x²+9≥9所以log1/3(x²+9)≤log1/3(9)=-2;

x²+9≥9
所以log1/3(x²+9)≤log1/3(9)=-2;

f(x)= log1/3(x)在（0,+正无穷）为单调递减函数,
x越大,对应的f(x)值就越小
所以此处可有设 x1=x² +9 x2=9
x1≥X2 那么log1/3(x1 )≤log1/3(x2)
log1/3(9)=-log3(9)=-2
答案理解过程就是这样,
有用请采纳!