\(\begin{array}{l}
Ba_2O\to BaO\\
Ba_2(OH)_3\to Ba(OH)_2\\
Mg_2O\to MgO\\
K(OH)_2\to KOH\\
AgO\to Ag_2O
\end{array}\)

\(\left\{{}\begin{matrix}x_0+y_0=3\\mx_0-y_0=3\\x_0=2y_0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}3y_0=3\\x_0=2y_0\\mx_0-y_0=3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y_0=1\\x_0=2\\2m-1=3\end{matrix}\right.\Leftrightarrow m=2\\ \to B\)

\(a,A=\sqrt{x}\ge0\\ "="\Leftrightarrow x=0\\ b,B=x+\sqrt{x}+1\ge0+0+1=1\\ "="\Leftrightarrow x=0\\ c,C=\sqrt{x-3}+2\ge2\\ "="\Leftrightarrow x-3=0\Leftrightarrow x=3\\ d,D=\left(x-2\sqrt{x}+1\right)+2=\left(\sqrt{x}-1\right)^2+2\ge2\\ "="\Leftrightarrow\sqrt{x}=1\Leftrightarrow x=1\)

\(x^2-6x+2022\\ =\left(x^2-6x+9\right)+2013\\ =\left(x-3\right)^2+2013\ge2013>0,\forall x\)

\(A=\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{9}+...+\dfrac{1}{2021}-\dfrac{1}{2023}+\dfrac{1}{2023}-\dfrac{1}{2025}\\ A=\dfrac{1}{3}-\dfrac{1}{2025}=\dfrac{674}{2025}\)

\(M=\dfrac{\sqrt{x}-3}{\sqrt{x}+3}\left(x\ge0\right)\\ M\left(\sqrt{x}+3\right)\ge x+5\sqrt{x-1}-\sqrt{x}-2\left(x\ge1\right)\\ \leftrightarrow\sqrt{x}-3\ge x+5\sqrt{x-1}-\sqrt{x}-2\\ \leftrightarrow x+5\sqrt{x-1}-2\sqrt{x}+1\le0\\ \leftrightarrow\left(\sqrt{x}-1\right)^2+5\sqrt{x-1}\le0\)

Mặt khác \(\left(\sqrt{x}-1\right)^2\ge0;5\sqrt{x-1}\ge0\)

Dấu \("="\leftrightarrow\left\{{}\begin{matrix}\sqrt{x}-1=0\\\sqrt{x-1}=0\end{matrix}\right.\leftrightarrow x=1\left(\text{nhận}\right)\)

Vậy \(x=1\)

\(8\left(x-2009\right)^2=25-y^2\ge0\\ \to y^2\le25\\ \to-5\le y\le5\\ \to y\in\left\{0;1;2;3;4;5\right\}\left(y\in N\right)\)

Với \(y=0\to\left(x-2009\right)^2=\dfrac{25}{8}\left(\text{loại}\right)\)

Với \(y=1\to\left(x-2009\right)^2=\dfrac{25-1}{8}=3\left(\text{loại}\right)\)

Với \(y=2\to\left(x-2009\right)^2=\dfrac{25-4}{8}=\dfrac{21}{8}\left(\text{loại}\right)\)

Với \(y=3\to\left(x-2009\right)^2=\dfrac{25-9}{8}=2\left(\text{loại}\right)\)

Với \(y=4\to\left(x-2009\right)^2=\dfrac{25-16}{8}=\dfrac{9}{8}\left(\text{loại}\right)\)

Với \(y=5\to\left(x-2009\right)^2=0\to x=2009\left(\text{nhận}\right)\)

Vậy \(\left(x;y\right)=\left(2009;5\right)\)