2个回答
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解:设原来有油x千克。
x=(1/5x+20)×2+8·································理条件,总油=(第一次+第二次)×2【第三次】
x=2/5x+40+8············································运用乘法分配律拆括号
3/5x=48·······················································两边同时减2/5x,并将40、8合成一体
x=80························································x=48÷3/5,解出方程
答:原来这桶油有80千克.
x=(1/5x+20)×2+8·································理条件,总油=(第一次+第二次)×2【第三次】
x=2/5x+40+8············································运用乘法分配律拆括号
3/5x=48·······················································两边同时减2/5x,并将40、8合成一体
x=80························································x=48÷3/5,解出方程
答:原来这桶油有80千克.
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