Решите систему уравнений: Ребят, фото внизу будет.

Пошаговое объяснение:

$\left\{\begin{array}{ccc}(\frac{1}{4})^{\sqrt{y}-2\sqrt{x} } }=\frac{1}{2^{-4}} \\\\lg\sqrt{x*y} =1\end{array}\right=\left\{\begin{array}{ccc}(\frac{1}{4})^{\sqrt{y}-2\sqrt{x} } }=(\frac{1}{4})^{-2}} \\\\\sqrt{x*y} =10^1\end{array}\right=\left\{\begin{array}{ccc}(\frac{1}{4})^{\sqrt{y}-2\sqrt{x} } }=(\frac{1}{4})^{-2}} \\\\\sqrt{x}*\sqrt{y} =10\end{array}\right=\\=\left\{\begin{array}{ccc}\sqrt{y}-2*\sqrt{x}=-2 \\\\\sqrt{y} =\frac{10}{\sqrt{x} } \end{array}\right .$

$\frac{10}{\sqrt{x} } -2*\sqrt{x} =-2|*(-1)\\2*\sqrt{x} -\frac{10}{\sqrt{x} } =2\\2*\sqrt{x} -\frac{10}{\sqrt{x} } -2=0|:2\\\sqrt{x} -\frac{5}{\sqrt{x} } -1=0\\x-\sqrt{x} -5=0\\\sqrt{x}=t \geq 0\Rightarrow\\ t^2-t-5=0\\D=21;\sqrt{D}=\sqrt{21}\\ t_1=\frac{1-\sqrt{21} }{2} \notin;t_2=\frac{1+\sqrt{21} }{2} \in.\\$

$t=\sqrt{x} =\frac{\sqrt{21}+1 }{2} \\x=t^2=(\frac{\sqrt{21} +1}{2})^2 \\\sqrt{y}=\frac{10}{\sqrt{x} } =\frac{2*10}{\sqrt{21}+1 }=\frac{20*(\sqrt{21}-1) }{(\sqrt{21}+1)*(\sqrt{21}-1)} =\frac{20*(\sqrt{21}-1) }{21-1}=\frac{20*(\sqrt{21}-1) }{20} =\sqrt{21}-1 . \\y=(\sqrt{21}-1 )^2.$

Ответ: $x=(\frac{\sqrt{21}+1 }{2} )^2;y=(\sqrt{21}-1)^2.$