Question

Решить систему уравнений, 10ый класс

Answer 1

$(\frac{1}{4};\frac{1}{2} );(2;4)$

Объяснение:

$\left \{ {{x^{\log_{2}y}+y^{log_{2}x}=8} \atop {log_{2}y-log_{2}x=1}} \right.$

ОДЗ:  $x0, ~~~ y0$

$\left \{ {{(2^{\log_2{x}})^{\log_{2}y}+(2^{\log_2{y}})^{log_{2}x}=8} \atop {\log_{2}y=\log_{2}2+\log_2{x}}} \right. \\ \\ \left \{ {{2^{\log_2{x}\log_{2}y}+2^{\log_2{x}\log_{2}y}=8} \atop {\log_{2}y=\log_{2}(2x)}} \right. \\ \\ \left \{ {{2\cdot 2^{\log_2{x}\log_{2}y}=8} \atop {y=2x}} \right. \\ \\ \left \{ {{2^{\log_2{x}\log_{2}(2x)}=4} \atop {y=2x}} \right. \\ \\ \left \{ {{2^{\log_2{x}(\log_{2}2+\log_{2}x)}=2^2} \atop {y=2x}} \right.\\ \\ \left \{ {{\log_2{x}(1+\log_{2}x)}=2} \atop {y=2x}} \right.$

$\log_{2}x=t,~~t\in R \\ \\ t(1+t)=2 \\ \\ t+t^2=2 \\ \\t^2+t-2=0$

по теореме Виета:  $t_1+t_2=-1,~~t_1\cdot t_2=-2$

$\left[\begin{array}{c}{t_1=-2}&{t_2=1}\end{array}$

$\left[\begin{array}{c}{\log_{2}x=-2}&{\log_2{x}=1}\end{array}$

$\left[\begin{array}{c}{x=2^{-2}}&{x=2^1}\end{array}$

$\left[\begin{array}{c}{x=\frac{1}{4}}&{x=2}\end{array}$

$\left \{ {{x=\frac{1}{4}} \atop {y=2\cdot\frac{1}{4}}} \right.$  или $\left \{ {{x=2} \atop {y=2\cdot 2}} \right.$  

$\left \{ {{x=\frac{1}{4}} \atop {y=\frac{1}{2}}} \right.$  или $\left \{ {{x=2} \atop {y=4}} \right.$  

ответ:  $(\frac{1}{4};\frac{1}{2} );(2;4)$