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数学
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设X1,X2是方程2X^2+4X-3=0的两个根,利用根与系数的关系求:(1)(X1+X2)^2;(2)(X1+1)(X2+1)
还有两个 (3)X1^2X2+X1X2^2;(4)(X1-X2)^2
人气:199 ℃ 时间:2020-02-04 09:35:02
解答
x1+x2=-2,x1x2=-3/2
所以
(1)(X1+X2)^2=(-2)^2=4
(2)(X1+1)(X2+1)
=x1x2+(x1+x2)+1
=-3/2-2+1
=-5/2
(3)X1^2X2+X1X2^2
=x1x2(x1+x2
=-3/2*(-2)
=3
(4)(X1-X2)^2
=(x1+x2)^2-4x1x2
=4-4*(-3/2)
=4+6
=10
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