Question

25 x^2-3 I3-5xI<30x-9 (там модуль)

Answer 1

1.
$left { {{3-5x geq 0} atop {25x^2-3(3-5x) textless 30x-9}} right. \ \ left {3-5x geq 0} atop {25x^2-9+15x textless 30x-9}} right. \ \ left { {{3-5x geq 0} atop {25x^2-15x textless 0}} right. \ \ left { {{5x-3 leq 0} atop {5x(5x-3) textless 0}} right. \ \ left { {{5x-3 leq 0} atop {x textgreater 0}} right. \ \ left { {{xleq frac{3}{5} } atop {x textgreater 0}} right.$
x ∈ (0; 3/5]
2.
$left { {{3-5x textless 0} atop {25x^2+3(3-5x) textless 30x-9}} right. \ \ left { {{3-5x textless 0} atop {25x^2+9-15x textless 30x-9}} right. \ \ left { {{3-5x textless 0} atop {25x^2-45x+18 textless 0}} right. \ \ 25x^2-45x+18=0 \ D = 2025-1800=225 \ x_1= frac{45-15}{50} = frac{3}{5} \ x_2= frac{45+15}{50} = frac{6}{5} \ left { {{3-5x textless 0} atop {25(x- frac{3}{5} )(x- frac{6}{5}) textless 0}} right.$
$left { {{5x-3 textgreater 0} atop {5(5x- 3)(x- frac{6}{5}) textless 0}} right. \ \left { {{5x-3 textgreater 0} atop {x- frac{6}{5} textless 0}} right. \ \left { {{x textgreater frac{3}{5} } atop {x textless frac{6}{5}}} right.$
x ∈ (3/5; 6/5)
Объединяя два решения, окончательно получаем: x ∈ (0; 6/5).