Question

P = \(\left(\frac{\sqrt{b}}{\sqrt{b}+2}-\frac{\sqrt{b}}{\sqrt{b}-2}+\frac{4\sqrt{b}-1}{b-4}\right):\frac{1}{\sqrt{b}+2}\)

a) rút gọn P b) Tính P tại b = 6 + \(4\sqrt{2}\)

Answer 1

a) ĐKXĐ: \(\left\{{}\begin{matrix}b\ge0\\b\ne4\end{matrix}\right.\)

\(P=\left(\frac{\sqrt{b}}{\sqrt{b}+2}-\frac{\sqrt{b}}{\sqrt{b}-2}+\frac{4\sqrt{b}-1}{b-4}\right):\frac{1}{\sqrt{b}+2}\)

\(=\frac{\sqrt{b}\left(\sqrt{b}-2\right)-\sqrt{b}\left(\sqrt{b}+2\right)+4\sqrt{b}-1}{\left(\sqrt{b}-2\right)\left(\sqrt{b}+2\right)}:\frac{1}{\sqrt{b}+2}\)

\(=\frac{-\left(\sqrt{b}+2\right)}{\left(\sqrt{b}+2\right)\left(\sqrt{b}-2\right)}\)\(=\frac{-1}{\sqrt{b}-2}\)

b) Tại b = \(6+4\sqrt{2}\)

\(P=\frac{-1}{\sqrt{6+4\sqrt{2}}-2}=\frac{-1}{\sqrt{\left(2+\sqrt{2}\right)^2}-2}\)\(=\frac{-1}{\left|2+\sqrt{2}\right|-2}=\frac{-1}{\sqrt{2}}\)