Question

Answer 1

1) Ta có: \(B=\left(\dfrac{15-\sqrt{x}}{x-25}+\dfrac{2}{\sqrt{x}+5}\right):\dfrac{\sqrt{x}+3}{\sqrt{x}-5}\)

\(=\dfrac{15-\sqrt{x}+2\sqrt{x}-10}{\left(\sqrt{x}+5\right)\left(\sqrt{x}-5\right)}\cdot\dfrac{\sqrt{x}-5}{\sqrt{x}+3}\)

\(=\dfrac{\sqrt{x}+5}{\sqrt{x}+5}\cdot\dfrac{1}{\sqrt{x}+3}\)

\(=\dfrac{1}{\sqrt{x}+3}\)

2) Ta có: P=A+B

\(=\dfrac{2\sqrt{x}}{\sqrt{x}+3}+\dfrac{1}{\sqrt{x}+3}\)

\(=\dfrac{2\sqrt{x}+1}{\sqrt{x}+3}\)

Để P nguyên thì \(2\sqrt{x}+1⋮\sqrt{x}+3\)

\(\Leftrightarrow-5⋮\sqrt{x}+3\)

\(\Leftrightarrow\sqrt{x}+3=5\)

\(\Leftrightarrow\sqrt{x}=2\)

hay x=4(thỏa ĐK)