Упростите выражение: [tex](log_{a}{b}+log_{b}{a}+2)(log_{a}{b}-log_{ab}{b})log_{b}{a}-1[/tex]. Created by Kатюша69. algebra-ru

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Kатюша69 @Kатюша69

July 2021 1 23 Report

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

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Artem112

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$(\log_{a}{b}+\log_{b}{a}+2)(\log_{a}{b}-\log_{ab}{b})\log_{b}{a}-1 = \\\ = \left(\log_{a}{b}+ \dfrac{1}{\log_{a}{b}} +2\right) \left(\log_{a}{b}- \dfrac{1}{\log_{b}{ab}} \right) \dfrac{1}{\log_{a}{b}} -1 = \\\ =\left(\dfrac{\log^2_{a}{b}}{\log_{a}{b}}+ \dfrac{1}{\log_{a}{b}} +\dfrac{2\log_{a}{b}}{\log_{a}{b}}\right) \left(\log_{a}{b}- \dfrac{1}{\log_{b}{a}+\log_{b}{b}} \right)\dfrac{1}{\log_{a}{b}}-1 =$
$=\dfrac{\log^2_{a}{b}+2\log_{a}{b}+1}{\log_{a}{b}} \left(\log_{a}{b}- \dfrac{1}{\log_{b}{a}+1} \right)\dfrac{1}{\log_{a}{b}}-1 = \\\ 
=\dfrac{(\log_{a}{b}+1)^2}{\log^2_{a}{b}} \left(\log_{a}{b}- \dfrac{1}{ \frac{1}{\log_{a}{b}} +1} \right)-1 = \\\
=\dfrac{(\log_{a}{b}+1)^2}{\log^2_{a}{b}} \left(\log_{a}{b}- \dfrac{1}{ \frac{1+\log_{a}{b}}{\log_{a}{b}} } \right)-1 = \\\ 
=\dfrac{(\log_{a}{b}+1)^2}{\log^2_{a}{b}} \left(\log_{a}{b}- \dfrac{\log_{a}{b}}{1+\log_{a}{b} } \right)-1 =$
$=\dfrac{(\log_{a}{b}+1)^2}{\log^2_{a}{b}} \cdot \dfrac{\log_{a}{b}+\log^2_{a}{b}-\log_{a}{b}}{1+\log_{a}{b} } -1 = \\\ = 
\dfrac{(\log_{a}{b}+1)^2}{\log^2_{a}{b}} \cdot \dfrac{\log^2_{a}{b}}{1+\log_{a}{b} } -1 =(\log_{a}{b}+1)-1=\log_{a}{b}$

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