①f(x)=x/(x+2)=(x+2-2)/(x+2)=1-2/x
∵-2/x在（負(fù)無窮,-2）單調(diào)遞增,+1為常數(shù)
∴f(x)在（負(fù)無窮,-2）單調(diào)遞增
證畢
②當(dāng)x∈[1,正無窮)∪（-1,負(fù)無窮）時(shí),
f(x)=x²-1+x
則對(duì)稱軸可知：x∈（負(fù)無窮,-1】為減函數(shù),x∈【1,正無窮）為增函數(shù)
當(dāng)x∈（-1,1）時(shí)
f(x)=-x²+x+1
則對(duì)稱軸可知：x∈（-1,1/2）為增函數(shù),x∈【1/2,1)為減函數(shù)
綜上所述：x∈（負(fù)無窮,-1】為減函數(shù),x∈【1,正無窮）為增函數(shù)
x∈（-1,1/2）為增函數(shù),x∈【1/2,1)為減函數(shù)
樓上第一題顯然證的不完整,若f（x）為無規(guī)律曲線函數(shù),則方法不可取
請(qǐng)采納,謝謝