12 и 13 прошу решите

Question

Алгебра

Burilanim

5 год назад

12 и 13 прошу решите

ОТВЕТЫ

Устин

Jul 1, 2019

№ 12
$\left \{ {{x+y=34} \atop {log_{2}x+log_{2}y=6}} \right.$

$\left \{ {{x+y=34} \atop {log_{2}(xy)=log_{2}64}} \right.$

$\left \{ {{x+y=34} \atop {xy=64} \right.$

$\left \{ {{x=34-y} \atop {xy=64} \right.$

$y(34-y)=64$
$34y-y^{2}-64=0$
$y^{2}-34y+64=0, D=34^{2}-4*64=1156-256=900=30^{2}$
$y_{1}= \frac{34-30}{2}=2\ \textgreater \ 0$
$y_{2}= \frac{34+30}{2}=32\ \textgreater \ 0$

$\left \{ {{y=2} \atop {x=34-2=32}} \right.$

$\left \{ {{y=32} \atop {x=34-32=2}} \right.$

Ответ: (2; 32) и (32; 2)

№ 13
$\left \{ {{log_{ \frac{1}{3} }(x+y)=2} \atop {log_{3}(x-y)=2}} \right.$

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

$\left \{ {{log_{3}(x+y)=log_{3}( \frac{1}{9} )} \atop {log_{3}(x-y)=log_{3}9}} \right.$

$\left \{ {{x+y= \frac{1}{9}} \atop {x-y=9}} \right.$

$\left \{ {{x+y= \frac{1}{9}} \atop {x=9+y}} \right.$

$9+y+y= \frac{1}{9}$
$2y= \frac{1}{9}-9= \frac{1-81}{9}=- \frac{80}{9}$
$y=- \frac{80}{18}=- \frac{40}{9}=-4 \frac{4}{9}$
$x=9- \frac{40}{9}= \frac{81-40}{9}= \frac{41}{9}=4 \frac{5}{9}$

Ответ: $(4 \frac{5}{9}; -4 \frac{4}{9})$

109