∵a的平方+b的平方-10a-6b+34=0
∴(a-5)²+(b-3)²=0
∴a=5,b=3
(a+b分之a-b)²除以ab分之(a²-b²)[a²-b²分之a²+2ab+b²]
=[(a-b)²ab]/[(a+b)²(a²-b²)][(a+b)²/(a²-b²)]
=(a-b)²ab
=4*15
=60
∵a的平方+b的平方-10a-6b+34=0
∴(a-5)²+(b-3)²=0
∴a=5,b=3
(a+b分之a-b)²除以ab分之(a²-b²)[a²-b²分之a²+2ab+b²]
=[(a-b)²ab]/[(a+b)²(a²-b²)][(a+b)²/(a²-b²)]
=(a-b)²ab
=4*15
=60