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设x1和x2是方程2x²-6x+3=0,求下列各式的值.
x1²x2+x1x2² 1/x1-1/x2
x1-x2用韦达定理怎么变形?
人气:379 ℃ 时间:2020-05-04 07:26:09
解答
∵x1,x2是方程的·两个根
∴x1x2=c/a=-3/2
x1+x2=-b/a=3
(1)x1x2(x1+x2)
=-3/2*3
=-9/2
(2)1/x1-1/x2=(x2-x1)/x1x2
=√(x2+x1)²-4x1x2/x1x2
=√9+6/(-3/2)
=√15/(-3/2)
=-2√15/2或2√15/2【1】题,c等于3吧?应该是3/2吧。额。看错(1)x1x2/(x1+x2)=3/2*3=9/2(2)同理√(x2+x1)²-4x1x2/x1x2=√9-4*3/2/(3/2)=√2/(3/2)=2√2/3或-2√2/3
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