Question

Решите систему уравнений

Answer 1

$\begin{cases}\left(\frac1{36}\right)^{-y^2}=6^{2x+4}\\\log_4(2y^2-2x+4)=6^{\log_6(3\lg\sqrt[3]{10})}+\log_4(x+1)\end{cases}\Rightarrow\\\\\\\Rightarrow\begin{cases}6^{y^2}=6^{2x+4}\\\log_4(2y^2-2x+4)-\log_4(x+1)=3\lg\sqrt[3]{10}\end{cases}\Rightarrow\\\\\\\Rightarrow\begin{cases}y^2=2x+4\\\log_4\left(\frac{2y^2-2x+4}{x+1}\right)=\frac13\cdot3\lg{10}\end{cases}\Rightarrow\begin{cases}y^2=2x+4\\\log_4\left(\frac{2y^2-2x+4}{x+1}\right)=1\end{cases}\Rightarrow$

$\Rightarrow\begin{cases}y^2=2x+4\\\frac{2y^2-2x+4}{x+1}=4\end{cases}\Rightarrow\begin{cases}y^2=2x+4\\2y^2-2x+4=4x+4\end{cases}\Rightarrow\Rightarrow\begin{cases}y^2=2x+4\\2y^2=6x\end{cases}\Rightarrow\\\\\\\Rightarrow\begin{cases}y^2=2x+4\\y^2=3x\end{cases}\\\\\\2x+4=3x\\x=4\\\\\begin{cases}x=4\\y^2=12\end{cases}\Rightarrow\begin{cases}x=4\\y=\pm2\sqrt3\end{cases}$