x(y+1)=y-2
xy+x=y-2
y-xy=x+2
y(1-x)=x+2
y=f(x)=(x+2)/(1-x)
分母1-x≠0
定义域(-∞,1)∪(1,+∞)
f(√3)=(√3+2)/(1-√3)
=-(√3+2)(√3+1)/(√3-1)(√3+1)
=-(3+3√3+2)/(3-1)
=-(3√3+5)/2
x(y+1)=y-2
xy+x=y-2
y-xy=x+2
y(1-x)=x+2
y=f(x)=(x+2)/(1-x)
分母1-x≠0
定义域(-∞,1)∪(1,+∞)
f(√3)=(√3+2)/(1-√3)
=-(√3+2)(√3+1)/(√3-1)(√3+1)
=-(3+3√3+2)/(3-1)
=-(3√3+5)/2