计算下列算式
934÷[216+{413−(112+134)}]
解答
934÷[216+{413−(112+134)}]
=9×4+34÷[2×6+16+{4×3+13−(1×2+12+4×1+34)}]
=394÷[136+{133−(32+74)}]
4和2的最小公倍数是2
=394÷[136+{133−(6+74)}]
=394÷[136+{133−(134)}]
=394÷[136+{133−134}]
3和4的最小公倍数是12
=394÷[136+{13×4−13×312}]
=394÷[136+{52−3912}]
=394÷[136+1312]
6和12的最小公倍数是12
=394÷[13×2+1312]
=394÷[26+1312]
=394÷[3912]
=394÷3912
=394×1239
=394×1239 (39×1=39,4×3=12)
=3
因此答案是 3
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