已知3x²-7x+2=0有两根为x1x2,则x1x2²+x1²x2=?
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已知3x²-7x+2=0有两根为x1x2,则x1x2²+x1²x2=?
已知3x²-7x+2=0有两根为x1x2,则x1x2²+x1²x2=?
已知3x²-7x+2=0有两根为x1x2,则x1x2²+x1²x2=?
根据题意得
x1+x2=7/3
x1x2=2/3
于是
x1x2²+x1²x2
=x1x2(x1+x2)
=2/3*7/3
=14/9
根据“韦达定理”得:
x1+x2=7/3……(1)
x1*x2=2/3……(2)
由(1)得:
x1^2+2x1x2+x2^2=49/9……(3)
(3)*(2)得:
x1^2x2+(x1x2)^2+x1x2^2=49/9*2/3=98/27
x1^2x2+x1x2^2=98/27-2(x1x2)=98/27-2*2/3=62/27
则x1x2^2+x1^2x2=62/27 .