Question

1)2√24+√20-3√6+4√5:2=
2)[√40-3√2•√5+√ 5•(-√ 2)la puterea 3] :√ 10=
3)2•[(2√ 7-√ 28•3)•4supra√ 7+ 7supra2 (4+3√ 8)]=

Answer 1

1)2√24+√20-3√6+4√5:2=2*2√6+2√5-3√6+2√5=4√6+2√5-3√6+2√5=√6+4√5 2)[√40-3√2*√5+√5*(-√2)la puterea a 3]:10=[2√10-3√10+√5*(-2√2)]:10=(2√10-3√10-2√19):10=-3√10:10=-0,3√10 3)2*[(2√7-√28*3)*4/√7+7/2(4+3√8)]=2[(2√7-2√7*3)*4/√7+7/2(4+3*2√2)]=[(2√7-6√7)*4/7+7/2(4+6√2)]=[-4√7*4/7+28/2+42√2/2]=-16√7/7+28+42√2/2 amplifici prima fractie cu 2 iar a doua cu 7=-32√7+196+294√2/14