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设x1,x2是方程3x2+2x-6=0的两个根,求(1)x1的三次方+x2的三次方的值 (2)3x1的平方乘(6-2x2)的值
人气:222 ℃ 时间:2020-02-03 17:06:52
解答
(1)由韦达定理,x1+x2 = -2/3 ,x1x2 = -2于是,x1^3+x2^3 = (x1+x2)(x1²-x1x2+x2²) = -2/3[(x1+x2)²-3x1x2]= -116/27(2)3x2²+2x2-6=0,于是6-2x2 = 3 x2²∴3x1²(6-2x2) = 9(x1x2)&su...
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