Question

Помо7ите решить по алгебре номер 741 (2,4)

Answer 1

#2

$\left \{ {{0\ \textless \ 1-3x\ \textless \ 1} \atop {3-4x\ \textless \ 2}} \right.$

$\left \{ {{x\ \textgreater \ 0.25} \atop {-1\ \textless \ -3x\ \textless \ 0}} \right.$

$\left \{ {{x\ \textgreater \ \frac{1}{4} } \atop {0\ \textless \ x\ \textless \ \frac{1}{3} }} \right.$

x∈($\frac{1}{4}$;$\frac{1}{3}$)

#4


$\left \{ {{3x-2 \leq 5x-8} \atop { \frac{2x-1}{2-x}\ \textless \ 4 }} \right.$

$\left \{ {{x \geq 3} \atop { \frac{2x-1}{2-x}\ \textgreater \ 4 }} \right.$

$\left \{ {{x \geq 3} \atop { \frac{2x-1}{2-x}-4\ \textless \ 0 }} \right.$

$\left \{ {{x \geq 3} \atop { \frac{2x-1-8+4x}{2x}\ \textless \ 0 }} \right.$

$\left \{ {{x \geq 3} \atop { \frac{6x-9}{2x}\ \textless \ 0 }} \right.$

Найдём нули

6x-9=0
2x=0

x=1.5
x=0

$\left \{ {{x \geq 3} \atop {{1.5\ \textgreater \ x\ \textgreater \ 0} }} \right.$

x∈(1.5;3]