Question

{(9^(x-1)=27),(|x-5|<=5):}

Answer 1

Объяснение:

$\displaystyle\\\left \{ {{9^{x-1}=27} \atop {|x-5|\leq 5}} \right.\ \ \ \ \ \ \left\{\begin{array}{ccc}(3^2)^{x-1}=3^3\\x-5\leq 5\\-(x-5)\leq 5\end{array}\right \ \ \ \ \ \ \left\{\begin{array}{ccc}3^{2x-2}=3^3\\x\leq 10\\-x+5\leq 5\end{array}\right \ \ \ \ \ \ \left\{\begin{array}{ccc}2x-2=3\\x\leq 10\\x\geq 0\end{array}\right\\\\\\\left \{ {{2x=5\ |:2} \atop {x\in[0;10]}} \right. \ \ \ \ \ \ \left \{ {{x=2,5} \atop {x\in[0;10]}\right. \ \ \ \ \ \ \Rightarrow\ \ \ \ \ \ x=2,5.$

Ответ: x=2,5.