x^2+4x+7)/(x+1)+2
≥2×√[(x+1)×4/(x+1)
=(x+1)+4/(x+1)]+2=4+2=6,等号成立的条件是x+1=4/,(x+1)/(x+1);(x^2+4x+7)的最大值是1/,即x=1。
所以;(x^2+4x+7)≤1/6,即(x+1)/(x+1)
=[(x+1)^2+2(x+1)+4]/
4
=(x+1)Ǘ/[x+1+1/[x+3+1/[(x+1)x+3(x+1)+6+1]
=1/(2+2)
=1/=1/(x+1)+2]
>(x+1)]
=1/
a = x + 1 > 0;
y = 1 / (a + 2 + 4/a) <= 1/6
当x=1时,最大值是1/6
晕,只要X>-1 的话,y 就边成无限大了。。。。。
还要回答,问题都看不清