解：原式=a^2-2ab+(a+b)[2(a-b)+a+b]
=a^2-2ab+(a+b)(3a-b).
=a^2-2ab+3a^2-ab+3ab-b^2.
=4a^2-b^2.
=4*(-1/2)^2-1.
=0.
a(a-2b)+2(a+b)(a-b)+(a+b)²=a²-2ab+2(a²-b²)+a²+2ab+b²
=2a²+b²+2a²-2b²
=4a²-b²
=4×(-1/2)²-1²
=1-i
=0
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