Question

\(\sqrt{6,5+\sqrt{12}}\)+\(\sqrt{6,5-\sqrt{12}}\)+\(2\sqrt{6}\)

Answer 1

Đặt \(A=\sqrt{6,5+\sqrt{12}}+\sqrt{6,5-\sqrt{12}}\)

<=> \(A^2=\left(\sqrt{6,5+\sqrt{12}}+\sqrt{6,5-\sqrt{12}}\right)^2\)

<=> \(A^2=6,5+\sqrt{12}+2\sqrt{\left(6,5+\sqrt{12}\right)\left(6,5-\sqrt{12}\right)}+6,5-\sqrt{12}\)

<=> \(A^2=13+2\sqrt{42,25-12}\)

<=> \(A^2=13+2\sqrt{\frac{121}{4}}\)

<=> \(A^2=13+2\cdot\frac{11}{2}=13+11=24\)

=> \(A=2\sqrt{6}\)

Vậy \(\sqrt{6,5+\sqrt{12}}+\sqrt{6,5-\sqrt{12}}+2\sqrt{6}=4\sqrt{6}\)