Verified answer

Ответ:

[tex]\boldsymbol{ \boxed{ x \in (-\infty; - 1) }}[/tex]

Объяснение:

[tex]3^{2x + 1} + 5 \cdot 3^{x} - 2 < 0[/tex]

[tex]3 \cdot 3^{2x} + 5 \cdot 3^{x} - 2 < 0[/tex]

[tex]3 \cdot (3^{x})^{2} + 5 \cdot 3^{x} - 2 < 0[/tex]

Замена: [tex]3^{x} = t; t > 0[/tex]

[tex]3t^{2} + 5t - 2 < 0[/tex]

[tex]3t^{2} + 5t - 2 = 0[/tex]

[tex]D = 25 - 4 \cdot 3 \cdot (-2) = 25 + 24 = 49 = 7^{2}[/tex]

[tex]t_{1} = \dfrac{-5 + 7}{2 \cdot 3} = \dfrac{2}{2 \cdot 3} =\dfrac{1}{3}[/tex]

[tex]t_{2} = \dfrac{-5 - 7}{2 \cdot 3} = -\dfrac{12}{6} = - 2[/tex]

[tex]3t^{2} + 5t - 2 = 3 \bigg(t - \dfrac{1}{3} \bigg)\bigg(t + 2 \bigg)[/tex]

[tex]\displaystyle 3 \cdot (3^{x})^{2} + 5 \cdot 3^{x} - 2 < 0 \Longleftrightarrow \left \{ {{3 \bigg(t - \dfrac{1}{3} \bigg)\bigg(t + 2 \bigg) < 0 \bigg|:3} \atop {t > 0}} \right.[/tex]

[tex]\displaystyle\left \{ {{ \bigg(t - \dfrac{1}{3} \bigg)\bigg(t + 2 \bigg) < 0} \atop {t > 0}} \right \Longleftrightarrow \left \{ {{ -2 < t < \dfrac{1}{3} } \atop {t > 0}} \right \Longleftrightarrow \left \{ {{ -2 < 3^{x} < \dfrac{1}{3} } \atop {3^{x} > 0}} \right[/tex]

[tex]-2 < 3^{x} < \dfrac{1}{3}[/tex]  - так как [tex]3^{x} > -2[/tex], так как по свойствам показательной функции [tex]3^{x} > 0[/tex] при [tex]x \in \mathbb R[/tex] и [tex]0 > -2[/tex].

[tex]3^{x} < \dfrac{1}{3}[/tex]

[tex]3^{x} < 3^{-1} \Longleftrightarrow x < -1 \Longleftrightarrow\boldsymbol{ \boxed{ x \in (-\infty; - 1) }}[/tex]