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:
| \(\sqrt{x}+1\) | 1 | -1 | 5 | -5 |
| \(\sqrt{x}\) | 0 | -2 | 4 | -6 |
| \(x\) | 0 (loại) | loại | 16 | loại |
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....
