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一元二次方程ax2+bx+c=0的两根为x1和x2
求(1)Ix1-x2I和x1+x2/2
(2)x1的三次方+x2的三次方
人气:297 ℃ 时间:2019-10-17 03:00:06
解答
(1)
Ix1-x2I=I[-b+√(b²-4ac)]/(2a)-[-b-√(b²-4ac)]/(2a)I=I√(b²-4ac)/aI=√(b²-4ac)/IaI;
(x1+x2)/2?=(-b/a)/2=-b/(2a)
(2)
x1的三次方+x2的三次方
=(x1+x2)(x1²-x1x2+x2²)
=(x1+x2)(x1²+2x1x2+x2²-3x1x2)
=(x1+x2)[(x1+x2)²-3x1x2]
=(-b/a)[(-b/a)²-3c/a]
=3bc/a²-b³/a³
=(3abc-b³)/a³
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