Question

\(B=\dfrac{3}{\sqrt{x+5}}+\dfrac{20-2\sqrt{x}}{x-25}\)

Answer 1

ĐKXĐ: \(x\ge0;x\ne25\)

\(B=\dfrac{3}{\sqrt{x}+5}+\dfrac{20-2\sqrt{x}}{x-25}\) (sửa đề)

\(=\dfrac{3\left(\sqrt{x}-5\right)}{\left(\sqrt{x}-5\right)\left(\sqrt{x}+5\right)}+\dfrac{20-2\sqrt{x}}{\left(\sqrt{x}-5\right)\left(\sqrt{x}+5\right)}\)

\(=\dfrac{3\sqrt{x}-15+20-2\sqrt{x}}{\left(\sqrt{x}-5\right)\left(\sqrt{x}+5\right)}\)

\(=\dfrac{\sqrt{x}+5}{\left(\sqrt{x}-5\right)\left(\sqrt{x}+5\right)}\)

\(=\dfrac{1}{\sqrt{x}-5}\)