设3x²+2x-3=0,两根为x1,x2,求①x2/x1 + x1/x2 ②x1^2+x2^2-4x1x2
人气:369 ℃ 时间:2020-06-30 11:47:30
解答
根据韦达定理有
X1+X2=-b/a=-2/3,X1*X2=c/a=-3/3=-1
①x2/x1 + x1/x2
=(x2²+x1²)/(x1x2)
=【(x1+x2)²-2x1x2】/(x1x2)
=【4/9-2×(-1)】/(-1)
=-22/9
②x1^2+x2^2-4x1x2
=(x1+x2)²-2x1x2-4x1x2
=(x1+x2)²-6x1x2
=4/9-6×(-1)
=6又4/9
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