Question

求十道二次根式要过程

Answer 1

①5√8-2√32+√50
=5*3√2-2*4√2+5√2
=√2(15-8+5)
=12√2
②√6-√3/2-√2/3
=√6-√6/2-√6/3
=√6/6

③(√45+√27)-(√4/3+√125)
=(3√5+3√3)-(2√3/3+5√5)
=-2√5+7√5/3
④(√4a-√50b)-2(√b/2+√9a)
=(2√a-5√2b)-2(√2b/2+3√a)
=-4√a-6√2b

⑤√4x*(√3x/2-√x/6)
=2√x(√6x/2-√6x/6)
=2√x*(√6x/3)
=2/3*|x|*√6

⑥(x√y-y√x)÷√xy
=x√y÷√xy-y√x÷√xy
=√x-√y

⑦(3√7+2√3)(2√3-3√7)
=(2√3)^2-(3√7)^2
=12-63
=-51
⑧(√32-3√3)(4√2+√27)
=(4√2-3√3)(4√2+3√3)
=(4√2)^2-(3√3)^2
=32-27
=5
⑨(3√6-√4)²
=(3√6)^2-2*3√6*√4+(√4)^2
=54-12√6+4
=58-12√6

⑩（1+√2-√3）（1-√2+√3）
=[1+(√2-√3)][1-(√2-√3)]
=1-(√2-√3)^2
=1-(2+3+2√6)
=-4-2√6