{x|x+y=1}={y|x+y=1}?为什么?f(X+2)=X+7,求f(7)=?是否f（）括号里面的代数一定是自变量?,那f（x+2）中的x与f(x)中的x相同?怎样代数啊?

y'/y=1/x y'-(1/x)y=xe^-x x-y/x-x+y/y-(x+y)(x-y)/y² y/x=2 y(x+y)+(x-y)²-（x+y）（x-y）,其中x=-1/3,y=3 （x-y)(x+y)-(x+y)^2+2y(y-x),其中x=1,y=3. 2y(x+二分之一y)-[(x+y)(x-y)+2y(y+x)],其中|x-1|=2 x>=1,x-y y=x-1(x 化简 x / y(x+y) - y / x(x+y) = 已知x+y=a,2x-y=-2a,求[（x/y-y/x)/(x+y)-x(1/x-1/y)]/[(x+1)/y]的值 x.y都是自然数,且x(x-u)-y(y-x)+12,求x.y得值快1且x(x-y)-y(y-x)=12 若2/x-1/y=3,求[y/x-y/x-y(x-y/x-x+y)]/x-2y/x的值 [(y*y*y-y)/(xy+1)*(xy+1)-(x+y)*(x+y)]/[(x+y)*y/(x+1)]其中x=-11,y=1/12,化简再求值. y=[x-1], 设x,y满足约束条件x+y>=1,x-y>=-1,2x-y 设x,y满足约束条件x+y>=1,x-y>=-1,2x-y {3(x+y)+3(y+x)=1,3(x+y)+4(y-x)=-1 常微分方程y'=(x+y)ln(x+y)-1