a²+b²-1/2a²-1/2[a+b]×b=1/2a²+1/2b²-1/2ab化简

a²+b²-1/2a²-1/2[a+b]×b=1/2a²+1/2b²-1/2ab化简
a²+b²-1/2a²-1/2[a+b]×b=1/2a²+1/2b²-1/2ab化简

a²+b²-1/2a²-1/2[a+b]×b=1/2a²+1/2b²-1/2ab化简
这是个恒等式啊,是要证明吧.
a²+b²-(1/2)a²-(1/2)(a+b)b
=a²+b²- a²/2-ab/2 -b²/2
=(1/2)a²/2+(1/2)b²-(1/2)ab
等式成立.

a²+b²-(1/2)a²-(1/2)(a+b)b
=a²+b²- a²/2-ab/2 -b²/2
=(1/2)a²+(1/2)b²-(1/2)ab