已知2x^2-x-2014=0,求(2x-7)(x-3)(2x+5)(x+3)-91/(x-4)(2x+7)的值
2个回答
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由2x^2-x-2014=0得2x^2-x=2014, 2x^2=x+2014
(2x-7)(x-3)(2x+5)(x+3)-91/(x-4)(2x+7)
=(2x^2-13x+21)(2x^2+11x+15)-91/(2x^2-x-28)
=(x+2014-13x+21)(x+2014+11x+15)-91/(2014-28)
=(2035-12x)(12x+2029)-91/1986
=(2035*2029+72x-144x^2)-91/1986
=(2035*2029-72*2014)-91/1986
=3984007-91/1986
如果 (2x-7)(x-3)(2x+5)(x+3)-91整个是分母,则化简为
(3984007-91)/1986=2006
(2x-7)(x-3)(2x+5)(x+3)-91/(x-4)(2x+7)
=(2x^2-13x+21)(2x^2+11x+15)-91/(2x^2-x-28)
=(x+2014-13x+21)(x+2014+11x+15)-91/(2014-28)
=(2035-12x)(12x+2029)-91/1986
=(2035*2029+72x-144x^2)-91/1986
=(2035*2029-72*2014)-91/1986
=3984007-91/1986
如果 (2x-7)(x-3)(2x+5)(x+3)-91整个是分母,则化简为
(3984007-91)/1986=2006
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