|2a-1|+(3b+2)²=0,求a²-3b²÷(a+b)这题怎么算啊?
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|2a-1|+(3b+2)²=0,求a²-3b²÷(a+b)这题怎么算啊?
|2a-1|+(3b+2)²=0,求a²-3b²÷(a+b)这题怎么算啊?
|2a-1|+(3b+2)²=0,求a²-3b²÷(a+b)这题怎么算啊?
根据题意,得:
2a-1=0且3b+2=0
得:a=1/2、b=-2/3,则:
a²-3b²÷(a+b)
=(1/2)²-3×(-2/3)²÷[(1/2)+(-2/3)]
=(1/4)-3×(4/9)÷[-(1/6)]
=(1/4)-(4/3)×[-6]
=(1/4)+8
=33/4
因为|2a-1|和(3b+2)²都是非负数,所以他们的和为0,那就只能每一个式子都为0
所以a=1/2,b=-2/3且a²=1/4, 3b²=4/3, a+b=-1/6
所以a²-3b²÷(a+b)=1/4-4/3/(-1/6)=1/4+8=33/4