\(\begin{array} \\ \bullet\;\;+KOH:\\ ZnO+2KOH\to K_2ZnO_2+H_2O\\ KOH+HCl\to KCl+H_2O\\ 2KOH+SiO_2\to K_2SiO_3+H_2O \\ Al_2O_3+2KOH\to 2KAlO_2+H_2O\\ 6KOH+P_2O_5\to 2K_3PO_4+3H_2O\\ MgCO_3+KOH\to Mg(OH)_2\downarrow+K_2CO_3\\ 2KOH+H_2SO_4\to K_2SO_4+2H_2O\\ SO_2+2KOH\to K_2SO_3+H_2O\\ Fe_2(SO_4)_3+6KOH\to 2Fe(OH)_3\downarrow+3K_2SO_4\\ \bullet\;\;+Ca(OH)_2:\\ Ca(OH)_2+ZnO\to CaZnO_2+H_2O\\ Ca(OH)_2+2HCl\to CaCl_2+2H_2O\\ Ca(OH)_2+Al_2O_3\to Ca(AlO_2)_2+H_2O\\ Ca(OH)_2+Na_2SO_3\to CaSO_3\downarrow+2NaOH\\ Ca(OH)_2+K_2SO_4\to CaSO_4\downarrow+2KOH\\ 3Ca(OH)_2+P_2O_5\to Ca_3(PO_4)_2+3H_2O\\ Ca(OH)_2+MgCO_3\to CaCO_3\downarrow+Mg(OH)_2\downarrow\\ Ca(OH)_2+K_2CO_3\to CaCO_3\downarrow+2KOH\\ Ca(OH)_2+H_2SO_4\to CaSO_4\downarrow+2H_2O\\ Ca(OH)_2+SO_2\to CaSO_3\downarrow+H_2O\\ 3Ca(OH)_2+Fe_2(SO_4)_3\to 3CaSO_4\downarrow+2Fe(OH)_3\downarrow\end{array}\)

\(\begin{array} {l} a)\\ Fe+2HCl\to FeCl_2+H_2\\ n_{Fe}=\dfrac{11,2}{56}=0,2(mol)\\ n_{FeCl_2}=n_{Fe}=0,2(mol)\\ m_{FeCl_2}=0,2.127=25,4(g)\\ b)\\ n_{H_2}=n_{Fe}=0,2(mol)\\ V_{H_2}=0,2.22,4=4,48(l)\\ c)\\ n_{O_2}=\dfrac{4,48}{22,4}=0,2(mol)\\ 2H_2+O_2\xrightarrow{t^o}2H_2O\\ \dfrac{n_{H_2}}{2}<n_{O_2}\to O_2\text{ dư}\\ n_{H_2O}=n_{H_2}=0,2(mol)\\ m_{H_2O}=0,2.18=3,6(g) \end{array}\)

\(\begin{array} {l} n_{Al}=\dfrac{5,4}{27}=0,2(mol)\\ n_{H_2SO_4}=\dfrac{39,2}{98}=0,4(mol)\\ 2Al+3H_2SO_4\to Al_2(SO_4)_3+3H_2\\ \dfrac{n_{Al}}{2}<\dfrac{n_{H_2SO_4}}{3}\to H_2SO_4\text{ dư}\\ n_{H_2}=\dfrac{3}{2}n_{Al}=0,3(mol)\\ V_{H_2(đktc)}=0,3.22,4=6,72(l) \end{array}\)

\(\begin{array} {l} Fe+2HCl\to FeCl_2+H_2\\ n_{Fe}=\dfrac{14}{56}=0,25(mol)\\ n_{HCl}=2n_{Fe}=0,5(mol)\\ 200ml=0,2l\\ C_{M\,HCl}=\dfrac{0,5}{0,2}=2,5M \end{array}\)

Theo Cauchy:

\(3\sqrt{2a-1}=3\sqrt{1\left(2a-1\right)}\le\dfrac{3\left(1+2a-1\right)}{2}=3a\)

\(a\sqrt{5-4a^2}\le\dfrac{a^2+5-4a^2}{2}=\dfrac{5-3a^2}{2}\)

\(A\le3a+\dfrac{5-3a^2}{2}=\dfrac{5-3a^2+6a}{2}=\dfrac{-3\left(a-1\right)^2}{2}+4\le4\)

Vậy \(A_{max}=4\Leftrightarrow x=1\)

$S+Fe\xrightarrow{t^o}FeS$

PT có 2 nghiệm khi:

\(\Delta=\left(m-1\right)^2-4\left(m-1\right)=\left(m-1\right)\left(m-5\right)\ge0\\ \Rightarrow\left[{}\begin{matrix}m< 1\\m>5\end{matrix}\right.\)

