Question

Radical din 20 plus radical din 45plus radical din 5
​

Answer 1

 

$\displaystyle\bf\\\sqrt{20}+\sqrt{45}+\sqrt{5}=\\\\=\sqrt{4\times5}+\sqrt{9\times5}+\sqrt{5}=\\\\=2\sqrt{5}+3\sqrt{5}+\sqrt{5}=\boxed{\bf6\sqrt{5}}$