Question

Решить систему уравнений. (0,2)^x - 2^0,5y = 3 (0,04)^x - 2^y = 21

Answer 1

$\displaystyle \left \{ {{0.2^x-2^{0.5y}=3} \atop {0.04^x-2^y=21}} \right. \rightarrow \left \{ {{0.2^x-\sqrt{2^{y}}=3} \atop {0.2^{2x}-2^y=21}} \right. \rightarrow \left[\begin{array}{ccc}2^y=t,\,\,\,t\ \textgreater \ 0\\0.2^x=k\,\,\,k\ \textgreater \ 0\end{array}\right]\rightarrow\\\\\\\rightarrow \left \{ {{k-\sqrt{t}=3} \atop {k^2-t=21}} \right. \rightarrow\left \{ {{k-\sqrt{t}=3} \atop {t=k^2-21}} \right. \rightarrow k-\sqrt{k^2-21}=3\\\\\\k-3=\sqrt{k^2-21}\\\\(k-3)^2=\sqrt{k^2-21}\\\\k^2-6k+9=k^2-21\\\\-6k=-30\\\\k=\frac{30}6\\\\k=5$

$\displaystyle k^2-t=21\\\\25-t=21\\\\t=25-21=5\\\\\\2^y=t\\\\2^y=4\\\\y=2\\\\\\0.2^x=5\\\\\bigg(\frac{1}5\bigg)^x=5\\\\5^{-x}=5\\\\-x=1\\\\x=1$

ответ: (1;2)