设3x²+2x-3=0,两根为x1,x2,求①x2/x1 + x1/x2 ②x1^2+x2^2-4x1x2
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![设3x²+2x-3=0,两根为x1,x2,求①x2/x1 + x1/x2 ②x1^2+x2^2-4x1x2](/uploads/image/z/10180742-14-2.jpg?t=%E8%AE%BE3x%26%23178%3B%2B2x-3%3D0%2C%E4%B8%A4%E6%A0%B9%E4%B8%BAx1%2Cx2%2C%E6%B1%82%E2%91%A0x2%2Fx1+%2B+x1%2Fx2+%E2%91%A1x1%5E2%2Bx2%5E2-4x1x2)
设3x²+2x-3=0,两根为x1,x2,求①x2/x1 + x1/x2 ②x1^2+x2^2-4x1x2
设3x²+2x-3=0,两根为x1,x2,求①x2/x1 + x1/x2 ②x1^2+x2^2-4x1x2
设3x²+2x-3=0,两根为x1,x2,求①x2/x1 + x1/x2 ②x1^2+x2^2-4x1x2
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
由韦达定理,得 X1+X2=-2/3,X1*X2=-1
①=(X2^2+X1^2)/(X1*X2)
=[(X1+X2)^2-2*X1*X2]/X1*X2=22/9
②=(X1+X2)^2-6*X1*X2=58/9