Question

a=根号3—根号2/根号3+根号2,b=根号3+根号2/根号3—根号2 求 根号a*b+(a+b)²/根号a*b-(a+b)²

2.(2X-5)²-(X+4)²=0

Answer 1

因为a=(√3-√2)/(√3+√2)
=(√3-√2)^2/(√3+√2)(√3-√2)
=(√3-√2)^2
=5-2√6
b=(√3+√2)/(√3-√2)
=(√3+√2)^2/(√3+√2)(√3-√2)
=(√3+√2)^2
=5+2√6
a+b=5-2√6+5+2√6=10,
ab=(5-2√6)(5+2√6)=1
所以[√ab+(a+b^2)]/[√ab-(a+b)^2]
=(1+100)/(1-100)
=-101/99

2.(2X-5)²-(X+4)²=0
(2X-5)²=(X+4)²，
2x-5=x+4,或2x-5=-(x+4)
解得x1=9,x2=1/3

Answer 2

a=﹙√3－√2﹚／﹙√3＋√2﹚==﹙√3－√2﹚²=5-2√6
b=5＋2√6,ab=1,a＋b=10
√﹙ab﹚=1,﹙a＋b﹚=100
∴原式=﹙1＋100﹚／﹙1-100﹚=-101／99

Answer 3

a=根号3-根号2 1/a=1/(根号3-根号2) =(根号3+根号2)/(根号3+根号2)(根号3-根号2) =根号3+根号2 则 a+1/a=2根号3 两边平方 (a+1/a)