Question

решите систему показательных уравнений с модулем

Answer 1

Ответ:

$(6; 1); \quad (9; 4);$

Объяснение:

$\displaystyle \left \{ {{2^{x-y}-6|3y-x|=14} \atop {|3y-x|-2^{x-y-1}=-13}} \right. \Leftrightarrow \left \{ {{2^{x-y}-6|3y-x|=14} \atop {2|3y-x|-2^{x-y}=-26}} \right. \bigg |+ \Leftrightarrow$

$\displaystyle \Leftrightarrow \left \{ {{-4|3y-x|=-12} \atop {2|3y-x|-2^{x-y}=-26}} \right. \Leftrightarrow \left \{ {{|3y-x|=3} \atop {2 \cdot 3-2^{x-y}=-26}} \right. \Leftrightarrow \left \{ {{3y-x=-3} \atop {2^{x-y}=6-(-26)}} \right. \vee$

$\displaystyle \vee \left \{ {{3y-x=3} \atop {2^{x-y}=6-(-26)}} \right. \Leftrightarrow \left \{ {{3y-x=-3} \atop {2^{x-y}=32}} \right. \vee \left \{ {{3y-x=3} \atop {2^{x-y}=32}} \right. \Leftrightarrow \left \{ {{3y-x=-3} \atop {2^{x-y}=2^{5}}} \right. \vee$

$\displaystyle \vee \left \{ {{3y-x=3} \atop {2^{x-y}=2^{5}}} \right. \Leftrightarrow \left \{ {{3y-x=-3} \atop {x-y=5}} \right. \bigg |+ \vee \left \{ {{3y-x=3} \atop {x-y=5}} \right. \bigg |+ \Leftrightarrow \left \{ {{x=3y+3} \atop {2y=2}} \right. \vee$

$\displaystyle \vee \left \{ {{x=3y-3} \atop {2y=8}} \right. \Leftrightarrow \left \{ {{x=6} \atop {y=1}} \right. \vee \left \{ {{x=9} \atop {y=4}} \right. ;$

$(6; 1); \quad (9; 4);$

V означает "или".

Answer 2

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