关于题目中的一个算式怎么理解一列一字形队伍长120米,匀速前进,通讯员C以恒定的速率由队尾B走到队首A,又立刻走回队尾,

2个回答

  • 120u/(v-u)+120u/(v+u)=288

    120u[1/(v-u)+1/(v+u)]=288

    5u[1/(v-u)+1/(v+u)]=12

    5u{2v/[(v-u)(v+u)]}=12

    10uv/[(v-u)(v+u)]=12

    10uv=12[(v-u)(v+u)]

    5uv=6[(v-u)(v+u)]

    5uv=6(v²-u²)

    6v²-5uv-6V²=0

    (2u-3v)*(3u+2v)=0

    u=3v/2 u=-2v/3(舍去)

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