Question

решите систему уравнений срочно пожалуйста​

Answer 1

Ответ:

$(4; 1); \quad (\frac{1}{9}; 1\frac{7}{9});$

Объяснение:

а) ОДЗ:

$x \geq 0, \quad y \geq 0;$

Решение:

$\left \{ {{2^{x} \cdot 4^{y}=64} \atop {\sqrt{x}+\sqrt{y}=3}} \right. \Leftrightarrow \left \{ {{2^{x} \cdot 2^{2y}=2^{6}} \atop {x+2\sqrt{xy}+y=9}} \right. \Leftrightarrow \left \{ {{x+2y=6} \atop {2\sqrt{xy}=9-(x+y)}} \right. \Leftrightarrow \left \{ {{x+2y=6} \atop {4xy=81-18(x+y)+(x+y)^{2}}} \right. \Leftrightarrow$

$\Leftrightarrow \left \{ {{x=6-2y} \atop {4y(6-2y)=81-18(6-2y+y)+(6-2y+y)^{2}}} \right. \Leftrightarrow \left \{ {{x=6-2y} \atop {24y-8y^{2}=81-18(6-y)+(6-y)^{2}}} \right. \Leftrightarrow$

$\Leftrightarrow \left \{ {{x=6-2y} \atop {24y-8y^{2}=81-108+18y+36-12y+y^{2}}} \right. \Leftrightarrow \left \{ {{x=6-2y} \atop {24y-8y^{2}=9+6y+y^{2}}} \right. \Leftrightarrow \left \{ {{x=6-2y} \atop {9y^{2}-18y+9=0}} \right. \Leftrightarrow$

$\Leftrightarrow \left \{ {{x=6-2y} \atop {y^{2}-2y+1=0}} \right. \Leftrightarrow \left \{ {{x=6-2y} \atop {(y-1)^{2}=0}} \right. \Leftrightarrow \left \{ {{x=6-2y} \atop {y-1=0}} \right. \Leftrightarrow \left \{ {{x=6-2y} \atop {y=1}} \right. \Leftrightarrow \left \{ {{x=6-2 \cdot 1} \atop {y=1}} \right. \Leftrightarrow \left \{ {{x=4} \atop {y=1}} \right. ;$

$(4; 1);$

б) ОДЗ:

$x \geq 0, \quad y \geq 0, \quad y > x;$

Решение:

$\left \{ {{9^{x} \cdot 3^{y}=9} \atop {\sqrt{y}-\sqrt{x}=1}} \right. \Leftrightarrow \left \{ {{3^{2x} \cdot 3^{y}=3^{2}} \atop {y-2\sqrt{xy}+x=1}} \right. \Leftrightarrow \left \{ {{2x+y=2} \atop {2\sqrt{xy}=(x+y)-1}} \right. \Leftrightarrow \left \{ {{y=2-2x} \atop {4xy=(x+y)^{2}-2(x+y)+1}} \right. \Leftrightarrow$

$\Leftrightarrow \left \{ {{y=2-2x} \atop {4x(2-2x)=(x+2-2x)^{2}-2(x+2-2x)+1}} \right. \Leftrightarrow \left \{ {{y=2-2x} \atop {8x-8x^{2}=(2-x)^{2}-2(2-x)+1}} \right. \Leftrightarrow$

$\Leftrightarrow \left \{ {{y=2-2x} \atop {8x-8x^{2}=4-4x+x^{2}-4+2x+1} \right. \Leftrightarrow \left \{ {{y=2-2x} \atop {8x-8x^{2}=1+x^{2}-2x}} \right. \Leftrightarrow \left \{ {{y=2-2x} \atop {9x^{2}-10x+1=0}} \right. \Leftrightarrow$$\Leftrightarrow \left \{ {{y=2-2x} \atop {x^{2}-1\frac{1}{9}x+\frac{1}{9}=0}} \right. \Leftrightarrow \left \{ {{y=2-2x} \atop {x=\frac{1}{9}}} \right. \quad \vee \quad \left \{ {{y=2-2x} \atop {x=1}} \right. \Leftrightarrow \left \{ {{y=2-2 \cdot \frac{1}{9}} \atop {x=\frac{1}{9}}} \right. \quad \vee \quad \left \{ {{y=2-2 \cdot 1} \atop {x=1}} \right. \Leftrightarrow$

$\Leftrightarrow \left \{ {{x=\frac{1}{9}} \atop {y=1\frac{7}{9}}} \right. \quad \vee \quad \left \{ {{x=1} \atop {y=0}} \right. ;$

Вторая пара корней не удовлетворяет ОДЗ.

$(\frac{1}{9}; 1\frac{7}{9});$