Question

Помогите пж с алгеброй

Answer 1

Ответ:

{(1; -1)}

Объяснение:

Дана система

$\displaystyle \tt \left \{ {{\sqrt{18-16 \cdot x-y^2} =y+2} \atop {y=x-2}} \right. .$

Область допустимых значений: 18-16·х-у²≥0 и у+2≥0.

$\displaystyle \tt \left \{ {{\sqrt{18-16 \cdot x-y^2} =y+2} \atop {y=x-2}} \right. \Leftrightarrow \left \{ {{\sqrt{18-16 \cdot x-(x-2)^2} =x-2+2} \atop {y=x-2}} \right. \Leftrightarrow \\\\\Leftrightarrow \left \{ {{\sqrt{18-16 \cdot x-(x-2)^2} =x,\; x\geq 0} \atop {y=x-2}} \right. \Leftrightarrow \\\\ \Leftrightarrow \left \{ {{18-16 \cdot x-(x-2)^2=x^2,\; x\geq 0} \atop {y=x-2}} \right. \Leftrightarrow$

$\displaystyle \tt \Leftrightarrow \left \{ {{18-16 \cdot x-x^2+4 \cdot x -4=x^2,\; x\geq 0} \atop {y=x-2}} \right. \Leftrightarrow \\\\\Leftrightarrow \left \{ {{2 \cdot x^2+12 \cdot x-14=0,\; x\geq 0} \atop {y=x-2}} \right. \Leftrightarrow \left \{ {{x^2+6 \cdot x-7=0,\; x\geq 0} \atop {y=x-2}} \right. \Leftrightarrow$

$\displaystyle \tt \Leftrightarrow \left \{ {{x^2-1+6 \cdot x-6=0,\; x\geq 0} \atop {y=x-2}} \right. \Leftrightarrow \\\\\Leftrightarrow \left \{ {{(x-1) \cdot (x+1)+6 \cdot (x-1)=0,\; x\geq 0} \atop {y=x-2}} \right. \Leftrightarrow \\\\\Leftrightarrow \left \{ {{(x-1) \cdot (x+1+6)=0,\; x\geq 0} \atop {y=x-2}} \right. \Leftrightarrow \left \{ {{(x-1) \cdot (x+7)=0,\; x\geq 0} \atop {y=x-2}} \right. \Leftrightarrow$

$\displaystyle \tt \Leftrightarrow \left \{ {{x=1>0, \; x = -7 <0,\; x\geq 0} \atop {y=x-2}} \right. \Leftrightarrow \left \{ {{x=1} \atop {y=1-2=-1}} \right. .$

Проверим ОДЗ: 18-16·1-(-1)²=1>0 и -1+2=1>0 - условия выполняются.