Question

Решить систему 2^x+2^y=3; lg(x+y)=0
Решить неравенство
Log(x^3-x^2-2x) по основанию x =<3
Помогите пожалуйста,очень срочно надо

Answer 1

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

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