Question

СРочно!!! пж очень срочно​

Answer 1

Ответ: (4;4), (2;6)

Объяснение:

Во вложении

Answer 2

Ответ:

$(2; 6)$

Объяснение:

ОДЗ:

$y-6 \geq 0 \Rightarrow y \geq 6;$

Решение:

$\left \{ {{(x-4)\sqrt{y-6}=0} \atop {x+y=8}} \right. \Leftrightarrow \left \{ {{(x-4)\sqrt{y-6}=0} \atop {x=8-y}} \right. \Leftrightarrow \left \{ {{(8-y-4)\sqrt{y-6}=0} \atop {x=8-y}} \right. \Leftrightarrow \left \{ {{(4-y)\sqrt{y-6}=0} \atop {x=8-y}} \right. \Leftrightarrow$

$\Leftrightarrow \left \{ {{((4-y)\sqrt{y-6})^{2}=0^{2}} \atop {x=8-y}} \right. \Leftrightarrow \left \{ {{(4-y)^{2} \cdot (\sqrt{y-6})^{2}=0} \atop {x=8-y}} \right. \Leftrightarrow \left \{ {{(4-y)^{2} \cdot (y-6)=0} \atop {x=8-y}} \right. \Leftrightarrow$

$\Leftrightarrow \left \{ {{(4-y)^{2}=0} \atop {x=8-y}} \right. \quad \vee \quad \left \{ {{y-6=0} \atop {x=8-y}} \right. \Leftrightarrow \left \{ {{4-y=0} \atop {x=8-y}} \right. \quad \vee \quad \left \{ {{y=6} \atop {x=8-y}} \right. \Leftrightarrow \left \{ {{y=4} \atop {x=4}} \right. \quad \vee \quad \left \{ {{y=6} \atop {x=2}} \right. ;$

Пара (4; 4) не удовлетворяет ОДЗ, так как

$y=4<6;$

Остаётся пара

$(2; 6);$