已知x1、x2是方程x^2-2x+a=0的两个实数根,且x1+2x2=3-根号2.求：x1、x2和a的值.x1^3-3x

x1+x2=2
所以x1+2x2=2+x2=3-√2
x2=1-√2
则x1=2-x2=1+√2
a=x1x2=-1
x²-2x-1=0
所以x1²-2x1-1=0
x1²=2x1+1
x1³=x1(2x1+1)=2x1²+x1
所以原式=2x1²+x1-3x1²+2x1+x2
=-x1²+3x1+x2
=-(2x1+1)+3x1+x2
=x1+x2-1
=2-1
=1