Question

Thực hiện phép tính\(\sqrt{\frac{1}{\left(2-\sqrt{6}\right)^2}}-\sqrt{\frac{1}{\left(2+\sqrt{6}\right)^2}}\)

Answer 1

dòng đầu bị lỗi tí, sửa lại nè:

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

Answer 2

$\sqrt{\frac{1}{{\left(2-\sqrt{6}\right)}^2}}-\sqrt{\frac{1}{{\left(2+\sqrt{6}\right)}^2}}$

=\(\frac{1}{\sqrt{6}-2}\)-\(\frac{1}{2+\sqrt{6}}\)

=\(\frac{2+\sqrt{6}-\sqrt{6}+2}{\left(\sqrt{6}-2\right)\left(\sqrt{6}+2\right)}\)

\(\frac{4}{6-4}=\frac{4}{2}=2\)