设:每箱x千克,一共y千克 10x=3/5 y 3x+10=2/5 y 解得:x=
设x.y都是有理数,并且x,y满足x的平方-2y+√5y=10+3√5 x+y的值
已知x-y/x+y=3,求代数式5(x-y)/x+y-x+y/2(x-y)
每千克芒果卖x元,每千克苹果卖y元,妈妈买了5千克芒果和3千克苹果,一共付多少元?
(4x-2y-z)-{5x[8y-2y-(x+y)]-x+(3y-10z)]=? kuai
设变量xy满足 x+y-3 x-y+1 3x-y+5 z=x+y/x
先化简再求值(x-y)(x+y)-(x-2y) 的完全平方+x(3x-5y)-(x-y)(x-2y),其中x=1/2 y
解方程组:(0.2x+0.1y)/2- (4x-y)/10=(x-y)/5+ (3x+0.5y)/30 (3x+2y-1
6(x-y)-2(x+y)=14 3(x-y)+(x+y)=5
3(x+y)-4(x-y)=11,2(x-y)+5(x+y)=27
(1)x+y=15 (2)5x-2(10-x)=6y-3(10-y)+19 解得x=9
已知x,y满足约束条件{2x+5y>=10,2x-3y>=-6,2x+y
{ x-y/7-x+y/10=1/2 {2(x-y)+5(x+y)=3