25 x^2-3 I3-5xI

February 2022 1 30 Report

25 x^2-3 I3-5xI<30x-9 (там модуль)

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

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