dạ e cảm ơn ạ :3

C1 

\(\left\{{}\begin{matrix}4x-y=5\\2x+3y=-1\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}4x-y=5\\4x+6y=-2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}-7y=7\\4x-y=5\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=-1\\x=1\end{matrix}\right.\)

C2

\(\left\{{}\begin{matrix}4x-y=5\\2x+3y=-1\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}y=4x-5\\2x+3.\left(4x-5\right)=-1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=4x-5\\14x-15=-1\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}y=4.1-5=-1\\x=1\end{matrix}\right.\)

Ta có 

\(p+n+e=126\Rightarrow2p+n=126\left(p=e\right)\)

\(\left\{{}\begin{matrix}2p+n=126\\-p+n=12\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}p=38\\n=50\end{matrix}\right.\)

=> Ô 38 , nguyên tử khối 88

=> Sr

\(7x=3+39\Leftrightarrow7x=42\Leftrightarrow x=6\)

Vậy x = 6

\(5.a+a-5=43\Leftrightarrow6a=43+5\Leftrightarrow6a=48\Leftrightarrow a=6\)

Vậy a = 6.

\(CaCO3\rightarrow CaO+CO2 \)

2mol       

\(nCaO=90\%.2=1,8\left(mol\right)\Rightarrow mCaO=1,8.56=100,8\left(g\right)\)

\(nCO2=1,8\left(mol\right)\Rightarrow mCO2=1,8.44=79,2\left(g\right)\)

Số NST môi trường cung cấp 

\(a.2n.\left(2^k-1\right)=5.50.\left(2^4-1\right)=3750\left(NST\right)\)