Question

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

a) Tìm ĐKXĐ và rút gọn

b) Tính A khi a \(=\frac{\sqrt{6}}{2+\sqrt{6}}\)

Answer 1

ĐKXĐ:\(\left\{{}\begin{matrix}x\ne1\\x\ne0\end{matrix}\right.\)

\(\left(\frac{\sqrt{a}+1}{\sqrt{a}-1}-\frac{\sqrt{a}-1}{\sqrt{a}+1}+4\sqrt{a}\right).\left(\sqrt{a}-\frac{1}{\sqrt{a}}\right)\\ =>\left(\frac{\left(\sqrt{a}+1\right)^2-\left(\sqrt{a}-1\right)^2}{a-1}+\frac{4\left(a-1\right)\sqrt{a}}{a-1}\right).\frac{a-1}{\sqrt{a}}\)

=>\(\frac{4\sqrt{a}+4\sqrt{a}\left(a-1\right)}{a-1}.\frac{a-1}{\sqrt{a}}\\ =>\frac{4\sqrt{a}.a}{\sqrt{a}}\\ =>4a\)

b, \(a=\frac{\sqrt{6}}{2+\sqrt{6}}\)

suy ra A=4.a=...