\((x+1)=(x+1)^{2}\\ \Leftrightarrow x+1=x^{2}+2x+1\\ \Leftrightarrow x+1-x^{2}-2x-1=0\\ \Leftrightarrow -x^{2}-x=0\\ \Leftrightarrow x(-x-1)=0\\ \Leftrightarrow \begin{cases} x =0\\ -x-1=0 \end{cases} \to\begin{cases} x =0\\ x=-1 \end{cases} \)

\(Mg+H_2SO_4 \to MgSO_4+H_2\\ n_{Mg}=\frac{9,6}{24}=0,4mol\\ n_{Mg}=n_{H_2}=n_{H_2SO_4}=n_{MgSO_4}=0,4mol\\ a/\\ V_{H_2}=0,4.22,4=8,96(l)\\ b/\\ m_{H_SO_4(dd)}=\frac{0,4.98}{9,8\%}=400g\\ C\%=\frac{0,4.120}{408,8}.100=11,74\%\)

\(x^{3}-3=0\\ \Leftrightarrow x^{2}=3\\ \Leftrightarrow \left[\begin{array}{} x_1=\sqrt{3}\\ x_2=-\sqrt{3} \end{array} \right.\)

\(2/ \\ \text{Tổng hạt}: 2p+n=46 (1)\\ \text{Hạt mang điẹn nhiều hơn không mạng điện}: 2p-n=14(2)\\ (1)(2)\\ p=e=15 n=16\)

\(4/ \\ n_{H_2}=0,48mol\\ \to n_{O(X)}=2,n_{H_2}=0,48.2=0,96mol\\ n_{O_2}=1,69mol\\ n_{H_2O}=1,7mol\\ BT O:\\ \to n_{CO_2}=1,32mol\\ BTKL:\\ \to m_X=34,6g\\ CH_3OH; C_2H_4(OH)_2; C_3H_5(OH)_3\\ \to n_{C_3H_5(OH)_2}=0,18mol\\ \to \%_{C_3H_5(OH)_3}=\frac{0,18.58}{34,6}=30,17\% \)

\(n_{Al_2O_3}=\frac{10,2}{102}=0,1mol\\ n_{HCl}=\frac{219.5\%}{36,5}=0,3mo\\\ Al_2O_3+6HCl \to 2 AlCl_3+3H_2O\\ Al_2O_3: 0,1 > HCl: \frac{0,3}{6}=0,05\\ \Rightarrow \text{Al2O3 du}\\ n_{Al_2O_3}=0,05mol\\ m_{Al_2O_3}=(0,1-0,05).1025,1g\ b/ C\%_{AlCl_3}=\frac{0,1.133,5}{0,05.102+219}.100=5,96\%\)

\(a/ 4Zn+2HCl \to ZnCl_2+H_2 \\ n_{HCl}=\frac{250.7,3\%}{36,5}=0,5(mol)\\ b/ \\ n_{Zn}=n_{H_2}=n_{ZnCl_2}=\frac{1}{2}.n_{HCl}=\frac{1}{2}.0,5=0,25(mol)\\ m_{Zn}=0,25.65=16,25(g)\\ V_{H_2}=0,25.22,4=5,6(l)\\ c/ \\ C\%_{ZnCl_2}=\frac{0,25.136}{16,25+250-0,25.2}.100=12,8\% \)

\(a/ \\ Fe+2HCl \rightarrow FeCl_2+H_2\\ n_{Fe}=\frac{5,6}{56}=0,1(mol)\\ n_{HCl}=0,1.1=0,1(mol)\\ Fe: 0,1>HCl: \frac{0,1}{2}=0,05\\ \Rightarrow \text{Fe dư, HCl hết} n_{H_2}=\frac{1}{2}.n_{HCl}=0,05(mol)\\ V_{H_2}=1,12l\\ b/ \\ \text{Fe dư}\\ n_{Fe}=\frac{1}{2}.n_{HCl}=0,05(mol)\\ m_{Fe}=(0,1-0,05).56=2,8g\\ C/ \\ n_{FeCl_2}=\frac{1}{2}.n_{HCl}=0,05(mol)\\ CM_{FeCl_2}=0,5M \)

\(CT: RHN_2\\ BTKL: \\ \Rightarrow m_{HNO_3}=310-121=189g\\ n_{HNO_3}=3(mol)\\ n_{HNO_3}=n_{amin}=3(mol)\\ M_{amin}=40>31\\ \Rightarrow CH_3NH_2\\ n=1,5mol\\ m=46,5g\)

\(Fe: \text{Có hoá trị III}\)