用简便方法计算
(1)0.9999×0.7+0.1111×2.7;
(2)(1+0.228﹣0.21)×(0.228﹣0.21+0.2003)﹣(1+0.228﹣0.21+0.2003)×(0.228﹣0.21)
解:(1)0.9999×0.7+0.1111×2.7
=(0.1111×9)×0.7+0.1111×2.7
=0.1111×6.3+0.1111×2.7
=0.1111×(6.3+2.7)
=0.1111×9
=0.9999
(2)设1+0.228﹣0.21=a,0.228﹣0.21=b,
(1+0.228﹣0.21)×(0.228﹣0.21+0.2003)﹣(1+0.228﹣0.21+0.2003)×(0.228﹣