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2018*2012
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2018*2012
答
左式=(10x+y)[10x+(10-y)]=(10x+y)[(10x-y)+10]=(10x+y)(10x-y)+(10x+y)10=100x^2-y^2+100x+10y右式=100x(x+1)+y(10-y)=100x^2+100x+10y-y^2∴左式=右式∵求2018*2012利用此恒等式计算∴10x+y=2018 ① 10x+10-y=2012...