Log x^3-9x^2+27x-27(9-x)≥0 нужно подробно! )

kotgemer kotgemer    2   11.07.2019 03:20    2

Ответы
Ангел89178931527 Ангел89178931527  03.10.2020 00:51
log_{ x^{3} - 9 x^{2} +27x-27}(9 - x) \geq 0 \\ x^{3} - 9 x^{2} +27x-27 = (x-3)^{3} =\ \textgreater \ \\ log_{ (x-3)^{3}}(9 - x) \geq 0 \\
 \frac{1}{3} log_{ (x-3)}(9 - x) \geq 0 \\ 

log_{ (x-3)}(9 - x) \geq 0 \\ 
 log_{ (x-3)}(9 - x) - 0 \geq 0 \\ 
 log_{ (x-3)}(9 - x) - log_{ (x-3)}1 \geq 0 \\ 


\left \{ {{(x-3 -1)(9-x-1)\geq 0 \\} \atop {x-3\ \textgreater \ 0, x-3 \neq 1,9 - x\ \textgreater \ 0} } \right. \\
 \left \{ {{(x-4)(8-x)\geq 0 \\} \atop {x\ \textgreater \ 3, x \neq 4,x\ \textless \ 9} } \right. \\
 \left \{ {{(x-4)(x-8) \leq 0 \\} \atop {x\ \textgreater \ 3, x \neq 4,x\ \textless \ 9} } \right. \\
\left \{ {{4 \leq x \leq 8 \\} \atop {x\ \textgreater \ 3, x \neq 4,x\ \textless \ 9} } \right. \\
4\ \textless \ x \leq 8 \\

ответ:  ( 4 ; 8 ].
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