\(=\left(a-2b\right)\left(a+2b\right)-5\left(a+2b\right)=\left(a+2b\right)\left(a-2b-5\right)\)

\(a,A=\dfrac{x+x+2}{\left(x-2\right)\left(x+2\right)}\cdot\dfrac{\left(x-2\right)\left(x+2\right)}{x+3}=\dfrac{2x+2}{x+3}\\ b,A=\dfrac{2\left(x+3\right)-4}{x+3}=2-\dfrac{4}{x+3}\in Z\\ \Leftrightarrow x+3\inƯ\left(4\right)=\left\{-4;-2;-1;1;2;4\right\}\\ \Leftrightarrow x\in\left\{-7;-5;-4;-1;1\right\}\left(x\ne-2\right)\)

Sửa: \(x^2+y^2=369\)

\(x:y=4:5\Rightarrow\dfrac{x}{4}=\dfrac{y}{5}\Rightarrow\dfrac{x^2}{16}=\dfrac{y^2}{25}=\dfrac{x^2+y^2}{16+25}=\dfrac{369}{41}=9\\ \Rightarrow\left\{{}\begin{matrix}x^2=144\\y^2=225\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=\pm12\\y=\pm15\end{matrix}\right.\)

Vậy \(\left(x;y\right)=\left(12;15\right);\left(-12;-15\right)\)

\(a,A=\left\{4;5;6;7;8;9\right\}\\ A=\left\{x\in N|3< x\le9\right\}\\ b,B=\left\{1;2;3;4;5;6;7;8;9;10;11\right\}\\ B=\left\{x\in N\text{*}|x\le11\right\}\\ c,C=\left\{15;16;17;...;48;49\right\}\\ C=\left\{x\in N|15\le x< 50\right\}\)

\(B=\dfrac{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)-\left(\sqrt{x}-2\right)\left(\sqrt{x}+1\right)}{\left(\sqrt{x}+1\right)^2\left(\sqrt{x}-1\right)}\cdot\dfrac{\sqrt{x}+1}{\sqrt{x}}\\ B=\dfrac{x+\sqrt{x}-2-x+\sqrt{x}+2}{\left(\sqrt{x}+1\right)^2\left(\sqrt{x}-1\right)}\cdot\dfrac{\sqrt{x}+1}{\sqrt{x}}\\ B=\dfrac{2\sqrt{x}}{\sqrt{x}\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}=\dfrac{2}{x-1}\)

\(=\dfrac{\left(3^3\right)^{15}\left(3^2\right)^{20}}{\left(3^4\right)^{12}\cdot3^{36}}=\dfrac{3^{45}\cdot3^{40}}{3^{48}\cdot3^{36}}=3\)

\(ĐK:\left\{{}\begin{matrix}2x-4\ne0\\2x+4\ne0\\4-x^2\ne0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\ne2\\x\ne-2\end{matrix}\right.\)

\(\Rightarrow\left|x+\dfrac{3}{4}\right|=\dfrac{1}{6}\Rightarrow\left[{}\begin{matrix}x+\dfrac{3}{4}=\dfrac{1}{6}\\x+\dfrac{3}{4}=-\dfrac{1}{6}\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=-\dfrac{7}{12}\\x=-\dfrac{11}{12}\end{matrix}\right.\)

Ta có \(a^2+\dfrac{1}{b+c}=a^2+\dfrac{1}{6-a}\)

Mà \(a+b+c=6\Rightarrow0\le a,b,c\le2\)

\(\Rightarrow a^2+\dfrac{1}{6-a}\ge2^2+\dfrac{1}{6-2}=\dfrac{17}{4}\)

\(\Rightarrow P=\sum\sqrt{a^2+\dfrac{1}{b+c}}=\sum\sqrt{a^2+\dfrac{1}{6-a}}\ge\sqrt{\dfrac{17}{4}}+\sqrt{\dfrac{17}{4}}+\sqrt{\dfrac{17}{4}}=\dfrac{3\sqrt{17}}{2}\)

Dấu \("="\Leftrightarrow a=b=c=2\)