Question

x1x2是方程x²-2x+a=0的两个实数根，且x1+2x2=3-根号2

1求x1,x2及a的值 2求x1³-3x1²+2x1+x2的值

Answer 1

1、
x1+x2=2
x1+2x2=3-√2
所以x2=3-√2-2=1-√2
x1=2-x2=2-1+√2=1+√2
a=x1x2=(1-√2)(1+√2)=1-2=-1
2、
x1³-3x1²+2x1+x2
=x1(x1²-3x1+2)+x2
=x1(x1-1)(x1-2)+x2
=(1+√2)(√2)(-1+√2)+1-√2
=√2+1-√2
=1

Answer 2

由根与系数的关系知道
x1+x2=2
又x1+2x2=3-√2
所以x2=1-√2，x1=1+√2
那么x1x2=1-2=-1=a
所以a=-1

x1³-3x1²+2x1+x2
=（1+√2）³-3（1+√2）²+2（1+√2）+（1-√2）
=（3+2√2）（-2+√2）+√2+3
=-2-√2+√2+3
=1

Answer 3

1、x1+x2=2
x1+2x2=3-√2
故x1=√2+1，x2=1-√2
a=x1x2=-1
2、1

Answer 4

x1+x2=-b/a=2
x2=1-根号2
x1=2-(1-根号2)=1加根号2
x1x2=c/a=a
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