Question

Cho B=\(\left(\dfrac{\sqrt{x}}{\sqrt{x}+4}+\dfrac{4}{\sqrt{x}-4}\right):\dfrac{x+16}{\sqrt{x}+2}\)

a Tìm điều kiện của x để B xác định

b Hãy rút gọn biểu thức B

Answer 1

a) ĐKXĐ: \(x\ge0;x\ne16\)

b) \(\left(\dfrac{\sqrt{x}}{\sqrt{x}+4}+\dfrac{4}{\sqrt{x}-4}\right):\dfrac{x+16}{\sqrt{x}+2}\)

\(\Leftrightarrow\dfrac{\sqrt{x}\left(\sqrt{x}-4\right)+4\left(\sqrt{x}+4\right)}{x-16}:\dfrac{x+16}{\sqrt{x}+2}\)

\(\Leftrightarrow\dfrac{x-4\sqrt{x}+4\sqrt{x}+16}{x-16}.\dfrac{\sqrt{x}+2}{x+16}\)

\(\Leftrightarrow\dfrac{\sqrt{x}+2}{x-16}\)

Answer 2

a) ĐKXĐ: \(x\ge0,x\ne16\)

b) \(B=\left(\dfrac{\sqrt{x}}{\sqrt{x}+4}+\dfrac{4}{\sqrt{x}-4}\right):\dfrac{x+16}{\sqrt{x}+2}\)

\(B=\dfrac{\left(\sqrt{x}-4\right)\sqrt{x}+4\left(\sqrt{x}+4\right)}{\left(\sqrt{x}+4\right)\left(\sqrt{x}-4\right)}.\dfrac{\sqrt{x}+2}{x+16}\\ B=\dfrac{x-4\sqrt{x}+4\sqrt{x}+16}{x-16}.\dfrac{\sqrt{x}+2}{x+16}\\ B=\dfrac{1}{x-16}.\left(\sqrt{x}+2\right)\\ B=\dfrac{\sqrt{x}+2}{x-16}\)