解不等式log1/2[a^2x+(ab)^x+2·b^2x+1]

解不等式log1/2[a^2x+(ab)^x+2·b^2x+1]
解不等式log1/2[a^2x+(ab)^x+2·b^2x+1]

解不等式log1/2[a^2x+(ab)^x+2·b^2x+1]
log1/2[a^2x+(ab)^x-2·b^2x+1]0
可以十字相乘：(a^x+2b^x)(a^x-b^x)>0
因为a^x+2b^x恒为正,所以：a^x-b^x>0
a^x>b^x
(a/b)^x>1
lg[(a/b)^x]>lg1
xlg(a/b)>0
当a>b时,a/b>1,则lg(a/b)>0,所以：x>0
当a