Question

Bài 1: Rút gọn

a) \(\frac{\sqrt{10}+10}{1+\sqrt{10}}-\frac{5\sqrt{2}-2\sqrt{5}}{\sqrt{5}-\sqrt{2}}\)

b)\(\sqrt{14-6\sqrt{5}}+\sqrt{6+2\sqrt{5}}\)

(mink đag cần gấp)

Answer 1

a. \(\frac{\sqrt{10}+10}{1+\sqrt{10}}-\frac{5\sqrt{2}-2\sqrt{5}}{\sqrt{5}-\sqrt{2}}\)

= \(\frac{\sqrt{10}\left(1+\sqrt{10}\right)}{1+\sqrt{10}}-\frac{\sqrt{10}\left(\sqrt{5}-\sqrt{2}\right)}{\sqrt{5}-\sqrt{2}}\)

= \(\sqrt{10}-\sqrt{10}\)

= 0

b. \(\sqrt{14-6\sqrt{5}}+\sqrt{6+2\sqrt{5}}\)

= \(\sqrt{\left(3-\sqrt{5}\right)^2}+\sqrt{\left(1+\sqrt{5}\right)^2}\)

= \(3-\sqrt{5}+1+\sqrt{5}\)

= 4

Answer 2

a\(\frac{\sqrt{10}+10}{1+\sqrt{10}}-\frac{5\sqrt{2}-2\sqrt{5}}{\sqrt{5}-\sqrt{2}}= \frac{\sqrt{10}\left(1+\sqrt{10}\right)}{1+\sqrt{10}}-\frac{\sqrt{5}\sqrt{2}\left(\sqrt{5}-\sqrt{2}\right)}{\sqrt{5}-\sqrt{2}}=\sqrt{10}-\sqrt{10}=0\)

b

\(\sqrt{14-6\sqrt{5}}+\sqrt{6+2\sqrt{5}}=\sqrt{9-6\sqrt{5}+5}+\sqrt{5+2\sqrt{5}+1}=\sqrt{\left(3-\sqrt{5}\right)^2}+\sqrt{\left(\sqrt{5}+1\right)^2}=\left|3-\sqrt{5}\right|+\left|\sqrt{5}+1\right|=3-\sqrt{5}+\sqrt{5}+1=4\)