Question

Решите пожалуйста систему уравнений:
$1)\left \{ {{2^{x}+2^{y}=12} \atop {x+y=5}} \right.$

$2)\left \{ {{2^{x+3}-2^{y+1}=8} \atop {2x-y=0}} \right.$

Answer 1

Ответ:

$1)(3;2)(2;3)\\2)(1;2)$

Объяснение:

$1)\left \{ {{2^{x}+2^{y}=12} \atop {x+y=5}} \right. \\\\\left \{ {{2^{x}+2^{y}=12} \atop {x=5-y}} \right. \\\\\\2^{5-y}+2^{y}=12\\\\\frac{32}{2^{y}} +2^{y}=12\\\\2^{y}=b\\\\\frac{32}{b} +b-12=0\\\\b^{2}-12b+32=0\\D=144-128=16\\x_{1}=4\\x_{2}=8\\\\2^{y}=4=2^{2}\\y_{1}=2\\2^{y}=8=2^{3}\\y_{2}=3\\\\\\x=5-y\\x_{1}=5-2\\x_{1}=3\\x_{2}=5-3\\x_{2}=2\\Otvet:(3;2)(2;3)$

$2)\left \{ {{2^{x+3}-2^{y+1}=8} \atop {2x-y=0}} \right. \\\\\left \{ {{2^{x+3}-2^{y+1}=8} \atop {y=2x}} \right. \\\\\\2^{x+3}-2^{2x+1}=8\\\\2^{x}+8-2^{2x}+2=8\\\\2^{x}=g\\\\g+8-g^{2}+2-8=0\\\\-g^{2}+g+2=0\\\\D=9\\x_{1}=2\\x_{2}=-1\\\\\\2^{x}=g\\2^{x}=2^{1}\\x=1\\\\y=2x\\y=2*1\\y=2\\\\Otvet:(1;2)$