25 x^2-3 I3-5xI<30x-9 (там модуль)

Ответы 1

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