Question

\(\sqrt{4-\sqrt{7}}-\sqrt{4+\sqrt{7}}\)

Answer 1

\(=\frac{\sqrt{16-4\sqrt{7}}}{2}-\frac{\sqrt{16+4\sqrt{7}}}{2}\)

\(=\frac{\sqrt{2\left(8-2\sqrt{7}\right)}}{2}-\frac{\sqrt{2\left(8+2\sqrt{7}\right)}}{2}\)

\(=\frac{\sqrt{2\left(1-2\sqrt{7}+7\right)}}{2}-\frac{\sqrt{2\left(1+2\sqrt{7}+7\right)}}{2}\)

\(=\frac{\sqrt{2\left(1-\sqrt{7}\right)^2}}{2}-\frac{\sqrt{2\left(1+\sqrt{7}\right)^2}}{2}\)

\(=\frac{\sqrt{2}.\left(1-\sqrt{7}\right)}{2}-\frac{\sqrt{2}.\left(1+\sqrt{7}\right)}{2}\)

\(=\frac{\sqrt{2}\left(1-\sqrt{7}-1-\sqrt{7}\right)}{2}=\frac{\sqrt{2}.-2\sqrt{7}}{2}=\frac{-2\sqrt{14}}{2}=-\sqrt{14}\)

Answer 2

Nhầm tý :)), dòng thứ 4 nhé!!

\(=\frac{\sqrt{2\left(\sqrt{7}-1\right)^2}}{2}-\frac{\sqrt{2\left(\sqrt{7}+1\right)^2}}{2}\)

\(=\frac{\sqrt{2}\left(\sqrt{7}-1-\sqrt{7}-1\right)}{2}=\frac{\sqrt{2}.\left(-2\right)}{2}=-\sqrt{2}\)