Mình làm mẫu 2 ý đầu thôi nhé 

a)

\(CO:\left\{{}\begin{matrix}\%m_C=\dfrac{12}{28}.100\%=42,86\%\\\%m_O=100\%-42,86\%=57,14\%\end{matrix}\right.\\ FeS_2:\left\{{}\begin{matrix}\%m_{Fe}=\dfrac{56}{120}.100\%=46,67\%\\\%m_S=100\%-46,67\%=53,33\%\end{matrix}\right.\)

\(MgCl_2:\left\{{}\begin{matrix}\%m_{Mg}=\dfrac{24}{95}.100\%=25,26\%\\\%m_{Cl}=100\%-25,56\%=74,74\%\end{matrix}\right.\\ Cu_2O:\left\{{}\begin{matrix}\%m_{Cu}=\dfrac{128}{144}.100\%=88,89\%\\\%m_O=100\%-88,89\%=11,11\%\end{matrix}\right.\)

\(CO_2:\left\{{}\begin{matrix}\%m_C=\dfrac{12}{44}.100\%=27,27\%\\\%m_O=100\%-27,27\%=72,73\%\end{matrix}\right.\\ C_2H_4:\left\{{}\begin{matrix}\%m_C=\dfrac{24}{28}.100\%=85,71\%\\\%m_O=100\%-85,71\%=14,29\%\end{matrix}\right.\\ C_6H_6:\left\{{}\begin{matrix}\%m_C=\dfrac{72}{78}.100\%=92,31\%\\\%m_H=100\%-92,31\%=7,69\%\end{matrix}\right.\)

b)

\(FeO:\left\{{}\begin{matrix}\%m_{Fe}=\dfrac{56}{72}.100\%=77,78\%\\\%m_O=100\%-77,78\%=22,22\%\end{matrix}\right.\\ Fe_3O_4\left\{{}\begin{matrix}\%m_{Fe}=\dfrac{168}{232}.100\%=72,41\%\\\%m_O=100\%-72,41\%=27,59\%\end{matrix}\right.\\ Fe_2O_3\left\{{}\begin{matrix}\%m_{Fe}=\dfrac{112}{160}.100\%=70\%\\\%m_O=100\%-70\%=30\%\end{matrix}\right.\)

\(Fe\left(OH\right)_2:\left\{{}\begin{matrix}\%m_{Fe}=\dfrac{56}{90}.100\%=62,22\%\\\%m_O=\dfrac{32}{90}.100\%=35,56\%\\\%m_H=\left(100-62,22-35,56\right)\%=2,22\%\end{matrix}\right.\)

\(Fe\left(OH\right)_3:\left\{{}\begin{matrix}\%m_{Fe}=\dfrac{56}{107}.100\%=52,34\%\\\%m_O=\dfrac{48}{107}.100\%=44,86\%\\\%m_H=\left(100-52,34-44,86\right)\%=2,8\%\end{matrix}\right.\)

`(x - y)^2 = x^2 - 2xy +y^2`

`(x + y)^2 = x^2 + 2xy + y^2`

\(1-\dfrac{45}{46}=\dfrac{1}{46};1-\dfrac{99}{100}=\dfrac{1}{100}\)

Vì \(\dfrac{1}{46}< \dfrac{1}{100}\)

Nên \(\dfrac{45}{46}< \dfrac{99}{100}\)

x + 12 = -24 + 21

x + 12 = -3

x + 12 = -3 - 12

x = -15