Question

Rút gọn biểu thức :

a) \(A=\sqrt{4+\sqrt{7}}+\sqrt{4-\sqrt{7}}\)

b) \(B=\frac{2+\sqrt{6}+\sqrt{10}+\sqrt{2}+\sqrt{3}+\sqrt{5}}{\sqrt{8}+\sqrt{12}+\sqrt{20}}.\frac{\sqrt{2}-1}{3}\)

Answer 1

a)Bình phương 2 vế ta đc

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

⇔ \(A^2=4+\sqrt{7}+2\sqrt{\left(4+\sqrt{7}\right)\left(4-\sqrt{7}\right)}+4-\sqrt{7}\)

⇔ \(A^2=8+2\sqrt{16-7}=8+6=14\)
Vì A luôn ≥ 0 => A = \(\sqrt{14}\)

Answer 2

b) B = \(\frac{\sqrt{2.2}+\sqrt{2.3}+\sqrt{2.5}+\sqrt{2}+\sqrt{3}+\sqrt{5}}{2\sqrt{2}+2\sqrt{3}+2\sqrt{5}}\) . \(\frac{\sqrt{2}-1}{3}\)

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

= \(\frac{\left(\sqrt{2}+1\right)\left(\sqrt{2}-1\right)}{2.3}\)

= \(\frac{1}{6}\)