试说明；{10x+y}[10x+{10-y}]=100x{x+1}+y{10-y}恒为等式,并利用此恒等式计算1998×1992
问题描述：
试说明；{10x+y}[10x+{10-y}]=100x{x+1}+y{10-y}恒为等式,并利用此恒等式计算1998×1992
答
{10x+y}[10x+{10-y}]
=100x^2+10xy+100x+10y-10xy-y^2
=100(x^2+x)+10y-y^2
=100x(x+1)+y(10-y)
1998×1992
=(10×199+8)[10×199+(10-8)]
=100×199(199+1)+8(10-8)
=3980016