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设X1、X2是一元二次方程2X^2-5X+1=0的两个根,利用根与系数的关系,求下列各式的值 (要过程)
(1)(x1-3)(x2-3)
(2)(x1+1)2+(x2+1)2
(3)(x2/x1)+(x1/x2)
(4)|x1-x2|
人气:180 ℃ 时间:2019-08-20 20:40:04
解答
根据根与系数关系可得:
x1x2=0.5x1+x2=2.5
1、(x1-3)(x2-3)
=x1x2-3(x1+x2)+9
=0.5-7.5+9
=2
2、(x1+1)²+(x2+1)²
=x1²+2x1+1+x2²+2x2+1
=(x1²+x2²)+2(x1+x2)+2
=(x1+x2)²-2x1x2+2(x1+x2)+2
=6.25-1+5+2
=12.25
3、x2/x1+x1/x2
=(x2²+x1²)/x1x2
=[(x1+x2)²-2x1x2]/x1x2
=(6.25-1)/0.5
=10.5
4、|x1-x2|²=(x1-x2)²
=(x1+x2)²-4x1x2
=6.25-2
=4.25
所以:|x1-x2|=√4.25=√17/2
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