X1,X2是方程2x²-6x+3=0的两个根,利用根与系数的关系求X1²X2+X1X2²和(X1+1/X2)(X2+1/X1)
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![X1,X2是方程2x²-6x+3=0的两个根,利用根与系数的关系求X1²X2+X1X2²和(X1+1/X2)(X2+1/X1)](/uploads/image/z/3138799-31-9.jpg?t=X1%2CX2%E6%98%AF%E6%96%B9%E7%A8%8B2x%26%23178%3B-6x%2B3%3D0%E7%9A%84%E4%B8%A4%E4%B8%AA%E6%A0%B9%2C%E5%88%A9%E7%94%A8%E6%A0%B9%E4%B8%8E%E7%B3%BB%E6%95%B0%E7%9A%84%E5%85%B3%E7%B3%BB%E6%B1%82X1%26%23178%3BX2%2BX1X2%26%23178%3B%E5%92%8C%28X1%2B1%2FX2%29%28X2%2B1%2FX1%29)
X1,X2是方程2x²-6x+3=0的两个根,利用根与系数的关系求X1²X2+X1X2²和(X1+1/X2)(X2+1/X1)
X1,X2是方程2x²-6x+3=0的两个根,利用根与系数的关系求X1²X2+X1X2²和(X1+1/X2)(X2+1/X1)
X1,X2是方程2x²-6x+3=0的两个根,利用根与系数的关系求X1²X2+X1X2²和(X1+1/X2)(X2+1/X1)
由韦达定理:
x1+x2=6/2=3,x1*x2=3/2,
x1^2x2+x1x2^2=x1x2(x1+x2)=(3/2)*3=9/2,
(x1+1/x2)(x2+1/x1)=x1x2+2+1/x1x2=3/2+2+2/3=25/6.
9/2和25/9