Theo Vi-ét: $\begin{cases} x_1+x_2=m-1\\ x_1x_2=m-1 \end{cases}$

Ta có $x_1+2x_2+x_1x_2=m$

\(\Leftrightarrow\left(x_1+ x_2\right)+x_1x_2+x_2=m\\ \Leftrightarrow m-1+x_2+m-1=m\\ \Leftrightarrow x_2=-m+2\)

Mà \(x_1+x_2=m-1\Leftrightarrow x_1=m-1+m-2=2m-3\)

Thay vào $x_1x_2=m-1$

\(\Leftrightarrow\left(2m-3\right)\left(-m+2\right)=m-1\\ \Leftrightarrow2m^2-6m+5=0\left(\text{vô nghiệm}\right)\)

Vậy không có giá trị của \(m\) thỏa mãn

PT có 2 nghiệm phân biệt \(\Leftrightarrow\Delta'=\left(m+1\right)^2+32>0\left(\text{đúng }\forall m\right)\)

Theo Vi-ét: \(\begin{cases} x_1+x_2=-2(m+1)=-2m-2\\ x_1x_2=-8 \end{cases}\)

Vì $x_1$ là nghiệm của PT nên  \(x_1^2=-2(m+1)x_1+8\)

Ta có \(x_1^2=x_2\)

\(\Leftrightarrow-2\left(m+1\right)x_1+8=x_2\\ \Leftrightarrow x_2+2mx_1+2x_1-8=0\\ \Leftrightarrow\left(x_1+x_2\right)+2mx_1+x_1-8=0\\ \Leftrightarrow x_1\left(2m+1\right)-2m-10=0\\ \Leftrightarrow x_1=\dfrac{2m+10}{2m+1}\)

Mà \(x_1+x_2=-2m-2\Leftrightarrow x_2=-2m-2-\dfrac{2m+10}{2m+1}=\dfrac{-4m^2-8m-12}{2m+1}\)

Ta có \(x_1x_2=-8\)

\(\Leftrightarrow\dfrac{2m+10}{2m+1}\cdot\dfrac{-4m^2-8m-12}{2m+1}=-8\\ \Leftrightarrow\left(2m+10\right)\left(m^2+2m+3\right)=2\left(2m+1\right)^2\\ \Leftrightarrow m^3+3m^2+9m+14=0\\ \Leftrightarrow m^3+2m^2+m^2+2m+7m+14=0\\ \Leftrightarrow\left(m+2\right)\left(m^2+m+7\right)=0\\ \Rightarrow m=-2\)

Vậy $m=-2$

PT có 2 nghiệm phân biệt \(\Leftrightarrow\Delta'=\left(m-1\right)^2+8>0\left(\text{đúng }\forall m\right)\)

Theo Vi-ét: \(\begin{cases} x_1+x_2=2(m-1)=2m-2\\ x_1x_2=-2 \end{cases}\)

Vì \(x_1,x_2\) là nghiệm của PT nên \(\left\{{}\begin{matrix}x_1^2=2\left(m-1\right)x_1+2\\x_2^2=2\left(m-1\right)x_2+2\end{matrix}\right.\)

\(A=x_1^2+4x_2^2=\left(x_1+2x_2\right)^2-4x_1x_2\\ A=\left(x_1+2x_2\right)^2+8\ge8\)

\(\text{Dấu }"="\Leftrightarrow\left\{{}\begin{matrix}x_1=-2x_2\\x_1+x_2=2m-2\\x_1x_2=-2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x_1=4m-4\\x_2=2-2m\\x_1x_2=-2\end{matrix}\right.\\ \Leftrightarrow\left(4m-4\right)\left(2-2m\right)=-2\\ \Leftrightarrow8\left(m-1\right)^2=2\\ \Leftrightarrow\left(m-1\right)^2=\dfrac{1}{4}\\ \Rightarrow\left[{}\begin{matrix}m=\dfrac{5}{4}\\m=\dfrac{3}{4}\end{matrix}\right.\)

Vậy \(m\in\left\{\dfrac{3}{4};\dfrac{5}{4}\right\}\)

\(\begin{array} {l} 13)\\ n_{HCHO}=\dfrac{1,2}{30}=0,04(mol)\\ HCHO\xrightarrow{+AgNO_3/NH_3,t^o}4Ag\\ n_{Ag}=4n_{HCHO}=0,16(mol)\\ m=0,16.108=17,28(g)\\ \to A\\ 14)\\ X:C_nH_{2n}\\ n_{Br_2}=\dfrac{8}{160}=0,05(mol)\\ C_nH_{2n}+Br_2\to C_nH_{2n}Br_2\\ n_{C_nH_{2n}}=n_{Br_2}=0,05(mol)\\ M_{C_nH_{2n}}=14n=\dfrac{1,4}{0,05}=28(g/mol)\\ n=2\\ X:C_2H_4\\ \to A \end{array}\)