\(B=\dfrac{\sqrt{36}+2}{\sqrt{36}-2}=\dfrac{8}{4}=2\)
\(A=\left(\dfrac{\sqrt{x}}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}+\dfrac{\sqrt{x}+2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\right):\dfrac{\sqrt{x}+2}{x-4}\)
\(=\dfrac{2\sqrt{x}+2}{x-4}:\dfrac{\sqrt{x}+2}{x-4}=\dfrac{\left(2\sqrt{x}+2\right)\left(x-4\right)}{\left(x-4\right)\left(\sqrt{x}+2\right)}\)
\(=\dfrac{2\sqrt{x}+2}{\sqrt{x}+2}\)
\(C=B\left(A-2\right)=\dfrac{\sqrt{x}+2}{\sqrt{x}-2}.\left(\dfrac{2\sqrt{x}+2}{\sqrt{x}+2}-2\right)=\dfrac{\sqrt{x}+2}{\sqrt{x}-2}.\left(\dfrac{-2}{\sqrt{x}+2}\right)=\dfrac{-2}{\sqrt{x}-2}\)
\(C\in Z\Rightarrow2⋮\sqrt{x}-2\Rightarrow\sqrt{x}-2=\left\{-2;-1;1;2\right\}\)
\(\Rightarrow\sqrt{x}=\left\{0;1;3;4\right\}\)
\(\Rightarrow x=\left\{0;1;9;16\right\}\)
a: Thay x=36 vào B, ta được:
\(B=\dfrac{6+2}{6-2}=\dfrac{8}{4}=2\)
b: Ta có: \(A=\left(\dfrac{\sqrt{x}}{x-4}+\dfrac{1}{\sqrt{x}-2}\right):\dfrac{\sqrt{x}+2}{x-4}\)
\(=\dfrac{\sqrt{x}+\sqrt{x}+2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\cdot\dfrac{\sqrt{x}-2}{1}\)
\(=\dfrac{2\sqrt{x}+2}{\sqrt{x}+2}\)
c: Ta có: \(C=B\cdot\left(A-2\right)\)
\(=\dfrac{\sqrt{x}+2}{\sqrt{x}-2}\cdot\dfrac{2\sqrt{x}+2-2\sqrt{x}-4}{\sqrt{x}+2}\)
\(=\dfrac{-2}{\sqrt{x}-2}\)
Để C nguyên thì \(\sqrt{x}-2\in\left\{-2;-1;1;2\right\}\)
\(\Leftrightarrow\sqrt{x}\in\left\{0;1;3;4\right\}\)
hay \(x\in\left\{0;1;9;16\right\}\)
