Question

1/√2-1+√3(√3-√6)+√8

这是二次根式的题

① 1/（√2-1）+√3(√3﹣√6)+√8
②[(2/3)x√9x＋6x√(y/x)]＋[(y√﹙x/y﹚﹣x²√(1/x)]
③(1/10) √200 - 2√0.08 - 4√0.5 + (2/5)√72

这三道题要求写明解答过程 谢谢各位了
要求21:30之前 回复完毕

Answer 1

① 1/（√2-1）+√3(√3﹣√6)+√8
=√2+1+3-3√2+2√2
=4

②[(2/3)x√9x＋6x√(y/x)]＋[(y√﹙x/y﹚﹣x²√(1/x)]
=2x√x＋6√xy+√xy-x√x
=7√xy+x√x

③(1/10) √200 - 2√0.08 - 4√0.5 + (2/5)√72
=√2-2√(8/100)-4√(1/2)+2/5√(2*36)
=√2-2/5√2-2√2+12/5√2
=√2