|2a-1|+(3b+2)²=0,求a²-3b²÷(a+b)这题怎么算啊?
问题描述:
|2a-1|+(3b+2)²=0,求a²-3b²÷(a+b)这题怎么算啊?
答
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...对不起啊,忘记了个东西了是(a²-3b²)÷(a+b),忘打括号了,能再解答一下吗?专家?²³²³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]=[-13/12]×[-6]=13/2