1 函数f(x)(x属于R)为奇函数,f(1)=1/2,f(x+2)=f(x)+f(2),则f(5)=-----

1。f(x),x∈R为奇函数,那么f(-x)=-f(x)f(-1)=-f(1)=-1/2f(1)=f(-1+2)=f(-1)+f(2)=-1/2+f(2)=1/2得f(2)=1f(3)=f(1+2)=f(1)+f(2)=1/2+1=3/2f(5)=f(3+2)=f(3)+f(2)=3/2+1=5/22。2p+1/p+2＜0当p＞0时,该不等式显然不成立当p＜0时,不等式两边同乘以p,得2p²+2p+1＞0得p＜0p的范围是p＜0要使z²-2z+5-P²=0有实根,那么Δ=4-4(5-P²)=4p²-16≥0得p≤-2或p≥2当p＜-2时,z²-2z+5-P²=0有实根;当-2＜p＜0时,z²-2z+5-P²=0无实根3。x²-2x+3=(x-1)²+2有最小值,由指数函数的单调性可知0＜a＜1log(x²-5x+7)＞00＜x²-5x+7＜1x²-5x+7＞0Δ=25-28=-3＜0x²-5x+7恒大于0x²-5x+7＜1x²-5x+6＜0(x-2)(x-3)＜02＜x＜3log(x²-5x+7)＞0的解集是x∈(2,3)4。f(x)=ln[(1+x)/(1-x)](1+x)/(1-x)＞0当x＜1时1+x＞0得-1＜x＜1当x＞1时1+x＜0x是空集得f(x)的定义域是(-1,1)-1＜x/2＜1得-2＜x＜2-1＜1/x＜1得x＞1取两者交集得g(x)的定义域(1,2)