Question

Решите прошууууууууууу

Answer 1

Объяснение:

1)

$\left \{ {{\frac{1}{3}x+\frac{2}{3} > 0\ |*3 } \atop {2x-\frac{1}{3} < 0\ |*3}} \right.\ \ \ \ \left \{ {{x+2 > 0} \atop {6x-1 < 0}} \right. \ \ \ \ \ \left \{ {{x > -2} \atop {6x < 1\ |:6}} \right.\ \ \ \ \left \{ {{x > -2} \atop {x < \frac{1}{6} }} \right.\ \ \ \ \Rightarrow\ \ \ \ x\in(-2;\frac{1}{6}).$

2)

$\left \{ {{0,5x+\frac{1}{10} < \frac{2}{3}x-0,7 } \atop {0,2x-\frac{1}{5} < 1}} \right. \ \ \ \ \left \{ {{0,5x+0,1 < \frac{2}{3}x-0,7 } \atop {0,2x-0,2 < 1}} \right. \ \ \ \ \left \{ {{\frac{2}{3}x-0,5x > 0,8\ |*3 } \atop {0,2x < 1,2\ |:0,2}} \right. \ \ \ \ \left \{ {{2x-1,5x > 2,4} \atop {x < 6}} \right.$

$\left \{ {{0,5x > 2,4\ |:0,5} \atop {x < 6}} \right. \ \ \ \ \left \{ {{x > 4,8} \atop {x < 6}} \right.\ \ \ \ \Rightarrow\ \ \ \ x\in(4,8;6).$