已知X1,X2,为方程X的平方加3X加1等于0的两实根,则X1的立方加8X2加20等于

如题所述

解:
x1,x2是方程x²+3x+1=0的根,由韦达定理,得
x1+x2=-3
又两根均满足方程,有
x1²+3x1+1=0
x1²=-3x1-1
x1²+3x=-1

x1³+8x2+20
=x1(x1²)+8x2+20
=x1(-3x1-1)+8x2+20
=-3x1²-x1+8x2+20
=-3x1²-9x1+8x1+8x2+20
=-3(x1²+3x1)+8(x1+x2)+20
=-3(-1)+8(-3)+20
=3-24+20
=-1
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第1个回答  2011-04-07
X^2+3X+1=0,X1^3+8X2+20=?
x1+x2=-3,x1x2=1
X1^3+8X2+20=X1^3+8(-X1-3)+20
=X1^3-8X1-4
=x1(X1^2+3x1+1)-3x1^2-9X1-4
=-3x1^2-9X1-4
=-3(x1^2+3x1+1)-1
=-1
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