Question

rút gọn  B= \(\dfrac{1}{\sqrt{x}+2}+\dfrac{3}{1-\sqrt{x}}+\dfrac{x+8}{x+\sqrt{x}-2}\)

Answer 1

Answer 2

ĐK: \(x\ge0;x\ne1\)

\(B=\dfrac{1}{\sqrt{x}+2}+\dfrac{3}{1-\sqrt{x}}+\dfrac{x+8}{x+\sqrt{x}-2}\)

\(=\dfrac{\sqrt{x}-1}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}-\dfrac{3\sqrt{x}+6}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}+\dfrac{x+8}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}\)

\(=\dfrac{x-2\sqrt{x}+1}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}\)

\(=\dfrac{\left(\sqrt{x}-1\right)^2}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}=\dfrac{\sqrt{x}-1}{\sqrt{x}+2}\)

Answer 3

Ta có: \(B=\dfrac{1}{\sqrt{x}+2}+\dfrac{3}{1-\sqrt{x}}+\dfrac{x+8}{x+\sqrt{x}-2}\)

\(=\dfrac{\sqrt{x}-1-3\sqrt{x}-6+x+8}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}\)

\(=\dfrac{x-2\sqrt{x}+1}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}\)

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