展開する (ln(-x+1)(-x+1)-xln(x))/(xln(x))

解

展開する

\frac{ln ( - x + 1 ) ( - x + 1 ) - x ln ( x )}{x ln ( x )}

- \frac{ln ( - x + 1 )}{ln ( x )} + \frac{ln ( - x + 1 )}{x ln ( x )} - 1

解答ステップ

\frac{ln ( - x + 1 ) ( - x + 1 ) - x ln ( x )}{x ln ( x )}

拡張

ln (- x + 1) (- x + 1) - x ln (x) : - x ln (- x + 1) + ln (- x + 1) - x ln (x)

= \frac{- x ln ( - x + 1 ) + ln ( - x + 1 ) - x ln ( x )}{x ln ( x )}

分数の規則を適用する:

\frac{a \pm b}{c} = \frac{a}{c} \pm \frac{b}{c}

\frac{- x ln ( - x + 1 ) + ln ( - x + 1 ) - x ln ( x )}{x ln ( x )} = - \frac{x ln ( - x + 1 )}{x ln ( x )} + \frac{ln ( - x + 1 )}{x ln ( x )} - \frac{x ln ( x )}{x ln ( x )}

= - \frac{x ln ( - x + 1 )}{x ln ( x )} + \frac{ln ( - x + 1 )}{x ln ( x )} - \frac{x ln ( x )}{x ln ( x )}

規則を適用

\frac{a}{a} = 1

\frac{x ln ( x )}{x ln ( x )} = 1

= - \frac{x ln ( - x + 1 )}{x ln ( x )} + \frac{ln ( - x + 1 )}{x ln ( x )} - 1

キャンセル

\frac{x ln ( - x + 1 )}{x ln ( x )} : \frac{ln ( - x + 1 )}{ln ( x )}

= - \frac{ln ( - x + 1 )}{ln ( x )} + \frac{ln ( - x + 1 )}{x ln ( x )} - 1