已知|2a-b+1|+(a+1⼀2b)^2=0,求代数式b^2⼀a+b除以[(a⼀a-b-1)(a-a⼀a+b)]的值

2025-01-07 04:32:04
推荐回答(1个)
回答1:

|2a-b+1|+(a+1/2b)^2=0左边两式均为非负数
则2a-b+1=0 ;a+1/2b=0解得a=b=-1
b^2/a+b=-1/2
(a/a-b-1)(a-a/a+b)=-3/2
结果为1/3