(1)√(x²+y²)²(x²-y²)化简 (2)若a²-4a+4+根号b-3=0,求a²-b的值
问题描述:
(1)√(x²+y²)²(x²-y²)化简 (2)若a²-4a+4+根号b-3=0,求a²-b的值
答
化简
√[(x²+y²)²(x²-y²)]
=(x²+y²)√(x²-y²)
若a²-4a+4+√(b-3)=0,则有:
(a-2)²+√(b-3)=0 两个非负数的和等于0,这两个非负数都等于0
a-2=0,b-3=0
a=2 ,b=3
a²-b=2²-3
=4-3
=1第一题就这样?你好:同学!第一题的被开方数是(x²+y²)²(x²-y²), (x²+y²)²≥0,(x²-y²)≥0结果当然等于 (x²+y²)√(x²-y²)