Question

求解 若a²+ab-b²=0 a,b均为正数 则a²-b²/(b-a)(b-2a)+2a²-ab/4a²-4ab+b&su

接2a²-ab/4a²-4ab+b后面是*2a+b/2a-b

Answer 1

错了，是a²-ab+b²=0
a²-ab=b²
4a²-4ab=4b²

原式=(a+b)(a-b)/(a-b)(2a-b)+a(2a-b)/(2a-b)²*(2a+b)/(2a-b)
=(a+b)/(2a-b)+a(2a+b)/(2a-b)²
=(2a²-ab+2ab-b²-2a²-ab)/(2a-b)²
=-b²/(4a²-4ab+b²)
=-b²/(4b²+b²)
=-1/5

Answer 2

f(x)=(2x+2+1)/(x+1)
=(2x+2)/(x+1)+1/(x+1)
=2+1/(x+1)

1/(x+1)≠0
所以2+1/(x+1)≠2
所以值域(-∞,2)∪(2,+∞)