Question

Решите систему пожалуйста очень надо

Answer 1

ОДЗ : x > 0

$\left \{ {x^{y}=4 } \atop {y+log_{2}x=3}} \right\\\\\left \{ {{x^{y}=2^{2}} \atop {log_{2}x=3-y}} \right.\\\\\left \{ {{log_{2} x^{y}=log_{2}2^{2}} \atop {log_{2}=3-y}} \right.\\\\\left \{ {{ylog_{2}x=2 } \atop {log_{2}x=3-y }} \right.\\\\\left \{ {{y(3-y)=2} \atop {log_{2}x=3-y }} \right.$

y(3 - y) = 2

3y - y² - 2 = 0

y² - 3y + 2 = 0

По теореме Виета :

y₁ = 1        y₂ = 2

$1)\left \{ {{y_{1}=1 } \atop {log_{2} x=2}} \right. \\\\\left \{ {{y_{1}=1 } \atop {x_{1}=4 }} \right.\\\\2)\left \{ {{y_{2}=2 } \atop {log_{2}x=1 }} \right. \\\\\left \{ {{y_{2}=2 } \atop {x_{2}=2 }} \right. \\\\Otvet:\boxed{(4;1),(2;2)}$