已知1/x-1/x=3,则代数式(2x-14xy-2y)/(x-2xy-y)的值为?求过程!!!!!!!!!!
已知1/x-1/y=3,则1/y-1/x=-3
(2x-14xy-2y)/(x-2xy-y)
=(2/y-14-2/x)/(1/y-2-1/x)
=(2/y-2/x-14)/(1/y-1/x-2)
=[2(1/y-1/x)-14)]/(1/y-1/x-2)
=[2(-3)-14)]/(-3-2)
=[-6-14]/(-3-2)
=(-20)/(-5)
=4
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