abc=1,解方程(2ax)/(ab+a+1)+(2bx)/(bc+b+1)+(2cx)/(ca+c+1)=4

1个回答

  • 原方程即:

    2x*[a/(ab+a+1)+b/(bc+b+1)+c/(ca+c+1)]=4

    x*[a/(ab+a+1)+b/(bc+b+1)+c/(ca+c+1)]=2····①

    关键是先求出中括号的值:

    a/(ab+a+1)+b/(bc+b+1)+c/(ca+c+1)·····注:不断将abc=1替换分母中的1;

    =a/(ab+a+abc)+b/(bc+b+1)+c/(ca+c+1)

    =1/(b+1+bc)+b/(bc+b+1)+c/(ca+c+1)

    =(1+b)/(bc+b+1)+c/(ca+c+1)

    =(abc+b)/(bc+b+abc)+c/(ca+c+1)

    =(ac+1)/(c+1+ac)+c/(ca+c+1)

    =(ac+c+1)/(ac+c+1)

    =1

    所以,方程①就变成:

    x=2

    即为所求.

    关于上面部分过程,可以参考我以前的回答: