Question

求函数y=x^2+4x-12/x^2+x-6的值域

Answer 1

值域是(-∞,1)∪(1,+∞).将分子,分母因式分解,则有y=(x+6)/(x+3).===>y=(x+6)/(x+3)=[(x+3)+3]/(x+3)=1+[3/(x+3)]===>(y-1)(x+3)=3.故y≠1.