1)\(A=\dfrac{1}{3+2\sqrt{2}}+\dfrac{1}{3-2\sqrt{2}}=\dfrac{3-2\sqrt{2}+3+2\sqrt{2}}{\left(3+2\sqrt{2}\right)\left(3-2\sqrt{2}\right)}\)
\(=\dfrac{6}{1}=6\)
\(\left\{{}\begin{matrix}3x-y=7\\5x+y=9\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}y=3x-7\\5x+3x-7=9\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}y=3x-7\\8x=9+7=16\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}y=3.2-7=-1\\x=\dfrac{16}{8}=2\end{matrix}\right.\)
`x^2-3x-10=0`
`<=> x^2 +2x - 5x -10 =0`
`<=> x(x+2)-5(x+2)=0`
`<=>(x-5)(x+2)=0`
\(\Leftrightarrow\left[{}\begin{matrix}x-5=0\\x+2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=5\\x=-2\end{matrix}\right.\)
Vậy `S={5;-2}`
