(1/3)^[log(1/2)^(x^2-3x+1)]
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(1/3)^[log(1/2)^(x^2-3x+1)]
(1/3)^[log(1/2)^(x^2-3x+1)]
(1/3)^[log(1/2)^(x^2-3x+1)]
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1=(1/3)^0,而且函数y=(1/3)^x是单调减的.
于是原不等式等价于 log_0.5 (x^2-3x+1)>0,0=log_0.5 (1) 函数y=log_0.5 (x)也是单调减的.故上述不等式等价于 0
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