12 и 13 прошу решите

№ 12 [latex] \left \{ {{x+y=34} \atop {log_{2}x+log_{2}y=6}} \right. [/latex] [latex] \left \{ {{x+y=34} \atop {log_{2}(xy)=log_{2}64}} \right. [/latex] [latex] \left \{ {{x+y=34} \atop {xy=64} \right. [/latex] [latex] \left \{ {{x=34-y} \atop {xy=64} \right. [/latex] [latex]y(34-y)=64[/latex] [latex]34y-y^{2}-64=0[/latex] [latex]y^{2}-34y+64=0, D=34^{2}-4*64=1156-256=900=30^{2}[/latex] [latex]y_{1}= \frac{34-30}{2}=2\ \textgreater \ 0[/latex] [latex]y_{2}= \frac{34+30}{2}=32\ \textgreater \ 0[/latex] [latex] \left \{ {{y=2} \atop {x=34-2=32}} \right. [/latex] [latex] \left \{ {{y=32} \atop {x=34-32=2}} \right. [/latex] Ответ: (2; 32) и (32; 2) № 13 [latex] \left \{ {{log_{ \frac{1}{3} }(x+y)=2} \atop {log_{3}(x-y)=2}} \right. [/latex] [latex] \left \{ {{-log_{3}(x+y)=2} \atop {log_{3}(x-y)=2}} \right. [/latex] [latex] \left \{ {{log_{3}(x+y)=log_{3}( \frac{1}{9} )} \atop {log_{3}(x-y)=log_{3}9}} \right. [/latex] [latex] \left \{ {{x+y= \frac{1}{9}} \atop {x-y=9}} \right. [/latex] [latex] \left \{ {{x+y= \frac{1}{9}} \atop {x=9+y}} \right. [/latex] [latex]9+y+y= \frac{1}{9}[/latex] [latex]2y= \frac{1}{9}-9= \frac{1-81}{9}=- \frac{80}{9}[/latex] [latex]y=- \frac{80}{18}=- \frac{40}{9}=-4 \frac{4}{9}[/latex] [latex]x=9- \frac{40}{9}= \frac{81-40}{9}= \frac{41}{9}=4 \frac{5}{9}[/latex] Ответ: [latex](4 \frac{5}{9}; -4 \frac{4}{9})[/latex]