A-A+
用递等式计算:(1)113×[21÷(4112-2.625)-412]÷335(2)(258
问题详情
用递等式计算:
(1)1
(2)(2
(3)
参考答案
(1)113×[21÷(4112-2.625)-412]÷335,
=113×[21÷3524-412]÷335,
=113×[725-412]÷335,
=113×9910×518,
=113;
(2)(258-147)÷[(3112+4.375)÷1989],
=1356÷[71124÷1989],
=5956÷38,
=5921;
(3)1013÷21922-1.4×1113+7+2263×20%,
=1013×2263-75×1113+7+2263×15,
=7+(1013+15)×2263-75×1113,
=7+6365×2263-75×1113,
=7+2265-7765,
=72265-11265,
=6213.
考点:等式