Помогите ПОЖАЛУЙСТА! Буду очень благодарен :). Created by Yupi2505. algebra-ru

Помогите ПОЖАЛУЙСТА! Буду очень благодарен :)...

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ОДЗ : x > 0 , y > 0

$1)+\left \{ {{2log_{2}x-log_{3}y}=2 \atop {{log_{2}x+log_{3}y}=1}} \right.\\--------\\ 3log_{2}x=3\\\\log_{2}x=1\\\\x=2\\\\log_{3}y=1-log_{2}x=1-1=0\\\\y=1\\\\Otvet:\boxed{(2 \ ; \ 1)}$

$2)\left \{ {{8^{3x-1}=1 } \atop {5^{4x-y}=5 }} \right. \\\\\left \{ {{8^{3x-1}=8^{0} } \atop {5^{4x-y}=5 }} \right.\\\\\left \{ {{3x-y=0} \atop {4x-y=1}} \right.\\\\\left \{ {{y=3x} \atop {4x-3x=1}} \right. \\\\\left \{ {{x=1} \atop {y=3}} \right.\\\\Otvet:\boxed{(1 \ ; \ 3)}\\\\\boxed{2*3+2*1=8}$

3 votes Thanks 1

MatemaT123

Ответ:

$(2; 1);$

$(1; 3) \quad ; \quad 8 \quad ;$

Объяснение:

1. ОДЗ:

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

Решение:

$\displaystyle \left \{ {{2log_{2}x-log_{3}y=2} \atop {log_{2}x+log_{3}y=1}} \right. \bigg | + \Leftrightarrow \left \{ {{3log_{2}x=3} \atop {log_{2}x+log_{3}y=1}} \right. \Leftrightarrow \left \{ {{log_{2}x=1} \atop {1+log_{3}y=1}} \right. \Leftrightarrow$

$\displaystyle \left \{ {{x=2^{1}} \atop {log_{3}y=0}} \right. \Leftrightarrow \left \{ {{x=2} \atop {y=3^{0}}} \right. \Leftrightarrow \left \{ {{x=2} \atop {y=1}} \right. ;$

$(2; 1);$

$2. $ \displaystyle \left \{ {{8^{3x-y}=1} \atop {5^{4x-y}=5}} \right. \Leftrightarrow \left \{ {{8^{3x-y}=8^{0}} \atop {5^{4x-y}=5^{1}}} \right. \Leftrightarrow \left \{ {{3x-y=0} \atop {4x-y=1}} \right. \Leftrightarrow \left \{ {{y=3x} \atop {4x-3x=1}} \right. \Leftrightarrow $$