\(\begin{array}{l} a,\ x^2-10x+25\\ =x^2-2.x.5+5^2\\ =(x-5)^2\\ b,\ x^3-2x+x-xy^2\\ =x^3-(2x-x)-xy^2\\ =x^3-x-xy^2\\ =x(x^2-1-y^2)\\ =x[(x^2-y^2)-1]\\ =x[(x-y)(x+y)-1]\end{array}\)

\(\begin{array}{l} 1,\ xFe_2O_3+(3x-2y)CO\xrightarrow{t^o} 2Fe_xO_y+(3x-2y)CO_2\uparrow\\ 2,\ C_xH_y+\bigg(x+\dfrac{y}{4}\bigg)O_2\xrightarrow{t^o} xCO_2\uparrow+\dfrac{y}{2}H_2O\end{array}\)

\(\begin{array}{l} a,\\ PTHH:\\ AlCl_3+3KOH\to Al(OH)_3\downarrow+3KCl\ (1)\\ 2Al(OH)_3\xrightarrow{t^o} Al_2O_3+3H_2O\ (2)\\ b,\\ n_{KOH}=\dfrac{3,36}{56}=0,06\ (mol)\\ Theo\ pt\ (1):\ n_{AlCl_3}=\dfrac{1}{3}n_{KOH}=0,02\ (mol)\\ \Rightarrow m_{AlCl_3}=0,02\times 133,5=2,67\ (g)\\ c,\\ Theo\ pt\ (1):\ n_{Al(OH)_3}=\dfrac{1}{3}n_{KOH}=0,02\ (mol)\\ Theo\ pt\ (2):\ n_{Al_2O_3}=\dfrac{1}{2}n_{Al(OH)_3}=0,01\ (mol)\\ \Rightarrow m_{Al_2O_3}=0,01\times 102=1,02\ (g)\end{array}\)

\(\begin{array}{l} 4Al+3O_2\xrightarrow{t^o} 2Al_2O_3\\ Al_2O_3+3H_2SO_4\to Al_2(SO_4)_3+3H_2O\\ Al_2(SO_4)_3+3BaCl_2\to 3BaSO_4\downarrow+2AlCl_3\\ AlCl_3+3NaOH\to Al(OH)_3\downarrow+3NaCl\\ Al(OH)_3+3HNO_3\to Al(NO_3)_3+3H_2O\\ 2Al(NO_3)_3+3Mg\to 3Mg(NO_3)_2+2Al\\ 2Al+3H_2SO_4\to Al_2(SO_4)_3+3H_2\uparrow\\ Al_2(SO_4)_3+6NaOH\to 2Al(OH)_3\downarrow+3Na_2SO_4\\ 2Al(OH)_3\xrightarrow{t^o} Al_2O_3+3H_2O\\ 2Al_2O_3\xrightarrow{\text{đpnc, Criolit}} 4Al+3O_2\uparrow\\ 2Al+2NaOH+2H_2O\to 2NaAlO_2+3H_2\uparrow\end{array}\)

\(\begin{array}{l} M_{XO}=20M_{H_2}=20\times 2=40\ (g/mol)\\ \Rightarrow M_{X}+16=40\\ \Rightarrow M_{X}=24\ (g/mol)\\ \Rightarrow \text{X là Magie (Mg)}\\ \Rightarrow \text{Chọn A: Mg}\end{array}\)

\(\begin{array}{l} 1)\\ m_{Na_2SO_4}=3\times 142=426\ (g)\\ 2)\\ n_{CO_2}=\dfrac{3,36}{22,4}=0,15\ (mol)\\ \Rightarrow m_{CO_2}=0,15\times 44=6,6\ (g)\end{array}\)

\(\begin{array}{l} a,\\ \text{Phương trình chữ:}\\ \text{Photpho + Oxi $\to$ Điphotpho pentaoxit}\\ b,\\ PTHH:\\ 4P+5O_2\xrightarrow{t^o} 2P_2O_5\\ n_{P}=\dfrac{3,1}{31}=0,1\ (mol)\\ Theo\ pt:\ n_{O_2}=\dfrac{5}{4}n_{P}=0,125\ (mol)\\ \Rightarrow m_{O_2}=0,125\times 32=4\ (g)\end{array}\)

\(\begin{array}{l} \text{a, Phương trình chữ: Canxi cacbonat $\to$ Canxi oxit + Cacbon đioxit}\\ b,\\ \text{Áp dụng ĐLBT khối lượng ta có:}\\ m_{CaCO_3}=m_{CaO}+m_{CO_2}\\ \Rightarrow m_{CaCO_3}=112+88=200\ (g)\end{array}\)