Question

Tính

A=\(\dfrac{4}{\sqrt{5}-2}-\dfrac{32}{\sqrt{5}+1}\)

B=\(\dfrac{10}{3\sqrt{2}-4}+\dfrac{28}{3\sqrt{2}+2}\)

Answer 1

\(A=\dfrac{4\left(\sqrt{5}+2\right)}{\left(\sqrt{5}-2\right)\left(\sqrt{5}+2\right)}-\dfrac{32\left(\sqrt{5}-1\right)}{\left(\sqrt{5}+1\right)\left(\sqrt{5}-1\right)}\)

\(=\dfrac{4\left(\sqrt{5}+2\right)}{1}-\dfrac{32\left(\sqrt{5}-1\right)}{4}=4\sqrt{5}+8-8\sqrt{5}+8\)

\(=16-4\sqrt{5}\)

\(B=\dfrac{10\left(3\sqrt{2}+4\right)}{\left(3\sqrt{2}-4\right)\left(3\sqrt{2}+4\right)}+\dfrac{28\left(3\sqrt{2}-2\right)}{\left(3\sqrt{2}+2\right)\left(3\sqrt{2}-2\right)}\)

\(=\dfrac{10\left(3\sqrt{2}+4\right)}{2}+\dfrac{28\left(3\sqrt{2}-2\right)}{14}=5\left(3\sqrt{2}+4\right)+2\left(3\sqrt{2}-2\right)\)

\(=16+21\sqrt{2}\)