关于x 的方程x^2+2(m-2)x+m^2-3m+3=0 有两个不相等的实数根 x1、x2 .,
∴△/4=（m-2)^2-(m^2-3m+3)=1-m>0,m<1;
x1+x2=2(2-m),x1x2=m^2-3m+3.
∴mx1^2/(1-x1)+mx2^2/ (1-x2)
=m[x1^2(1-x2)+x2^2(1-x1)]/[(1-x1)(1-x2)]
=m[(x1+x2)^2-2x1x2-x1x2(x1+x2)]/[1-(x1+x2)+x1x2]
=m[4(2-m)^2-(m^2-3m+3)(2+4-2m)]/[1-(4-2m)+m^2-3m+3]
=2m(m-1)(m^2-3m+1)/(m^2-m)
=2(m-3/2)^2-5/2,(-1<=m<1)有最大值10.