已知a²+2ab+b²=0,求a(a+4b)-(a+2b)(a-2b)
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已知a²+2ab+b²=0,求a(a+4b)-(a+2b)(a-2b)
已知a²+2ab+b²=0,求a(a+4b)-(a+2b)(a-2b)
已知a²+2ab+b²=0,求a(a+4b)-(a+2b)(a-2b)
已知a²+2ab+b²=0
(a+b)²=0;
a+b=0;
a(a+4b)-(a+2b)(a-2b)
=a²+4ab-a²+4b²
=4b(a+b)
=0
a²+2ab+b²=(a+b)²=0
所以a+b=0
原式=a²+4ab-a²+4b²
=4b(a+b)
=0
a^2+2ab+b^2=(a+b)^2=0
a+b=0
a(a+4b)-(a+2b)(a-2b)
=a^2+4ab-(a^2-4b^2)
=4ab+4b^2
=4b(a+b)
=0