If the equation m(x-1)=2001-n(x-2) for x has infiniteroots,then m的2001次方加n的2001次方等于几?

问题描述:

If the equation m(x-1)=2001-n(x-2) for x has infiniteroots,then m的2001次方加n的2001次方等于几?

If the equation(方程) m(x-1)=2001-n(x-2) for x has infinite roots(无限个根) ,then m的2001次方+n的2001次方= 0
因为X有无限个根,所以先令X=1,得出0=2001-n(-1),再令X=2,得出m=2001
所以m=2001,n=-2001.
所以m的2001次方+n的2001次方=0
参考:
m(x-1)=2001-n(x-2)
(m+n)x-(m+2n)=2001,
要方程有无数个解,则
m+n=0,-(m+2n)=2001
解得m=2001,n=-2001
他们互为相反数,所以M的2001次方+N的2001次方=0