解下列不等式:3·log3(log3x) +log 1/3 [log3(9·³√x)]≥1参考答案如下:须使log 3 x >0且log3(9·³√x)>0,解得x>1.此时原不等式化为(log3x + 2)(log3x - 3)/(log3x + 6)≥0,即 -6 <l

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解下列不等式:3·log3(log3x) +log 1/3 [log3(9·³√x)]≥1参考答案如下:须使log 3 x >0且log3(9·³√x)>0,解得x>1.此时原不等式化为(log3x + 2)(log3x - 3)/(log3x + 6)≥0,即 -6 <l
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解下列不等式:3·log3(log3x) +log 1/3 [log3(9·³√x)]≥1参考答案如下:须使log 3 x >0且log3(9·³√x)>0,解得x>1.此时原不等式化为(log3x + 2)(log3x - 3)/(log3x + 6)≥0,即 -6 <l
解下列不等式:
3·log3(log3x) +log 1/3 [log3(9·³√x)]≥1
参考答案如下:
须使log 3 x >0且log3(9·³√x)>0,解得x>1.此时原不等式化为(log3x + 2)(log3x - 3)/(log3x + 6)≥0,即 -6 <log3x≤-2或log3x≥3.原不等式的解为x≥27

解下列不等式:3·log3(log3x) +log 1/3 [log3(9·³√x)]≥1参考答案如下:须使log 3 x >0且log3(9·³√x)>0,解得x>1.此时原不等式化为(log3x + 2)(log3x - 3)/(log3x + 6)≥0,即 -6 <l
解下列不等式:3·log₃(log₃x) +log ‹1/3› [log₃(9·³√x)]≥1
3·log₃(log₃x) +log ‹1/3›[log₃9+log₃∛x]≥1
3·log₃(log₃x) +log ‹1/3›[2+(1/3)log₃x]≥1
3log‹1/3›[1/log₃x]+log‹1/3›[2+(1/3)log₃x]≥1【这里是把第一个对数的底数和真数都取倒数】
log‹1/3›{[1/log₃x]³[2+(1/3)log₃x]}≥1
故0

x》9

答案不应该是x≥27啊,我们来设log3(x)=y,那么有
3·log3(log3x) +log 1/3 [log3(9·³√x)]
=3log3(y)+log 1/3 [2+log3(x^(1/3))]
=3log3(y)-log3(2+y/3)
=log3[y^3/(2+y/3)]
≥1所以得到
y^3/(2+y/3)≥3那么有

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答案不应该是x≥27啊,我们来设log3(x)=y,那么有
3·log3(log3x) +log 1/3 [log3(9·³√x)]
=3log3(y)+log 1/3 [2+log3(x^(1/3))]
=3log3(y)-log3(2+y/3)
=log3[y^3/(2+y/3)]
≥1所以得到
y^3/(2+y/3)≥3那么有
y^3-y-6≥0推出
(y^3-2y^2)+(2y^2-y-6)
=y^2(y-2)+(2y+3)(y-2)
=(y-2)(y^2+2y+3)
≥0解得
y≥2即log3(x)≥2
所以x≥3^2=9

收起

x》9