Решить систему 2^x+2^y=3; lg(x+y)=0 решить неравенство log(x^3-x^2-2x) по основанию x =< 3 , надо

Ответ:

Про100карина666

03.10.2020 18:50

$\left \{ {{2^x+2^y=3,} \atop {\lg(x+y)=0;}} \right. \left \{ {{2^x+2^y=3,} \atop {x+y=1;}} \right. \left \{ {{y=1-x,} \atop 2^x+2^{1-x}=3;}} \right. \\2^{2x}-3\cdot2^x+2=0, \\ 2^x=t, \\ t^2-3t+2=0, \\ t_1=1, \ t_2=2, \\ \left [ {{2^x=1,} \atop {2^x=2;}} \right. \left [ {{x=0,} \atop {x=1;}} \right. \\ \left [ {{y=1,} \atop {y=0;}} \right. \\ (0;1), \ (1;0).$

$\log_x(x^3-x^2-2x) \leq 3, \\ \left [ {{ \left\{\begin{array}{c} 0\ \textless \ x\ \textless \ 1, \\ x^3-x^2-2x\ \textgreater \ 0, \\ x^3-x^2-2x \geq x^3; \end{array}\right.} \atop {\left\{\begin{array}{c} x\ \textgreater \ 1, \\ x^3-x^2-2x\ \textgreater \ 0, \\ x^3-x^2-2x \leq x^3; \end{array}\right.}} \right. \left [ {{ \left\{\begin{array}{c} 0\ \textless \ x\ \textless \ 1, \\ x(x^2-x-2)\ \textgreater \ 0, \\ -x^2-2x \geq 0; \end{array}\right.} \atop {\left\{\begin{array}{c} x\ \textgreater \ 1, \\ x(x^2-x-2)\ \textgreater \ 0, \\ -x^2-2x \leq 0; \end{array}\right.}} \right.$
$\left [ {{ \left\{\begin{array}{c} 0\ \textless \ x\ \textless \ 1, \\ x(x+1)(x-2)\ \textgreater \ 0, \\ x(x+2) \leq 0; \end{array}\right.} \atop {\left\{\begin{array}{c} x\ \textgreater \ 1, \\ x(x+1)(x-2)\ \textgreater \ 0, \\ x(x+2) \geq 0; \end{array}\right.}} \right. \left [ {{ \left\{\begin{array}{c} 0\ \textless \ x\ \textless \ 1, \\ -1\ \textless \ x\ \textless \ 0\cup x\ \textgreater \ 2, \\ -2\ \textless \ x\ \textless \ 0; \end{array}\right.} \atop {\left\{\begin{array}{c} x\ \textgreater \ 1, \\ -1\ \textless \ x\ \textless \ 0\cup x\ \textgreater \ 2, \\ x\ \textless \ -2 \cup x\ \textgreater \ 0; \end{array}\right.}} \right. \\ \left [ {{x\in\varnothing,} \atop {x\ \textgreater \ 2} \right.$
$x\in(2;+\infty)$