Question

怎样求y=(x²-5x+6)/(x²+x-6)的值域

Answer 1

y=(x²-5x+6)/(x²+x-6)
因式分解，得到
y=(x-2)(x-3)/(x+3)(x-2) 其中x不等于2或-3
分式上下同消去（x-2)
y=(x-3)/(x+3)
=(x+3-6)/(x+3)
=1-6/(x+3)
因为x不等于2，所以y不等于1-6/5=-1/5
又因为6/(x+3)不等于0，所以y不等于1
值域就是y属于R且y不等于1和-1/5

Answer 2

y=(x²-5x+6)/(x²+x-6)
=(x-2)(x-3)/(x+3)(x-2)
=(x-3)/(x+3)
=(x+3-6)/(x+3)
=1-6/(x+3)
所以y∈R且y≠1

Answer 3

y=(x²-5x+6)/(x²+x-6)
=(x-2)(x-3)/(x+3)(x-2)
=(x-3)/(x+3)
=(x+3-6)/(x+3)
=1-6/(x+3)
所以y∈R且y≠1
另外 由分母不等于0 x不等于2 y≠-1/5

Answer 4

Δ>=0