Nguyên tử khối của nguyên tử X là:

\(2,6568.10^{-23}:\left(1,9926.10^{-23}:12\right)\)

= 16

Vậy X là nguyên tử của nguyên tố Oxi 

Kí hiệu hóa học : O

\(\dfrac{x+1}{x-3}+\dfrac{x-1}{x+3}=\dfrac{x+6}{x^2-9}\)(a)

ĐKXĐ\(\left\{{}\begin{matrix}x-3\ne0\\x+3\ne0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\ne3\\x\ne-3\end{matrix}\right.\)

(a)\(\Leftrightarrow\dfrac{x+1}{x-3}+\dfrac{x-1}{x+3}=\dfrac{x+6}{\left(x-3\right)\left(x+3\right)}\)

\(\Leftrightarrow\left(x+1\right).\left(x+3\right)+\left(x-1\right).\left(x-3\right)=x+6\)

\(\Leftrightarrow x^2+3x+x+3+x^2-3x-x+3=x+6\)

\(\Leftrightarrow x^2+3x+x+x^2-3x-x-x=6-3-3\)

\(\Leftrightarrow2x^2-x=0\)

\(\Leftrightarrow x\left(2x-1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\2x-1=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\2x=1\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\left(thỏa-mãn-ĐKXĐ\right)\\x=\dfrac{1}{2}\left(thỏa-mãn-ĐKXĐ\right)\end{matrix}\right.\)

Vậy S = \(\left\{0;\dfrac{1}{2}\right\}\)

Sơ đồ cấu tạo nguyên tử cacbon

\(2x+4=4x-2\)

\(2x-4x=-2-4\)

\(-2x=-6\)

    \(x=\left(-6\right):\left(-2\right)\)

    \(x=3\)