Question

18 subpunctele b si d va rog

Answer 1

$b).3(2|2x-5|-7)-19\ \textless \ 14 \\ 3(2|2x-5|-7)\ \textless \ 14+19 \\ 3(2|2x-5|-7)\ \textless \ 33 \\ 2|2x-5|-7\ \textless \ 33:3 \\ 2|2x-5|-7\ \textless \ 11 \\ 2|2x-5|\ \textless \ 11+7 \\ 2|2x-5|\ \textless \ 18 \\ |2x-5|\ \textless \ 18:2 \\ |2x-5|\ \textless \ 9 \\ 2x-5\ \textgreater \ -9 \\ 2x\ \textgreater \ -4 \\ x\ \textgreater \ -2 \\ 2x-5\ \textless \ 9 \\ 2x\ \textless \ 14 \\ x\ \textless \ 7$
$x$ ∈ $\left \{-1;1;2;3;4;5;6 \left \}$
$d).|x+3|(9-|2x+3|)\ \textgreater \ 0 \\ x+3\ \textless \ 0 \\ x\ \textless \ -3 \\ 2x+3\ \textless \ 0 \\ x\ \textless \ - \frac{3}{2} \\ -(x+3)(2x+12)\ \textgreater \ 0 \\ -2x^2-18x-36\ \textgreater \ 0 \\ -x^2-9x-18\ \textgreater \ 0 \\ x^2+9x+18\ \textless \ 0 \\ (x+3)(x+6)\ \textless \ 0 \\ x+3\ \textless \ 0 \\ x\ \textless \ -3 \\ x+6\ \textless \ 0 \\ x\ \textless \ -6$
$-2x^2+18\ \textgreater \ 0 \\ 2x^2\ \textless \ 18 \\ x^2\ \textless \ 9 \\ x\ \textless \ 3$
$x$ ∈ $\left \{ -5;-4;-3;-1;1;2 \left \}$