Question

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

a.Rút gọn

b.Tính giá trị của B tại \(a=6+4\sqrt{2}\)

Answer 1

không có ĐK hửm???

Answer 2

a)\(ĐK:a\ge0\)

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

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

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

\(=\frac{a-2\sqrt{a}-a-2\sqrt{a}+4\sqrt{a}-1}{a-4}:\frac{1}{\sqrt{a}+2}\)\(\left(a\ge0\right)\)

\(=\frac{-1}{a-4}:\frac{1}{\sqrt{a}+2}=-\frac{\sqrt{a}+2}{a-4}=-\frac{1}{\sqrt{a}-2}\)

b) Thay \(a=6+4\sqrt{2}\) ta có:

\(-\frac{1}{\sqrt{6+4\sqrt{2}}-2}\approx-0,707\)

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