=\dfrac{\left(\sqrt{a}-\sqrt{b}\right)^2}{\sqrt{a}-\sqrt{b}}-\dfrac{\sqrt{ab}\left(\sqrt{a}+\sqrt{b}\right)}{\sqrt{ab}}\)
\(=\sqrt{a}-\sqrt{b}-\sqrt{a}-\sqrt{b}=-2\sqrt{b}\)
b: \(B=\dfrac{2\sqrt{x}-x-x-\sqrt{x}-1}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\cdot\dfrac{x+\sqrt{x}+1}{x-1}\)
\(=\dfrac{-2x+\sqrt{x}-1}{\sqrt{x}-1}\cdot\dfrac{1}{x-1}\)
c: \(C=\dfrac{x-9-x+3\sqrt{x}}{x-9}:\left(\dfrac{3-\sqrt{x}}{\sqrt{x}-2}+\dfrac{\sqrt{x}-2}{\sqrt{x}+3}+\dfrac{x-9}{x+\sqrt{x}-6}\right)\)
\(=\dfrac{3\left(\sqrt{x}-3\right)}{x-9}:\dfrac{9-x+x-4\sqrt{x}+4+x-9}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-2\right)}\)
\(=\dfrac{3}{\sqrt{x}+3}\cdot\dfrac{\left(\sqrt{x}+3\right)\left(\sqrt{x}-2\right)}{x-4\sqrt{x}+4}\)
\(=\dfrac{3}{\sqrt{x}-2}\)
\(M=\frac{3x+3\sqrt{x}-3}{x+\sqrt{x}-2}-\frac{\sqrt{x}+1}{\sqrt{x}+2}+\frac{\sqrt{x}-2}{\sqrt{x}}.\left(\frac{1}{1-\sqrt{x}}-1\right)\)
\(M=\frac{3x+3\sqrt{x}-3}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}-\frac{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}\) \(+\frac{\sqrt{x}-2}{\sqrt{x}}.\frac{\sqrt{x}}{\sqrt{x}-1}\)
\(M=\frac{3x+3\sqrt{x}-3}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}-\frac{x-1}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}\) \(+\frac{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}\)
\(M=\frac{3x+3\sqrt{x}-3-x+1+x-4}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}\)
\(M=\frac{3x+3\sqrt{x}-6}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}\)
\(M=\frac{3\left(x+\sqrt{x}-2\right)}{x+\sqrt{x}-2}\)
\(M=3\)
Sau khi rút gọn ,ta được A=\(\left(\sqrt{x}+1\right)^2\)\(\Rightarrow\frac{1}{A}=\frac{1}{\left(\sqrt{x}+1\right)^2}\). Để \(\frac{1}{A}\)là số tự nhiên \(\hept{\begin{cases}\left(\sqrt{x}+1\right)^2>0\\\left(\sqrt{x}+1\right)^2\in U\left(1\right)\end{cases}}\) \(\Rightarrow x=0\)( thỏa mãn ĐK).