Упростите выражение: [tex](log_{a}{b}+log_{b}{a}+2)(log_{a}{b}-log_{ab}{b})log_{b}{a}-1[/tex]

Ответы 1

$(\log_{a}{b}+\log_{b}{a}+2)(\log_{a}{b}-\log_{ab}{b})\log_{b}{a}-1 = \\\ = \left(\log_{a}{b}+ \dfrac{1}{\log_{a}{b}} +2ight) \left(\log_{a}{b}- \dfrac{1}{\log_{b}{ab}} ight) \dfrac{1}{\log_{a}{b}} -1 = \\\ =\left(\dfrac{\log^2_{a}{b}}{\log_{a}{b}}+ \dfrac{1}{\log_{a}{b}} +\dfrac{2\log_{a}{b}}{\log_{a}{b}}ight) \left(\log_{a}{b}- \dfrac{1}{\log_{b}{a}+\log_{b}{b}} ight)\dfrac{1}{\log_{a}{b}}-1 =$ $=\dfrac{\log^2_{a}{b}+2\log_{a}{b}+1}{\log_{a}{b}} \left(\log_{a}{b}- \dfrac{1}{\log_{b}{a}+1} ight)\dfrac{1}{\log_{a}{b}}-1 = \\\ =\dfrac{(\log_{a}{b}+1)^2}{\log^2_{a}{b}} \left(\log_{a}{b}- \dfrac{1}{ \frac{1}{\log_{a}{b}} +1} ight)-1 = \\\ =\dfrac{(\log_{a}{b}+1)^2}{\log^2_{a}{b}} \left(\log_{a}{b}- \dfrac{1}{ \frac{1+\log_{a}{b}}{\log_{a}{b}} } ight)-1 = \\\ =\dfrac{(\log_{a}{b}+1)^2}{\log^2_{a}{b}} \left(\log_{a}{b}- \dfrac{\log_{a}{b}}{1+\log_{a}{b} } ight)-1 =$ $=\dfrac{(\log_{a}{b}+1)^2}{\log^2_{a}{b}} \cdot \dfrac{\log_{a}{b}+\log^2_{a}{b}-\log_{a}{b}}{1+\log_{a}{b} } -1 = \\\ = \dfrac{(\log_{a}{b}+1)^2}{\log^2_{a}{b}} \cdot \dfrac{\log^2_{a}{b}}{1+\log_{a}{b} } -1 =(\log_{a}{b}+1)-1=\log_{a}{b}$
- Автор:
  ricky28
- 6 лет назад
- 0
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