~~~~~~~~~~~~~~~一道因式分解……………………

{a²b²-[(a²+b²-c²）/2]²}*1/4
=[ab+(a²+b²-c²）/2][ab-(a²+b²-c²）/2]*1/4
=(a²+b²+2ab-c²)(a²-2ab+b²-c²)*(-1/16)
=(-1/16)[(a+b)²-c²][(a-b)²-c²]
=(-1/16)(a+b+c)(a+b-c)(a-b+c)(a-b-c)

{a²b²-[(a²+b²-c²）/2]²}*1/4
=[ab-(a²+b²-c²)/2}*[ab+(a²+b²-c²)/2]*1/4
=[2ac-(a²+b²-c²)]/2*[2ab+(a²+b²-c²)]/2*1/4
=[(a-b)²-c²]*[(a+b)²-c²]*1/16
=(a-b+c)*(a-b-c)*(a+b-c)*(a+b+c)/16

{a²b²-[(a²+b²-c²）/2]²}*1/4
解：原式=[ab+(a²+b²-c²）/2][ab-(a²+b²-c²）/2]*1/4
=(a²+b²+2ab-c²)(a²-2ab+b²-c²)*(-1/16)
=(-1/16)[(a+b)²-c²][(a-b)²-c²]
=(-1/16)(a+b+c)(a+b-c)(a-b+c)(a-b-c)

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