Question

Алгебра. Решение показательных уравнений и неравенств

Answer 1

$1)\ \ \left\{\begin{array}{l}4^{x+y}=16\\5^{x+2y-1}=1\end{array}\right\ \ \left\{\begin{array}{l}4^{x+y}=4^2\\5^{x+2y-1}=5^0\end{array}\right\ \ \left\{\begin{array}{l}x+y=2\\x+2y-1=0\end{array}\right\ \ominus \\\\\\\left\{\begin{array}{l}x=2-y\\-y=1\end{array}\right\ \ \left\{\begin{array}{l}x=3\\y=-1\end{array}\right\ \ \ \to \ \ \ (3;-1)$

$2)\ \ \left\{\begin{array}{ccc}3^{2y-x}=\dfrac{1}{81}\\3^{x-y+2}=27\end{array}\right\ \ \left\{\begin{array}{ccc}2y-x=-4\\x-y+2=3\end{array}\right\ \ \left\{\begin{array}{l}-x+2y=-4\\x-y=1\end{array}\right\ \oplus \\\\\\\left\{\begin{array}{l}y=-3\\x=1+y\end{array}\right\ \ \left\{\begin{array}{l}y=-3\\x=-2\end{array}\right\ \ \ \to \ \ \ (-2\, ;\, -3)$

$3)\ \ \left\{\begin{array}{l}x+y=3\\2^{x}+2^{y}=6\end{array}\right\ \ \left\{\begin{array}{l}y=3-x\\2^{x}+2^{3-x}-6=0\end{array}\right\ \ \left\{\begin{array}{l}y=3-x\\\dfrac{2^{2x}-6\cdot 2^{x}+8}{2^{x}}=0\end{array}\right\\\\\\\left\{\begin{array}{l}y=3-x\\2^{x}=2\ ,\ 2^{x}=4\end{array}\right\ \ \left\{\begin{array}{l}y_1=2\ ,\ y_2=1\\x_1=1\ ,\ x_2=2\end{array}\right\ \ \to \ \ (1;2)\ ,\ (2,1)$

$4)\ \ \left\{\begin{array}{l}x-y=2\\3^{x}-3^{y}=24\end{array}\right\ \ \left\{\begin{array}{l}y=x-2\\3^{x}-3^{x-2}=24\end{array}\right\ \ \left\{\begin{array}{l}y=x-2\\3^{x}-3^{x}\cdot \dfrac{1}{9}=24\end{array}\right\\\\\\\left\{\begin{array}{ccc}y=x-2\\3^{x}\cdot \dfrac{8}{9}=24\end{array}\right\ \ \left\{\begin{array}{l}y=x-2\\3^{x}=3^3\end{array}\right\ \ \left\{\begin{array}{ccc}y=1\\x=3\end{array}\right\ \ \to \ \ (3;1)$