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已知x1,x2是方程3x²-2x-1=0的两个根,求下列式子的值;
(1)(x1-x2)²; (2)(x1-2x2)(x2-2x1)
人气:286 ℃ 时间:2020-04-23 20:03:26
解答
x1+x2=2/3
x1x2=-1/3
所以(x1-x2)²
=(x1+x2)-4x1x2
=4/9+4/3
=16/9
原式=5x1x2-2x1²-2x2²
=9x1x2-2(x1²+2x1x2+x2²)
=9x1x2-2(x1+x2)²
=-3-8/9
=-35/9
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