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x/2y)^2*(y/2x)-[(x/y^2)/(2y^2/x)]
=x^2/4y^2*(y/2x)-(x/y^2)*(x/2y^2)
=x/8y-x^2/2y^4
=xy^3/8y^4-4x^2/8y^4
=(xy^3-4x^2)/(8y^4);
[(x+1)/x]*[2x/(x+1)]^2-[1/(x-1)-1/(x+1)]
=[(x+1)/x]*[4x^2/(x+1)^2]-{[(x+1)/(x-1)(x+1)]-[(x-1)/(x-1)(x+1)]}
=4x/(x+1)-2/(x-1)(x+1)
=4x(x-1)/(x-1)(x+1)-2/(x-1)(x+1)
=(4x^2-4x-2)/(x-1)(x+1).最佳答案
设原来规定X个月
同时假定总工程量是:1。当然你设成Y也一样最后能约掉
所以甲的工程速度:1/X
乙的工程速度:1/(X+6)
(单位都是“量每月”)
然后列方程:
(1/X+1/(X+6))×4+(1/(X+6))×(X-4)=1
所以解得:X=12
答:原规定12月
=x^2/4y^2*(y/2x)-(x/y^2)*(x/2y^2)
=x/8y-x^2/2y^4
=xy^3/8y^4-4x^2/8y^4
=(xy^3-4x^2)/(8y^4);
[(x+1)/x]*[2x/(x+1)]^2-[1/(x-1)-1/(x+1)]
=[(x+1)/x]*[4x^2/(x+1)^2]-{[(x+1)/(x-1)(x+1)]-[(x-1)/(x-1)(x+1)]}
=4x/(x+1)-2/(x-1)(x+1)
=4x(x-1)/(x-1)(x+1)-2/(x-1)(x+1)
=(4x^2-4x-2)/(x-1)(x+1).最佳答案
设原来规定X个月
同时假定总工程量是:1。当然你设成Y也一样最后能约掉
所以甲的工程速度:1/X
乙的工程速度:1/(X+6)
(单位都是“量每月”)
然后列方程:
(1/X+1/(X+6))×4+(1/(X+6))×(X-4)=1
所以解得:X=12
答:原规定12月
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