Question

Помогите пожалуйста, очень срочно

Answer 1

Ответ:

$(1; \quad 2)$

Объяснение:

ОДЗ:

$2^{y}>0, \quad 3^{x}>0;$

Решение:

$\left \{ {{3^{x} \cdot 2^{y}=12} \atop {2^{y+1}-3^{x}=5}} \right. \Leftrightarrow \left \{ {{3^{x} \cdot 2^{y}=12} \atop {3^{x}=2^{y+1}-5}} \right. \Leftrightarrow \left \{ {{(2^{y+1}-5) \cdot 2^{y}=12} \atop {3^{x}=2^{y+1}-5}} \right. \Leftrightarrow \left \{ {{(2^{y} \cdot 2^{1}-5) \cdot 2^{y}=12} \atop {3^{x}=2^{y+1}-5}} \right. \Leftrightarrow$

$\Leftrightarrow \left \{ {{2 \cdot (2^{y})^{2}-5 \cdot 2^{y}=12} \atop {3^{x}=2^{y+1}-5}} \right. ;$

$2 \cdot (2^{y})^{2}-5 \cdot 2^{y}=12;$

$t=2^{y};$

$2t^{2}-5t=12;$

$2t^{2}-5t-12=0;$

$D=b^{2}-4ac;$

$D=(-5)^{2}-4 \cdot 2 \cdot (-12)=25+96=121;$

$t_{1,2}=\frac{-b \pm \sqrt{D}}{2a};$

$t_{1}=\frac{-(-5)+\sqrt{121}}{2 \cdot 2}=\frac{5+11}{4}=\frac{16}{4}=4;$

$t_{2}=\frac{-(-5)-\sqrt{121}}{2 \cdot 2}=\frac{5-11}{4}=\frac{-6}{4}=-1,5;$

Корень t₂ не имеет смысла.

$2^{y}=4;$

$2^{y}=2^{2};$

$y=2;$

$\left \{ {{y=2} \atop {3^{x}=2^{y+1}-5}} \right. \Leftrightarrow \left \{ {{y=2} \atop {3^{x}=2^{2+1}-5}} \right. \Leftrightarrow \left \{ {{y=2} \atop {3^{x}=2^{3}-5}} \right. \Leftrightarrow \left \{ {{y=2} \atop {3^{x}=8-5}} \right. \Leftrightarrow \left \{ {{y=2} \atop {3^{x}=3}} \right. \Leftrightarrow \left \{ {{x=1} \atop {y=2}} \right. ;$

$(1; \quad 2);$