Question

 giúp mình câu này với

 

Answer 1

đề là j bn :?

Answer 2

\(\dfrac{1}{1+\sqrt{2}+\sqrt{7}}=\dfrac{\sqrt{2}+1-\sqrt{7}}{3+2\sqrt{2}-7}\)

\(=\dfrac{\sqrt{2}-\sqrt{7}+1}{-4+2\sqrt{2}}=\dfrac{\left(\sqrt{2}-\sqrt{7}+1\right)\left(2+\sqrt{2}\right)}{-2\left(2-\sqrt{2}\right)\left(2+\sqrt{2}\right)}\)

\(=\dfrac{2\sqrt{2}+2-2\sqrt{7}-\sqrt{14}+2+2\sqrt{2}}{-4}\)

\(=\dfrac{4\sqrt{2}+4-2\sqrt{7}-\sqrt{14}}{-4}=\dfrac{-4\sqrt{2}-4+2\sqrt{7}+\sqrt{14}}{4}\)