Lời giải:
ĐK: \(x\geq 0; x\neq 1\)
\(A=\frac{x+\sqrt{x}}{x\sqrt{x}+x+\sqrt{x}+1}:\frac{\sqrt{x}-1}{x+1}\)
\(=\frac{\sqrt{x}(\sqrt{x}+1)}{x(\sqrt{x}+1)+(\sqrt{x}+1)}.\frac{x+1}{\sqrt{x}-1}\)
\(=\frac{\sqrt{x}(\sqrt{x}+1)}{(\sqrt{x}+1)(x+1)}.\frac{x+1}{\sqrt{x}-1}=\frac{\sqrt{x}}{\sqrt{x}-1}\)
Với \(x=4+2\sqrt{3}=3+1+2\sqrt{3}=(\sqrt{3}+1)^2\)
\(\Rightarrow \sqrt{x}=\sqrt{3}+1\)
Do đó: \(A=\frac{\sqrt{3}+1}{\sqrt{3}+1-1}=\frac{\sqrt{3}+1}{\sqrt{3}}\)
Mặt khác: \(A=\frac{\sqrt{x}}{\sqrt{x}-1}=\frac{\sqrt{x}-1+1}{\sqrt{x}-1}=1+\frac{1}{\sqrt{x}-1}\)
Để \(A\in\mathbb{Z}\Rightarrow \frac{1}{\sqrt{x}-1}\in\mathbb{Z}\Rightarrow 1\vdots \sqrt{x}-1\)
\(\Rightarrow \sqrt{x}-1\in\left\{\pm 1\right\}\)
\(\Rightarrow \sqrt{x}\left\{0;2\right\}\Rightarrow x\in\left\{0;4\right\}\) (đều thỏa mãn)
A = \(\dfrac{x+\sqrt{x}}{x\sqrt{x}+x+\sqrt{x}+1}:\dfrac{\sqrt{x}-1}{x+1}\)
= \(\dfrac{\sqrt{x}\left(\sqrt{x}+1\right)}{x\left(\sqrt{x}+1\right)+\left(\sqrt{x}+1\right)}.\dfrac{x+1}{\sqrt{x}-1}\)
= \(\dfrac{\sqrt{x}\left(\sqrt{x}+1\right)}{\left(\sqrt{x}+1\right)\left(x+1\right)}.\dfrac{x+1}{\sqrt{x}-1}\)
= \(\dfrac{\sqrt{x}}{x+1}.\dfrac{x+1}{\sqrt{x}-1}\)
= \(\dfrac{\sqrt{x}}{\sqrt{x}-1}\)
Với x = 4 + \(2\sqrt{3}\) ta có:
A = \(\dfrac{\sqrt{4+2\sqrt{3}}}{\sqrt{4+2\sqrt{3}}-1}\)
=\(\dfrac{\sqrt{3+2\sqrt{3}+1}}{\sqrt{3+2\sqrt{3}+1}-1}\)
= \(\dfrac{\sqrt{\left(\sqrt{3}+1\right)^2}}{\sqrt{\left(\sqrt{3}+1\right)^2}-1}\)
= \(\dfrac{\sqrt{3}+1}{\sqrt{3}+1-1}=\dfrac{\sqrt{3}+1}{\sqrt{3}}\)
Ta có:
A = \(\dfrac{\sqrt{x}}{\sqrt{x}-1}=\)\(\dfrac{\sqrt{x}-1+1}{\sqrt{x}-1}=\dfrac{\sqrt{x}-1}{\sqrt{x}-1}+\dfrac{1}{\sqrt{x}-1}=1+\dfrac{1}{\sqrt{x}-1}\)
ĐKXĐ là: \(\sqrt{x}-1\ne0\)
\(\Leftrightarrow\sqrt{x}\ne1\)
\(\Leftrightarrow x\ne1\)
Để biểu thức A nhận giá trị nguyên thì:
1 \(⋮\left(\sqrt{x}-1\right)\)
\(\Rightarrow\) \(\left(\sqrt{x}-1\right)\inƯ\left(1\right)\)
\(\Rightarrow\left\{{}\begin{matrix}\sqrt{x}-1=1\\\sqrt{x}-1=-1\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\sqrt{x}=2\\\sqrt{x}=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=4\\x=0\end{matrix}\right.\) (Thỏa mãn điều kiện)
Vậy biểu thức A nhận giá trị nguyên khi x = 4 hoặc x = 0
