Question

初二解方程 (a²-b²)x²-4abx=a²-b² （a²≠b²，a,b为常数） （用公式法

Answer 1

(a²-b²)x²-4abx-(a²-b²)=0
△=(4ab)²+4(a²-b²)²=4(a²+b²)²
根据求根公式
x1,2=(4ab±√△)/2(a²-b²)=[4ab±2(a²+b²)]/2(a²-b²)
=[2ab±(a²+b²)]/(a²慎世-b²)

所以x1=[2ab+(a²+b²册孙)]/(a²-b²)
=(a+b)²/(a²-b²)
=(a+b)/(a-b)
x2=[2ab-(a²+b²州孝链)]/(a²-b²)
=-(a-b)²/(a²-b²)
=-(a-b)/(a+b)=(b-a)/(a+b)

Answer 2

(a²-b²)x²-4abx=a²-b²团搏
(a²-b²)x²-4abx-(a²-b²)=0
[(a-b)x-(a+b)][(a+b)x+(a-b)]=0
x=(a+b)/(a-b)或塌没祥x=(b-a)/察明(a+b)