3B:
a) Ta có: \(B=\left(\dfrac{\sqrt{x}}{x-4}+\dfrac{1}{\sqrt{x}-2}\right):\dfrac{\sqrt{x}+2}{x-4}\)
\(=\dfrac{\sqrt{x}+\sqrt{x}+2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\cdot\dfrac{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}{\sqrt{x}+2}\)
\(=\dfrac{2\sqrt{x}+2}{\sqrt{x}+2}\)
b) Ta có: C=A(B-2)
\(=\dfrac{\sqrt{x}+2}{\sqrt{x}-2}\cdot\dfrac{2\sqrt{x}+2-2\sqrt{x}-4}{\sqrt{x}+2}\)
\(=\dfrac{-2}{\sqrt{x}-2}\)
Để C nguyên thì \(-2⋮\sqrt{x}-2\)
\(\Leftrightarrow\sqrt{x}-2\in\left\{-2;-1;1;2\right\}\)
\(\Leftrightarrow\sqrt{x}\in\left\{0;1;3;4\right\}\)
hay \(x\in\left\{0;1;9;16\right\}\)
3A:
a) Ta có: \(A=\left(\dfrac{1}{\sqrt{x}-1}+\dfrac{\sqrt{x}}{x-1}\right):\left(\dfrac{\sqrt{x}}{\sqrt{x}-1}-1\right)\)
\(=\left(\dfrac{\sqrt{x}+1+\sqrt{x}}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\right):\left(\dfrac{\sqrt{x}-\sqrt{x}+1}{\sqrt{x}-1}\right)\)
\(=\dfrac{2\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\cdot\dfrac{\sqrt{x}-1}{1}\)
\(=\dfrac{2\sqrt{x}+1}{\sqrt{x}+1}\)
b) Ta có: \(M=A\cdot\dfrac{\sqrt{x}+1}{2\sqrt{x}+1}+\dfrac{x-\sqrt{x}-5}{\sqrt{x}+3}\)
\(=\dfrac{2\sqrt{x}+1}{\sqrt{x}+1}\cdot\dfrac{\sqrt{x}+1}{2\sqrt{x}+1}+\dfrac{x-\sqrt{x}-5}{\sqrt{x}+3}\)
\(=\dfrac{x+\sqrt{x}-5}{\sqrt{x}+3}\)
Để M nguyên thì \(x+\sqrt{x}-5⋮\sqrt{x}+3\)
\(\Leftrightarrow x+3\sqrt{x}-2\sqrt{x}-6+1⋮\sqrt{x}+3\)
\(\Leftrightarrow\sqrt{x}+3\inƯ\left(1\right)\)
\(\Leftrightarrow\sqrt{x}+3\in\left\{1;-1\right\}\)
\(\Leftrightarrow\sqrt{x}\in\left\{-2;-4\right\}\)(vô lý)
Vậy: Không có giá trị nào của x để M có giá trị nguyên
