Question

Chứng minh đẳng thức

\(\sqrt{\frac{4}{\left(2-\sqrt{5}\right)^2}}-\sqrt{\frac{4}{\left(2+\sqrt{5}\right)^2}}=8\)

Giải đúng mk tick

Answer 1

Ta có VT: \(\sqrt{\frac{4}{\left(2-\sqrt{5}\right)^2}}-\sqrt{\frac{4}{\left(2+\sqrt{5}\right)^2}}\)=\(\frac{\sqrt{4}}{\sqrt{\left(2-\sqrt{5}\right)^2}}-\frac{\sqrt{4}}{\sqrt{\left(2+\sqrt{5}\right)^2}}\)

=\(\frac{2}{\left|2-\sqrt{5}\right|}-\frac{2}{\left|2+\sqrt{5}\right|}\)

=\(\frac{2}{\sqrt{5}-2}-\frac{2}{2+\sqrt{5}}\)

=\(\frac{2.\left(2+\sqrt{5}\right)-2.\left(\sqrt{5}-2\right)}{\left(\sqrt{5}-2\right).\left(2+\sqrt{5}\right)}\)

=\(2.\left(2+\sqrt{5}\right)-2.\left(\sqrt{5}-2\right)\)

=\(4+2\sqrt{5}-2\sqrt{5}+4\)

=8 (bằng VP)

Answer 2

mình bt làm nhưng ko bt bấm phân số