a) ĐKXĐ: \(\left\{{}\begin{matrix}x>0\\\sqrt{x}-1\ne0\\x-1\ne0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x>0\\x\ne1\end{matrix}\right.\)
b) Ta có:
\(A=\left(\dfrac{\sqrt{x}+1}{\sqrt{x}-1}-\dfrac{\sqrt{x}-1}{\sqrt{x}+1}+\dfrac{x-4\sqrt{x}-1}{x-1}\right).\dfrac{\sqrt{x}+5}{\sqrt{x}}\\ =\dfrac{\left(\sqrt{x}+1\right)^2-\left(\sqrt{x}-1\right)^2+x-4\sqrt{x}-1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}.\dfrac{\sqrt{x}+5}{\sqrt{x}}\\ =\dfrac{x+2\sqrt{x}+1-x+2\sqrt{x}-1+x-4\sqrt{x}-1}{x-1}.\dfrac{\sqrt{x}+5}{\sqrt{x}}\\ =\dfrac{x-1}{x-1}.\dfrac{\sqrt{x}+5}{\sqrt{x}}=\dfrac{\sqrt{x}+5}{\sqrt{x}}\)
c) Do \(x\) là số chính phương \(\Rightarrow\sqrt{x}\) là số nguyên
Có \(A=\dfrac{\sqrt{x}+5}{\sqrt{x}}=1+\dfrac{5}{\sqrt{x}}\)
Để \(A\) nguyên \(\Leftrightarrow\sqrt{x}\inƯ^+\left(5\right)=\left\{1;5\right\}\)
\(\Rightarrow\left[{}\begin{matrix}\sqrt{x}=1\\\sqrt{x}=5\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=1\left(L\right)\\x=25\left(TM\right)\end{matrix}\right.\)
