Question

помогите решить пожалуйста​

Answer 1

Ответ:

$(1; \quad 3);$

Объяснение:

ОДЗ:

$\left \{ {{y>0} \atop {2y+3>0}} \right. \Leftrightarrow \left \{ {{y>0} \atop {y>-\frac{3}{2}}} \right. \Leftrightarrow y>0 \Leftrightarrow y \in (0; +\infty);$

Решение:

$\left \{ {{3^{x}=y}, \atop {9^{x}=2y+3}.} \right. \Leftrightarrow \left \{ {{3^{x}=y}, \atop {(3^{2})^{x}=2y+3}.} \right. \Leftrightarrow \left \{ {{3^{x}=y}, \atop {3^{2 \cdot x}=2y+3}.} \right. \Leftrightarrow \left \{ {{3^{x}=y}, \atop {3^{x \cdot 2}=2y+3}.} \right. \Leftrightarrow \left \{ {{3^{x}=y}, \atop {(3^{x})^{2}=2y+3}.} \right. \Leftrightarrow$

$\Leftrightarrow \left \{ {{(3^{x})^{2}=y^{2}}, \atop {(3^{x})^{2}=2y+3}.} \right. \Leftrightarrow \left \{ {{3^{x}=y}, \atop {y^{2}=2y+3}.} \right. \Leftrightarrow \left \{ {{3^{x}=y}, \atop {y^{2}-2y-3=0}.} \right. \Leftrightarrow \left \{ {{3^{x}=y}, \atop {(y+1)(y-3)=0}.} \right. \Leftrightarrow$

$\Leftrightarrow \left \{ {{3^{x}=y} \atop {y+1=0}} \right. \quad \vee \quad \left \{ {{3^{x}=y} \atop {y-3=0}} \right. \Leftrightarrow \left \{ {{3^{x}=y} \atop {y=-1}} \right. \quad \vee \quad \left \{ {{3^{x}=y} \atop {y=3}} \right. \Leftrightarrow \varnothing \quad \vee \quad \left \{ {{3^{x}=3} \atop {y=3}} \right. \Leftrightarrow$

$\Leftrightarrow \left \{ {{x=1} \atop {y=3}} \right. ;$

$(1; \quad 3);$