(sqrt (49-x^2) *log5 x)/ x -5 =>(больше или равно) 0. Created by diominkirill. algebra-ru

x -5 =>(больше или равно) 0...

diominkirill @diominkirill

June 2022 1 7 Report

(sqrt (49-x^2) *log5 x)/ x -5 =>(больше или равно) 0

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$\frac{\sqrt{49-x^2}\cdot log_5x}{x-5} \geq 0\\\\ODZ:\; \; \left \{ {{49-x^2 \geq 0} \atop {x\ \textgreater \ 0,\; x-5\ne 0}} \right. \; \left \{ {{(7-x)(7+x) \geq 0} \atop {x\ \textgreater \ 0,\; x\ne 5}} \right. \; \left \{ {{x\in [-7,7\, ]} \atop {x\ \textgreater \ 0,\; x\ne 5}} \right. \; x\in (0,5)\cup (5,7\, ]\\\\Tak\; kak\; \sqrt{49-x^2} \geq 0\; ,\; to\; \; \frac{log_5x}{x-5} \geq 0\; \; \; \Rightarrow \\\\ \left \{ {{log_5x \geq 0} \atop {x-5\ \textgreater \ 0}} \right. \; \; ili\; \; \left \{ {{log_5x \leq 0} \atop {x-5\ \textless \ 0}} \right.$

$\left \{ {{x \geq 1} \atop {x\ \textgreater \ 5}} \right. \; \; ili\; \; \; \left \{ {{0\ \textless \ x \leq 1} \atop {x\ \textless \ 5}} \right. \\\\x\ \textgreater \ 5\quad ili\quad 0\ \textless \ x \leq 1\\\\ \left \{ {{x\in (0,1\; ]\cup (5,+\infty )} \atop {x\in (0,5)\cup (5,7\, ]}} \right. \\\\Otvet:\; \; x\in (0,1\, ]\cup (5,7\; ]\; .$

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