Question

A=\(\sqrt{23+6\sqrt{10}}-\)\(\sqrt{23-6\sqrt{10}}\)

B=\(\left(\dfrac{2+\sqrt{2}}{\sqrt{2}+1}+1\right)\times\)\(\left(\dfrac{2-\sqrt{2}}{\sqrt{2}-1}-1\right)\)

giúp mik vs

Answer 1

a) Ta có: \(A=\sqrt{23+6\sqrt{10}}-\sqrt{23-6\sqrt{10}}\)

\(=\sqrt{18+2\cdot3\sqrt{2}\cdot\sqrt{5}+5}-\sqrt{18-2\cdot3\sqrt{2}\cdot\sqrt{5}+5}\)

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

\(=3\sqrt{2}+\sqrt{5}-3\sqrt{2}+\sqrt{5}\)

\(=2\sqrt{5}\)

Answer 2

b) Ta có: \(B=\left(\dfrac{2+\sqrt{2}}{\sqrt{2}+1}+1\right)\left(\dfrac{2-\sqrt{2}}{\sqrt{2}-1}-1\right)\)

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

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

=2-1=2