已知|a-1|+(b+4)²=0,求[(2a+b)²-(2a+b)(2a-b)-6a]÷2b的值
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![已知|a-1|+(b+4)²=0,求[(2a+b)²-(2a+b)(2a-b)-6a]÷2b的值](/uploads/image/z/3022068-12-8.jpg?t=%E5%B7%B2%E7%9F%A5%7Ca-1%7C%2B%28b%2B4%29%26%23178%3B%3D0%2C%E6%B1%82%5B%EF%BC%882a%2Bb%29%26%23178%3B-%EF%BC%882a%2Bb%29%282a-b%29-6a%5D%C3%B72b%E7%9A%84%E5%80%BC)
已知|a-1|+(b+4)²=0,求[(2a+b)²-(2a+b)(2a-b)-6a]÷2b的值
已知|a-1|+(b+4)²=0,求[(2a+b)²-(2a+b)(2a-b)-6a]÷2b的值
已知|a-1|+(b+4)²=0,求[(2a+b)²-(2a+b)(2a-b)-6a]÷2b的值
a-1=0
b+4=0
a=1,b=-4
[(2a+b)²-(2a+b)(2a-b)-6a]÷2b
=(4a^2+b^2+4ab-4a^2+b^2-6a)÷2b
=(2b^2+4ab-6a)÷2b
=b+2a-3a/b
=-4+2+3/4
=-5/4