y=(1-x∧2)/2-x为什么等于y=-1/2(x+1)∧2+1
设x>1,y>0,若x^y+x^-y=2根号2,则x^y-x^-y等于
y=(1-x∧2)/2-x为什么等于y=-1/2(x+1)∧2+1
1、x(x-y)(x+y)-x(x+y)^2
1/2X-1/X+Y*(X+Y/2X-X-Y)等于多少,
y=(x^2+x)/(x+1)
x-y+1大于等于0,x+y-2大于等于0,x小于等于1,求z=(x+2y)/(2x+y)的范围
若实数X Y满足{X-Y+1大于等于0 X+Y大于等于0 X小于等于0则Z=X+2Y的最大值?
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
(1)(x^2/x)-y-x-y
为什么|x-y+3|+(x+y-1)^2等于|x-y+3|+(x+y)^2-2x-2y+1?
(3x-y)^2+(3x+y)(3x-y),x=1,y=-2
已知4x=9y求(1)x+y/y (2)y-x/2x