作业帮y=log(4)(1-2x+x^2) =log(2)[(1-x)^2] /log(2)(4) =2[log(2)|1-x|] /2 这几步不太理解,注：题中是以4为底和以2为底
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y=log(4)(1-2x+x^2) =log(2)[(1-x)^2] /log(2)(4) =2[log(2)|1-x|] /2 这几步不太理解,注：题中是以4为底和以2为底
y=log(4)(1-2x+x^2) =log(2)[(1-x)^2] /log(2)(4) =2[log(2)|1-x|] /2 这几步不太理解,
注：题中是以4为底和以2为底
y=log(4)(1-2x+x^2) =log(2)[(1-x)^2] /log(2)(4) =2[log(2)|1-x|] /2 这几步不太理解,注：题中是以4为底和以2为底
因为：(1-2x+x^2)=(1-x)^2
所以：log(4)(1-2x+x^2)=log(4)[(1-x)^2]
对其使用换底公式（log(a)b=[log(c)b]/[log(c)a]）,将以4为底,换为以2为底,就有：
log(4)(1-2x+x^2)=log(4)[(1-x)^2]=log(2)[(1-x)^2]/log(2)4
然后对其进行计算,就有：
log(4)(1-2x+x^2)=log(4)[(1-x)^2]=log(2)[(1-x)^2]/log(2)4=2log(2)(1-x)/2
明白了吗?
继续计算,就有：2log(2)(1-x)/2=log(2)(1-x)
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y=log(4)(1-2x+x^2) =log(2)[(1-x)^2] /log(2)(4) =2[log(2)|1-x|] /2 这几步不太理解,注：题中是以4为底和以2为底