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若x1,x2是方程x2+2x-2007=0的两个根,试求下列各式的值:
(1)x12+x22
(2)
1
x1
+
1
x2

(3)(x1-5)(x2-5);
(4)|x1-x2|.
人气:381 ℃ 时间:2020-05-06 16:11:31
解答
∵x1,x2是方程x2+2x-2007=0的两个根,
∴x1+x2=-2,x1•x2=-2007.
(1)x12+x22=(x1+x22-2x1•x2=(-2)2-2×(-2007)=4018;
(2)
1
x1
+
1
x2
=
x1+x2
x1x2
=
−2
−2007
=
2
2007

(3)(x1-5)(x2-5)=x1•x2-5(x1+x2)+25=-2007-5×(-2)+25=-1972;
(4)|x1-x2|=
(x1x2)2
=
(x1+x2)2−4x1x2
=
(−2)2−4×(−2007)
=4
502
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