y=x2+2x+1、x-1,x-1>0求最值
求教 y={ x+1(x>0) x2 (x
函数y=x2+1/x(x
函数y=(x2-x)/(x2-x+1)的值域
求y=x-1/x2+2x的导数
x2-x-y2-y 解法:=(x2-y2)-(x+y) =(x+y)(x-y)-(x+y) =(x+y)(x-y-1)
已知x2-4x-1=0,求代数式(2x-3)2-(x+y)(x-y)-y2的值
因式分解 x2(x+1)-y(xy+x)
已知实数x,y满足(x2)+(y2)-xy+2x-y+1=0,求x,y的值
已知实数x、y满足X2+y2-xy+2x-y+1=0,试求x、y的值
用换元法解(2(x2+1)/x)+(6x/x2+1)=7其中(x2+1/x)=y
若x2-2xy-y2-x+Y-1=0 求x-y的值
(x+y)(x+y+1)=0 求x2+y2+2xy的值