Question

[Hiếu Tạ]

Rút gọn biểu thức
a) \(\sqrt{4-\sqrt{7}}-\sqrt{4+\sqrt{7}}+\sqrt{2}\)

b) \(\sqrt{6+\sqrt{11}}-\sqrt{6-\sqrt{11}}+3\sqrt{2}\)

Answer 1

[Nobi Nobita]

a) \(\sqrt{4-\sqrt{7}}-\sqrt{4+\sqrt{7}}+\sqrt{2}\)

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

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

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

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

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

\(=\frac{\sqrt{7}-1-\sqrt{7}-1+2}{\sqrt{2}}=\frac{0}{\sqrt{2}}=0\)

b) \(\sqrt{6+\sqrt{11}}-\sqrt{6-\sqrt{11}}+3\sqrt{2}\)

\(=\frac{\sqrt{2\left(6+\sqrt{11}\right)}-\sqrt{2\left(6-\sqrt{11}\right)}+3.2}{\sqrt{2}}\)

\(=\frac{\sqrt{12+2\sqrt{11}}-\sqrt{12-2\sqrt{11}}+6}{\sqrt{2}}\)

\(=\frac{\sqrt{11+2\sqrt{11}+1}-\sqrt{11-2\sqrt{11}+1}+6}{\sqrt{2}}\)

\(=\frac{\sqrt{\left(\sqrt{11}+1\right)^2}-\sqrt{\left(\sqrt{11}-1\right)^2}+6}{\sqrt{2}}\)

\(=\frac{\left|\sqrt{11}+1\right|-\left|\sqrt{11}-1\right|+6}{\sqrt{2}}\)

\(=\frac{\left(\sqrt{11}+1\right)-\left(\sqrt{11}-1\right)+6}{\sqrt{2}}\)

\(=\frac{\sqrt{11}+1-\sqrt{11}+1+6}{\sqrt{2}}=\frac{8}{\sqrt{2}}=4\sqrt{2}\)