Question

Помогите решить систему уравнений

Answer 1

$\displaystyle\tt \left \{ {{\log_4x+\log_4y=1+\log_4 9} \atop {x+y-20=0}} \right. \left \{ {{\log_4xy=\log_436} \atop {x=20-y}} \right. \left \{ {{xy=36} \atop {x=20-y}} \right. \\\left \{ {{(20-y)y=36} \atop {x=20-y}} \right. \left \{ {{y^2-20y+36=0 \;\;\; (1)} \atop {x=20-y}} \right. \\\\(1) \; y^2-20y+36=0\\\left \{ {{y_1y_2=36} \atop {y_1+y_2=20}} \right. \left [{ {{y=18} \atop {y=2}} \right.$

$\left[\begin{gathered}\left\{\begin{gathered} y=18 \hfill\\ x=20-y \hfill\\\end{gathered}\right. \hfill\\ \left\{ \begin{gathered} y=2 \hfill\\ x=20-y \hfill \\ \end{gathered}\right. \hfill \\\end{gathered}\right. \left[\begin{gathered}\left\{\begin{gathered} y=18 \hfill\\ x=2 \hfill\\\end{gathered}\right. \hfill\\ \left\{ \begin{gathered} y=2 \hfill\\ x=18 \hfill \\ \end{gathered}\right. \hfill \\\end{gathered}\right.$

Ответ: $\tt (2;18);(18;2)$

$\displaystyle \tt \left \{ {{\lg(x^2+y^2)=2 } \atop {\log_2x-4=\log_23-\log_2y}} \right. \left \{ {{x^2+y^2=100} \atop {\log_2x+\log_2y=\log_23+4}} \right. \left \{ {{x^2+y^2=100} \atop {\log_2xy=\log_248}} \right. \\\\\\ \left \{ {{x^2+y^2=100} \atop {xy=48 \;\;\;\;|\cdot 2}} \right. \left|+\left \{ {{x^2+2xy+y^2=196} \atop {xy=48}} \right. \left \{ {{(x+y)^2=14^2} \atop {xy=48}} \right. \left \{ {{x+y=14} \atop {xy=48}} \right.$

$\displaystyle \tt \left \{ {{x=14-y} \atop {y(14-y)=48}} \right. \left \{ {{x=14-y} \atop {y^2-14y+48=0}} \right. \left \{ {{x=14-y} \atop {(y-6)(y-8)=0 }} \right. \left \{ {{x=14-y} \atop {\left [{ {{y=6} \atop {y=8}} \right. }} \right.$

$\left[\begin{gathered} \left\{\begin{gathered} x=14-y \hfill \\ y=6 \hfill \\ \end{gathered} \right. \hfill \\ \left\{ \begin{gathered} x=14-y \hfill \\ y=8 \hfill \\ \end{gathered} \right. \hfill \\ \end{gathered} \right. \left[\begin{gathered} \left\{\begin{gathered} x=8 \hfill \\ y=6 \hfill \\ \end{gathered} \right. \hfill \\ \left\{ \begin{gathered} x=6 \hfill \\ y=8 \hfill \\ \end{gathered} \right. \hfill \\ \end{gathered} \right.$

Ответ: $\tt (6;8); (8;6)$

$\displaystyle \left \{ {{2^x\cdot 3^y=6} \atop {3^x\cdot 4^y=12}} \right. \left \{ {{2^x\cdot 3^y=2^1\cdot 3^1} \atop {3^x\cdot 4^y=3^1\cdot 4^1}} \right. \left \{ {{x=1} \atop {y=1}} \right.$

Ответ: $(1;1)$