a) Sửa đề: \(\sqrt{5}-\sqrt{\left(1+\sqrt{5}\right)^2}\)
Ta có: \(\sqrt{5}-\sqrt{\left(1+\sqrt{5}\right)^2}\)
\(=\sqrt{5}-\left|1+\sqrt{5}\right|\)
\(=\sqrt{5}-\left(1+\sqrt{5}\right)\)(Vì \(1+\sqrt{5}>0\))
\(=\sqrt{5}-1-\sqrt{5}=-1\)
b) Ta có: \(\sqrt{\left(2-\sqrt{3}\right)^2}+\sqrt{3}\)
\(=\left|2-\sqrt{3}\right|+\sqrt{3}\)
\(=2-\sqrt{3}+\sqrt{3}\)(Vì \(2>\sqrt{3}\))
\(=2\)
c) Sửa đề: \(\sqrt{\left(\sqrt{7}-3\right)^2}+\sqrt{7}\)
Ta có: \(\sqrt{\left(\sqrt{7}-3\right)^2}+\sqrt{7}\)
\(=\left|\sqrt{7}-3\right|+\sqrt{7}\)
\(=3-\sqrt{7}+\sqrt{7}\)(Vì \(\sqrt{7}< 3\))
\(=3\)
d) Ta có: \(\sqrt{\left(\sqrt{8}-7\right)^2}+\sqrt{8}\)
\(=\left|\sqrt{8}-7\right|+\sqrt{8}\)
\(=7-\sqrt{8}+\sqrt{8}\)(Vì \(\sqrt{8}< 7\))
\(=7\)
e) Ta có: \(\sqrt{\left(\sqrt{2}-\sqrt{7}\right)^2}+\sqrt{2}\)
\(=\left|\sqrt{2}-\sqrt{7}\right|+\sqrt{2}\)
\(=\sqrt{7}-\sqrt{2}+\sqrt{2}\)(Vì \(\sqrt{2}< \sqrt{7}\))
\(=\sqrt{7}\)
