Question

(radical32+radical128+radical288) :4​

Answer 1

$( \sqrt{32} + \sqrt{28} + \sqrt{288} ) \div 4 =$

$= (4 \sqrt{2} + 2 \sqrt{7} + 12 \sqrt{2} ) \div 4$

$= (16 \sqrt{2} + 2 \sqrt{7} ) \div 4$

$\frac{16 \sqrt{2} + 2 \sqrt{7} }{4}$

$\frac{2(8 \sqrt{2} + \sqrt{7)} }{4}$

$\frac{8 \sqrt{2} + \sqrt{7} }{2}$