Question

Решите пункт б плиз.

Answer 1

$left [ {{x(x+1) leq 0} atop {3x-9 textgreater 0}} right. ; ; left [ {{xin [-1,0; ]} atop {3x textgreater 9}} right. ; ; left [ {{xin [-1,0; ]} atop {x textgreater 3}} right. ; ; left [ {{xin [-1,0, ]} atop {xin (3,+infty )}} right. ; Rightarrow \\ xin [-1,0, ]cup (3,+infty )\\\x(x+1) leq 0quad +++[-1, ]---[, 0, ]+++quad xin [-1,0, ]$