\(2\cdot3\cdot12+4\cdot6\cdot42+8\cdot27\cdot3\\=(2\cdot12)\cdot3+(4\cdot6)\cdot42+(8\cdot3)\cdot27\\=24\cdot3+24\cdot42+24\cdot27\\=24\cdot(3+42+27)\\=24\cdot(45+27)\\=24\cdot72\\=1728\)
#\(Toru\)
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