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

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

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

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

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

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

\(M=3\)

b) \(\sqrt{x}=M\)

\(\Leftrightarrow x=M^2\)

thay vào ta có: 

\(x=3^2\)

\(x=9\)

c) \(M=3\in N\)

\(\Rightarrow x=3\)

d) \(M>1\Leftrightarrow x>1\)

\(\frac{1}{1-\sqrt{x}}-1\)sao bằng\(\frac{\sqrt{x}}{\sqrt{x}-1}\)đc

1: Ta có: \(P=\dfrac{x^2-\sqrt{x}}{x+\sqrt{x}+1}-\dfrac{2x+\sqrt{x}}{\sqrt{x}}+\dfrac{2\left(x-1\right)}{\sqrt{x}-1}\)

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

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

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

a, đk: \(x\ge0,x\ne9,x\ne4\)

\(Q=\dfrac{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)-\left(\sqrt{x}+1\right)\left(\sqrt{x}-3\right)-3\sqrt{x}+3}{\left(\sqrt{x}-3\right)\left(\sqrt{x}-2\right)}\)

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

\(=\dfrac{2-\sqrt{x}}{-\left(\sqrt{x}-3\right)\left(2-\sqrt{x}\right)}=\dfrac{-1}{\sqrt{x}-3}\)

b,\(Q< -1=>\dfrac{-1}{\sqrt{x}-3}+1< 0< =>\dfrac{-1+\sqrt{x}-3}{\sqrt{x}-3}< 0\)

\(< =>\dfrac{\sqrt{x}-4}{\sqrt{x}-3}< 0\)

\(=>\left\{{}\begin{matrix}\left[{}\begin{matrix}\sqrt{x}-4>0\\\sqrt{x}-3< 0\end{matrix}\right.\\\left[{}\begin{matrix}\sqrt{x}-4< 0\\\sqrt{x}-3>0\end{matrix}\right.\end{matrix}\right.\)\(< =>\left[{}\begin{matrix}\left\{{}\begin{matrix}x>16\\x< 9\end{matrix}\right.\\\left\{{}\begin{matrix}x< 16\\x>9\end{matrix}\right.\end{matrix}\right.\)\(< =>9< x< 16\)

c, \(=>2Q=\dfrac{-2}{\sqrt{x}-3}=1+\dfrac{1}{\sqrt{x}-3}\in Z\)

\(< =>\sqrt{x}-3\inƯ\left(1\right)=\left\{\pm1\right\}\)\(=>x\in\left\{16;4\right\}\)(loại 4)

=>x=16

a) \(Q=\dfrac{\sqrt{x}+2}{\sqrt{x}-3}-\dfrac{\sqrt{x}+1}{\sqrt{x}-2}-3\dfrac{\sqrt{x}-1}{x-5\sqrt{x}+6}\) 

Ta có \(x-5\sqrt{x}+6=\left(\sqrt{x}-3\right)\left(\sqrt{x}-2\right)\)

ĐKXĐ: \(\left\{{}\begin{matrix}x\ge0\\\sqrt{x}-3>0\\\sqrt{x}-2>0\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x\ge0\\x>9\\x>2\end{matrix}\right.\) \(\Leftrightarrow x>9\)

\(Q=\dfrac{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}-2\right)}-\dfrac{\left(\sqrt{x}+1\right)\left(\sqrt{x}-3\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}-3\dfrac{\sqrt{x}-1}{\left(\sqrt{x}-3\right)\left(\sqrt{x}-2\right)}\)

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

b) \(Q< -1\Leftrightarrow\dfrac{1}{3-\sqrt{x}}< -1\) \(\Leftrightarrow\dfrac{1}{3-\sqrt{x}}+1< 0\) \(\Leftrightarrow\dfrac{4-\sqrt{x}}{3-\sqrt{x}}< 0\)

\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}4-\sqrt{x}>0\\3-\sqrt{x}< 0\end{matrix}\right.\\\left\{{}\begin{matrix}4-\sqrt{x}< 0\\3-\sqrt{x}>0\end{matrix}\right.\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x< 16\\x>9\end{matrix}\right.\\\left\{{}\begin{matrix}x>16\\x< 9\end{matrix}\right.\end{matrix}\right.\) \(\Leftrightarrow9< x< 16\)

Vậy để \(Q< -1\) thì \(S=\left\{x/9< x< 16\right\}\)

c) \(2Q\in Z\Leftrightarrow\dfrac{2}{3-\sqrt{x}}\in Z\)

\(\Rightarrow3-\sqrt{x}\inƯ\left(2\right)\)\(\Leftrightarrow\left\{{}\begin{matrix}3-\sqrt{x}=2\\3-\sqrt{x}=-2\\3-\sqrt{x}=1\\3-\sqrt{x}=-1\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=1\\x=25\\x=4\\x=16\end{matrix}\right.\)

Kết hợp với ĐKXĐ,ta có để \(2Q\in Z\) thì \(x\in\left\{16;25\right\}\)

a) ĐKXĐ: \(\left\{{}\begin{matrix}x\ge0\\x\notin\left\{9;4\right\}\end{matrix}\right.\)

Ta có: \(Q=\dfrac{\sqrt{x}+2}{\sqrt{x}-3}-\dfrac{\sqrt{x}+1}{\sqrt{x}-2}-\dfrac{3\sqrt{x}-3}{x-5\sqrt{x}+6}\)

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

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

\(=\dfrac{-1}{\sqrt{x}-3}\)

c) Để 2Q là số nguyên thì \(-2⋮\sqrt{x}-3\)

\(\Leftrightarrow\sqrt{x}-3\in\left\{1;-1;2;-2\right\}\)

\(\Leftrightarrow\sqrt{x}\in\left\{4;2;5;1\right\}\)

\(\Leftrightarrow x\in\left\{16;25;1\right\}\)

a) \(P=\dfrac{3x+3\sqrt{x}-9}{x+\sqrt{x}-2}+\dfrac{\sqrt{x}+3}{\sqrt{x}+2}-\dfrac{\sqrt{x}-2}{\sqrt{x}-1}\left(x\ge0,x\ne1\right)\)

\(=\dfrac{3x+3\sqrt{x}-9}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}+\dfrac{\sqrt{x}+3}{\sqrt{x}+2}-\dfrac{\sqrt{x}-2}{\sqrt{x}-1}\)

\(=\dfrac{3x+3\sqrt{x}-9+\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)-\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}\)

\(=\dfrac{3x+5\sqrt{x}-8}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}=\dfrac{\left(\sqrt{x}-1\right)\left(3\sqrt{x}+8\right)}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}=\dfrac{3\sqrt{x}+8}{\sqrt{x}+2}\)

b) \(P=\dfrac{3\sqrt{x}+8}{\sqrt{x}+2}=\dfrac{3\sqrt{x}+6+2}{\sqrt{x}+2}=3+\dfrac{2}{\sqrt{x}+2}\)

Để \(P\in Z\Rightarrow2⋮\sqrt{x}+2\Rightarrow\sqrt{x}+2=2\left(\sqrt{x}+2\ge2\right)\)

\(\Rightarrow x=0\)

c) Ta có: \(\sqrt{x}\ge0\Rightarrow\sqrt{x}+2\ge2\Rightarrow\dfrac{2}{\sqrt{x}+2}\le1\Rightarrow3+\dfrac{2}{\sqrt{x}+2}\le4\)

\(\Rightarrow P_{max}=4\) khi \(x=0\)