Question

решите пожалуйста!!!

Answer 1

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

$\left \{ {{x=3y-1} \atop {y(2y-3)=0}} \right.\\\\\left \{ {{x=3y-1} \atop {\left[\begin{array}{ccc}y=0\\y=1,5\\\end{array}\right }} \right.\\\\1)\left \{ {{y=0} \atop {x=-1}} \right. \\\\2)\left \{ {{y=1,5} \atop {x=3,5}} \right.$

1) - не подходит, так как x + 1 = - 1 + 1 = 0 - недопустимо

Ответ : (3,5 ; 1,5)