Question

分母有理化1/(1+根号2+根号3)

Answer 1

1/(1+根号2+根号3)
=【 1 * {(根号2+1)-根号3}】/【{根号2+1)+根号3}* {(根号2+1)-根号3}】
= (根号2+1-根号3) /(3+2根号2-3)
= (根号2+1-根号3) /(2根号2)
= (根号2+1-根号3)*根号2 /(2根号2*根号2)
= (2+根号2-根号6) /4

追问

对不起，我抄错题了，应是1/(1+根号2-根号3)

追答

没问题，方法一样：
1/(1+根号2-根号3)
=【 1 * {(根号2+1)+根号3}】/【{根号2+1)-根号3}* {(根号2+1)+根号3}】
= (根号2+1+根号3) /(3+2根号2-3)
= (根号2+1+根号3) /(2根号2)
= (根号2+1+根号3)*根号2 /(2根号2*根号2)
= (2+根号2+根号6) /4

Answer 2

1/(1+√2-√3)

=[1-(√2-√3)]/[1-(√2-√3)][1+(√2-√3)]

=[1-(√2-√3)]/[1-(√2-√3)^2]

=[1-(√2-√3)]/[1-(5-2√6)]

=[1-(√2-√3)]/[1-5+2√6]

=[1-(√2-√3)]/[2√6-4]

=[1-(√2-√3)]/[2(√6-2)]

=[1-(√2-√3)](√6-2)/[2(√6-2)(√6+2)]

=[1-(√2-√3)](√6-2)/4

=(1-√2+√3)(√6-2)/4

=(√6-√12+√18-2+2√2-2√3)/4

=(√6-2√3+3√2-2+2√2-2√3)/4

=(√6-2√3-2√3+3√2+2√2-2)/4

=(√6-4√3+5√2)/4

Answer 3

分母变成：{（1+根号2）+根号3}乘以{（1+根号2）-根号3}化简为2根号2然后再乘个根号2就可以了…注意不要忘记分子哦～