Bài 1:
a) Ta có: \(\sqrt{3+\sqrt{5}}-\sqrt{3-\sqrt{5}}-\sqrt{2}\)
\(=\frac{\sqrt{6+2\sqrt{5}}-\sqrt{6-2\sqrt{5}}-2}{\sqrt{2}}\)
\(=\frac{\sqrt{5+2\cdot\sqrt{5}\cdot1+1}-\sqrt{5-2\cdot\sqrt{5}\cdot1+1}-2}{\sqrt{2}}\)
\(=\frac{\sqrt{\left(\sqrt{5}+1\right)^2}-\sqrt{\left(\sqrt{5}-1\right)^2}-2}{\sqrt{2}}\)
\(=\frac{\left|\sqrt{5}+1\right|-\left|\sqrt{5}-1\right|-2}{\sqrt{2}}\)
\(=\frac{\sqrt{5}+1-\left(\sqrt{5}-1\right)-2}{\sqrt{2}}\)(Vì \(\sqrt{5}>1>0\))
\(=\frac{\sqrt{5}+1-\sqrt{5}+1-2}{\sqrt{2}}=\frac{2-2}{\sqrt{2}}=\frac{0}{\sqrt{2}}=0\)
b) Ta có: \(\sqrt{4-\sqrt{7}}-\sqrt{4+\sqrt{7}}+\sqrt{7}\)
\(=\sqrt{\frac{7}{2}-2\cdot\sqrt{\frac{7}{2}}\cdot\sqrt{\frac{1}{2}}+\frac{1}{2}}-\sqrt{\frac{7}{2}+2\cdot\sqrt{\frac{7}{2}}\cdot\sqrt{\frac{1}{2}}+\frac{1}{2}}+\sqrt{7}\)
\(=\sqrt{\left(\sqrt{\frac{7}{2}}-\sqrt{\frac{1}{2}}\right)^2}-\sqrt{\left(\sqrt{\frac{7}{2}}+\sqrt{\frac{1}{2}}\right)^2}+\sqrt{7}\)
\(=\left|\sqrt{\frac{7}{2}}-\sqrt{\frac{1}{2}}\right|-\left|\sqrt{\frac{7}{2}}+\sqrt{\frac{1}{2}}\right|+\sqrt{7}\)
\(=\sqrt{\frac{7}{2}}-\sqrt{\frac{1}{2}}-\left(\sqrt{\frac{7}{2}}+\sqrt{\frac{1}{2}}\right)+\sqrt{7}\)(Vì \(\sqrt{\frac{7}{2}}>\sqrt{\frac{1}{2}}>0\))
\(=\sqrt{\frac{7}{2}}-\sqrt{\frac{1}{2}}-\sqrt{\frac{7}{2}}-\sqrt{\frac{1}{2}}+\sqrt{7}\)
\(=-2\sqrt{\frac{1}{2}}+\sqrt{7}\)
\(=-\sqrt{2}+\sqrt{7}\)
