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ĐKXĐ: \(x\ge0;x\ne4;x\ne9\)
a) \(A=\frac{2\sqrt{x}-9}{x-2\sqrt{x}-3\sqrt{x}+6}-\frac{\sqrt{x}+3}{\sqrt{x}-2}+\frac{2\sqrt{x}+1}{\sqrt{x}-3}\)
\(A=\frac{2\sqrt{x}-9-\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)+\left(\sqrt{x}-2\right)\left(2\sqrt{x}+1\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}\)
\(A=\frac{x-\sqrt{x}-2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}=\frac{\left(\sqrt{x}+1\right)\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}-2\right)}=\frac{\sqrt{x}+1}{\sqrt{x}-3}\)
\(ĐKXĐ:\)
\(\hept{\begin{cases}x-9\ne0\\\sqrt{x}-2\ne0\\\sqrt{x}+3\ne0;x\ge0\end{cases}}\)
\(\Leftrightarrow\hept{\begin{cases}x\ne9\\x\ne4\\x\ge0\end{cases}}\)
Vậy...................................................
\(A=\left(\frac{x-3\sqrt{x}}{x-9}-1\right):\left(\frac{9-x}{x+\sqrt{x}-6}+\frac{\sqrt{x}-3}{\sqrt{x}-2}-\frac{\sqrt{x}+2}{\sqrt{x}+3}\right)\)
\(=\left(\frac{\sqrt{x}\left(\sqrt{x}-3\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}-1\right):\left(\frac{9-x}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+3\right)}+\frac{\sqrt{x}-3}{\sqrt{x}-2}-\frac{\sqrt{x}+2}{\sqrt{x}+3}\right)\)
\(=\frac{\sqrt{x}-\sqrt{x}-3}{\left(\sqrt{x}+3\right)}:\left(\frac{9-x}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+3\right)}+\frac{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+3\right)}-\frac{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-2\right)}\right)\)
\(=\frac{-3}{\sqrt{x}+3}:\left(\frac{9-x}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+3\right)}+\frac{x-9}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+3\right)}-\frac{x-4}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-2\right)}\right)\)
\(=\frac{-3}{\sqrt{x}+3}:\frac{9-x+x-9-x+4}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+3\right)}\)
\(=\frac{-3}{\sqrt{x}+3}:\frac{-x+4}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+3\right)}\)
\(=\frac{-3}{\sqrt{x}+3}.\frac{\left(\sqrt{x}-2\right)\left(\sqrt{x}+3\right)}{4-x}\)
\(=\frac{3\left(2-\sqrt{x}\right)}{\left(2-\sqrt{x}\right)\left(2+\sqrt{x}\right)}\)
\(=\frac{3}{\left(2+\sqrt{x}\right)}\)
M = \(\frac{2\sqrt{x}-9x}{x-5\sqrt{x}+6}-\frac{\sqrt{x}+3}{\sqrt{x}-2}-\frac{2\sqrt{x}+1}{3-\sqrt{x}}\)
=\(\frac{2\sqrt{x}-9}{x-5\sqrt{x}+6}-\frac{\left(\sqrt{x}+3\right)\left(3-\sqrt{x}\right)+\left(\sqrt{x}-2\right)\left(2\sqrt{x}+1\right)}{\left(\sqrt{x}-2\right)\left(3-\sqrt{x}\right)}\)
=\(\frac{2\sqrt{x}-9}{x-5\sqrt{x}+6}+\frac{9-x+2x-3\sqrt{x}}{x-5\sqrt{x}+6}\)
=\(\frac{x-\sqrt{x}}{x-5\sqrt{x}+6}\)
a. ĐK: \(\left\{{}\begin{matrix}x\ge0\\x\ne1\end{matrix}\right.\)
\(A=\frac{2\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}+\frac{2\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}-\frac{5-\sqrt{x}}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\\ =\frac{2\sqrt{x}+2+2\sqrt{x}-2-5+\sqrt{x}}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\\ =\frac{5\sqrt{x}-5}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\\ =\frac{5\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}=\frac{5}{\sqrt{x}+1}\)
b. Khi x=9 ta có:
\(A=\frac{5}{\sqrt{9}+1}=\frac{5}{3+1}=\frac{5}{4}\)
c. Để A ∈ Z thì \(5⋮\sqrt{x}+1hay\sqrt{x}+1\inƯ\left(5\right)\)
Ta có bảng sau:
Vậy ...................
ĐKXĐ : \(x\ne1\)
a) \(A=\frac{2}{\sqrt{x}-1}+\frac{2}{\sqrt{x}+1}-\frac{5-\sqrt{x}}{x-1}\)
\(A=\frac{2\left(\sqrt{x}+1\right)+2\left(\sqrt{x}-1\right)-5+\sqrt{x}}{x-1}\)
\(A=\frac{-5+5\sqrt{x}}{x-1}\)
\(A=\frac{5\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
\(A=\frac{5}{\sqrt{x}+1}\)
b) Khi \(x=9\)ta có \(A=\frac{5}{\sqrt{9}+1}=\frac{5}{4}\)
c) \(A\in Z\Leftrightarrow5⋮\sqrt{x}+1\)
\(\Leftrightarrow\sqrt{x}+1\inƯ\left(5\right)=\left\{1;5\right\}\)( vì \(\sqrt{x}+1\ge1\forall x\))
\(\Leftrightarrow x\in\left\{0;16\right\}\)( thỏa )
Vậy....