Question

已知x1、x2是方程x^2+3x+1=0的两实数根，求x1^3+8x2+20的值。

已知x1、x2是方程x^2+3x+1=0的两实数根，求x1^3+8x2+20的值。
得数是-1,不能求出x1,x2后带入。

Answer 1

解：
由已知有
x1+x2 = -3
x1x2 = 1
原式 = x1³ + 8·1·x2 + 20
= x1³ + 8·x1x2·x2 + 20
= x1(x1²+8x2²) + 20
= x1[(x1²+x2²)+7x2²] + 20
= x1[(x1+x2)²-2x1x2+7x2²] + 20
= x1(9-2+7x2²) + 20
= x1(7+7x2²) + 20
= x1(7x1x2+7x2²) + 20
= 7x1x2(x1+x2) + 20
=7×1×(-3) + 20
= -1

Answer 2

x1、x2是方程x^2+3x+1=0的两实数根
韦达定理得:
x1+x2=-3
x1x2=1
x1^2+3x1+1=0
二边同乘x1:
x1^3+3x1^2+x1=0
x1^3+8x2+20
=-(3x1^2+x1)+8x2+20
=-3(-1-3x1)-x1+8x2+20
=3+9x1-x1+8x2+20
=8(x1+x2)+23
=8*(-3)+23
=-1
=

Answer 3

由题意 x1^2+3x1+1=0
x1^2=-1-3x1
原式=x1*x1^2+8x2+20
=x1(-1-3x1)+8x2+20
=-3x1^2-x1+8x2+20
=-3(-1-3x1)-x1+8x2+20
=8x1+8x2+23
=8(x1+x2)+23
=8(-3)+23
=-1

Answer 4

x1^2+3x1+1=0
x1^2=-(3x1+1)

x1+x2=-3

所以：
x1^3+8x2+20
=-x1*(3x1+1)+8x2+20
=-3x1^2-x1+8x2+20
=3*(3x1+1)-x1+8x2+20
=8(x1+x2)+23
=-24+23
=-1

Answer 5

问老师去吧