函数y=(x^2-1)/(x^2-4x+3)的值域为?

Question

函数y=(x^2-1)/(x^2-4x+3)的值域为?

Answer 1

定义域x²-4x+3≠0 得x≠1,x≠3
y=(x^2-1)/(x^2-4x+3)=【（x-1）（x+1）】/【（x-1）（x-3）】=(x+1)/(x-3)=1+4/(x-3)
当x=1,时y=-1
所以值域（负无穷,-1）∪（-1,1）∪（1,正无穷）.