[tex]\displaystyle\bf\\ODZ:\\\\x^{2} -25 > 0\\\\(x-5)(x+5) > 0[/tex]
+ + + + + (- 5) - - - - - (5) + + + + +
/////////// ///////////
[tex]\displaystyle\bf\\x\in(-\infty \ ; \ -5) \ \cup \ (5 \ ; \ +\infty)\\\\\\\log_{0,1} (x^{2} -25) > 0[/tex]
Представим правую часть как :
[tex]\displaystyle\bf\\0=\log_{0,1} 1[/tex]
Получим :
[tex]\displaystyle\bf\\\log_{0,1} (x^{2} -25) > \log_{0,1} 1\\\\0 < 0,1 < 1 \ \ \ \Rightarrow \ \ \ x^{2} -25 < 1\\\\x^{2}-26 < 0\\\\(x-\sqrt{26} )(x+\sqrt{26} ) < 0[/tex]
+ + + + + (- √26) - - - - - (√26) + + + + +
////////////
[tex]\displaystyle\bf\\x\in(-\sqrt{26} \ ; \ \sqrt{26} )[/tex]
С учётом ОДЗ , получим :
[tex]\displaystyle\bf\\x\in\Big(-\sqrt{26} \ ; \ -5\Big) \ \cup \ \Big(5 \ ; \ \sqrt{26}\Big )[/tex]
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Answers & Comments
[tex]\displaystyle\bf\\ODZ:\\\\x^{2} -25 > 0\\\\(x-5)(x+5) > 0[/tex]
+ + + + + (- 5) - - - - - (5) + + + + +
/////////// ///////////
[tex]\displaystyle\bf\\x\in(-\infty \ ; \ -5) \ \cup \ (5 \ ; \ +\infty)\\\\\\\log_{0,1} (x^{2} -25) > 0[/tex]
Представим правую часть как :
[tex]\displaystyle\bf\\0=\log_{0,1} 1[/tex]
Получим :
[tex]\displaystyle\bf\\\log_{0,1} (x^{2} -25) > \log_{0,1} 1\\\\0 < 0,1 < 1 \ \ \ \Rightarrow \ \ \ x^{2} -25 < 1\\\\x^{2}-26 < 0\\\\(x-\sqrt{26} )(x+\sqrt{26} ) < 0[/tex]
+ + + + + (- √26) - - - - - (√26) + + + + +
////////////
[tex]\displaystyle\bf\\x\in(-\sqrt{26} \ ; \ \sqrt{26} )[/tex]
С учётом ОДЗ , получим :
[tex]\displaystyle\bf\\x\in\Big(-\sqrt{26} \ ; \ -5\Big) \ \cup \ \Big(5 \ ; \ \sqrt{26}\Big )[/tex]