Question

Срочно осталось 10 мир !!!!

Answer 1

$sqrt{6+x-x^2} textgreater 1-x, \ left [ {{left { {{1-x textless 0,} atop {6+x-x^2geq0;}} right.} atop {left { {{1-xgeq0,} atop {6+x-x^2 textgreater (1-x)^2;}} right.}} right. left [ {{left { {{-x textless -1,} atop {x^2-x-6leq0;}} right.} atop {left { {{-xgeq-1,} atop {6+x-x^2 textgreater 1-2x+x^2;}} right.}} right. left { {{left { {{x textgreater 1,} atop {x^2-x-6leq0;}} right.} atop { left { {{xleq-1,} atop {2x^2-3x-5 textless 0;}} right.}} right.$
$x^2-x-6=0, \ x_1=-2, x_2=3; \ 2x^2-3x-5=0, \ D=49, \ x_1=-1, x_2=2,5; \ left [ {{left { {{x textgreater -1,} atop {(x+2)(x-3) leq 0;}} right} atop {left { {{xleq-1,} atop {(x+1)(x-2,5) textless 0;}} right}} right.left [ {{left { {{x textgreater -1,} atop {-2leq xleq3;}} right} atop {left { {{xleq-1,} atop {-1 textless x textless 2,5;}} right}} right. left [ {{-1 textless xleq3,} atop {xinvarnothing;}} right. \ -1 textless xleq3, \ xin(-1;3].$

Answer 2

По-моему x=-1,а x2=2,5