Question

已知：a=根号2+1分之根号2-1，b=根号2-1分之根号2+1，求a平方+b平方分之a+b+ab的值

a＝√2-1／√2+1
b＝√2+1／√2-1
求：a+b+ab／a²+b平方的值

Answer 1

∵a＝(√2-1)／(√2+1)=(√2-1)^2/(√2+1)(√2-1)=3-2√2,
b＝√2+1／√2-1=3+2√2,
a+b=6
ab=1
∴(a+b+ab)／(a²+b^2)
=(6+1)/[(a+b)^2-2ab]
=7/(36-2)
=7/34

Answer 2

a=(√2-1)/(√2+1）=(√2-1)^2=3-2√2
b=(√2+1)/(√2-1)=(√2+1)^2=3+2√2

(a+b+ab)/(a^2+b^2)
=[3-2√2+3+2√2+(3-2√2)(3+2√2)]/(9-12√2+8+9+12√2+8）
=（6+9-8)/34
=7/34