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

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

Thế (1) vào (2):

\(\Rightarrow\left\{{}\begin{matrix}x=\dfrac{-2\sqrt{2}+3y}{\sqrt{2}}\left(1\right)\\2\cdot\left(\dfrac{-2\sqrt{2}+3y}{\sqrt{2}}\right)+\sqrt{2}y=4\left(2\right)\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{-2\sqrt{2}+3y}{\sqrt{2}}\\\dfrac{-4\sqrt{2}}{\sqrt{2}}+\dfrac{6y}{\sqrt{2}}+\sqrt{2}y=4\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{-2\sqrt{2}+3y}{\sqrt{2}}\\-4+\dfrac{6y}{\sqrt{2}}+\sqrt{2}y=4\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{-2\sqrt{2}+3y}{\sqrt{2}}\\\dfrac{6y}{\sqrt{2}}+\dfrac{\sqrt{2}y\cdot\sqrt{2}}{\sqrt{2}}=4+4\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{-2\sqrt{2}+3y}{\sqrt{2}}\\\dfrac{6y+2y}{\sqrt{2}}=8\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{-2\sqrt{2}+3y}{\sqrt{2}}\\\dfrac{8y}{\sqrt{2}}=8\end{matrix}\right.\\\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{-2\sqrt{2}+3y}{\sqrt{2}}\\y=\dfrac{8}{\dfrac{8}{\sqrt{2}}}\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{-2\sqrt{2}+3y}{\sqrt{2}}\\y=\sqrt{2}\end{matrix}\right. \)

\(\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{-2\sqrt{2}+3\cdot\sqrt{2}}{\sqrt{2}}\\y=\sqrt{2}\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}x=1\\y=\sqrt{2}\end{matrix}\right.\)

c. \(\left\{{}\begin{matrix}-2x+y=-3\left(1\right)\\3x+4y=10\left(2\right)\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}y=-3+2x\left(1\right)\\3x+4y=10\left(2\right)\end{matrix}\right.\)

Thế (1) vào (2):

\(\Rightarrow\left\{{}\begin{matrix}y=-3+2x\left(1\right)\\3x+4\left(-3+2x\right)=10\left(2\right)\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}y=-3+2x\\3x+\left(-12\right)+8x=10\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}y=-3+2x\\3x+8x=10+12\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}y=-3+2x\\11x=22\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}y=-3+2x\\x=2\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}y=-3+2\cdot2\\x=2\end{matrix}\right.\\ \Rightarrow\left\{{}\begin{matrix}y=1\\x=2\end{matrix}\right.\)

b. \(\left\{{}\begin{matrix}2x+y=5\left(1\right)\\3x+2y=27\left(2\right)\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}y=5-2x\left(1\right)\\3x+2y=27\left(2\right)\end{matrix}\right.\)

Thế (1) vào (2):

\(\Rightarrow\left\{{}\begin{matrix}y=5-2x\left(1\right)\\3x+2\left(5-2x\right)=27\left(2\right)\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}y=5-2x\\3x+10-4x=27\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}y=5-2x\\3x-4x=27-10\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}y=5-2x\\-x=17\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}y=5-2x\\x=-17\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}y=5-2\cdot\left(-17\right)\\x=-17\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}y=39\\x=-17\end{matrix}\right.\)

a.\(\left\{{}\begin{matrix}x-2y=5\left(1\right)\\3x-5y=-18\left(2\right)\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x=5-2y\left(1\right)\\3x-5y=-18\left(2\right)\end{matrix}\right.\)

Thế (1) vào (2):

 \(\Rightarrow\left\{{}\begin{matrix}x=5-2y\left(1\right)\\3\cdot\left(5-2y\right)-5y=-18\left(2\right)\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}x=5-2y\left(1\right)\\15-6y-5y=-18\left(2\right)\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}x=5-2y\left(1\right)\\-11y=-18-15\left(2\right)\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}x=5-2y\left(1\right)\\-11y=-33\left(2\right)\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}x=5-2y\left(1\right)\\y=3\end{matrix}\right.\\\Leftrightarrow \left\{{}\begin{matrix}x=5-2\cdot3\\y=3\end{matrix}\right.\\ \Rightarrow\left\{{}\begin{matrix}x=-1\\y=3\end{matrix}\right.\)

\(4x-5=3x+2\\ \Leftrightarrow4x-3x=2+5\\ \Leftrightarrow x=7\)

y'=(\(\frac{\pi}{2}\)+2)'cos(\(\frac{\pi}{2}\)+2)=2cos(\(\frac{\pi}{2}\)+2)

\(y'=\left(\frac{\pi}{2}+2x\right)^'\cdot cos\left(\frac{\pi}{2}+2x\right)=2cos\left(\frac{\pi}{2}+2x\right)\)

\(n_X=\frac{6.72}{22.4}=0.3mol\)

\(C_3H_8+5O_2\rightarrow3CO_2+4H_2O\)

..x...........-->...........3x...-->.....4x

\(2C_3H_6+9O_2\rightarrow6CO_2+8H_2O\)

..y.........-->...............3y.....-->....4y

Ta có hệ:\(\left\{{}\begin{matrix}44x+42y=12.8\\x+y=0.3\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=0.1\\y=0.2\end{matrix}\right.\)

\(m_{CO_2}=3x+3y=3\cdot0.1+3\cdot0.2=0.9g\\ m_{H_2O}=4x+4y=4\cdot0.1+4\cdot0.2=1,2g\)

a).

\(Al+6HNO_3\rightarrow Al\left(NO_3\right)_3+3NO_2+3H_2O\\4 Al\left(NO_3\right)_3\underrightarrow{t^o}2Al_2O_3+12NO_2+3O_2\\ Al_2O_3+3H_2SO_4\rightarrow Al_2\left(SO_4\right)_3+3H_2O\)

b).

\(Cu\left(NO_3\right)_2+2NaOH\rightarrow Cu\left(OH\right)_2+2NaNO_3\\ Cu\left(OH\right)_2+2HNO_3\rightarrow Cu\left(NO_3\right)_2+2H_2O\)

c).

\(3Ca\left(OH\right)_2+2H_3PO_4\rightarrow Ca_3\left(PO_4\right)_2+H_2O\\ Ca_3\left(PO_4\right)_2+3H_2SO_4\rightarrow3CaSO_4+2H_3PO_4\)

d).

\(Al_4C_3\underrightarrow{t^o}4Al+3C\\ C+O_2\underrightarrow{t^o}CO_2\\CO_2+C⇌2CO\\ 4CO+Fe_3O_4\rightarrow3Fe+4CO_2\\ CaO+CO_2\underrightarrow{t^o}CaCO_3\\ CaCO_3+2HCl\rightarrow CaCl_2+CO_2+H_2O\)

e).

\(Si+O_2\underrightarrow{t^o}SiO_2\\ SiO_2+2NaOH\underrightarrow{t^o}Na_2SiO_3+H_2O\uparrow\\ Na_2SiO_3+H_2SO_4\rightarrow Na_2SO_4+H_2SiO_3\)

a. \(CuCl_2+2NaOH\rightarrow Cu\left(OH\right)_2\downarrow+2NaCl\)

0,125mol \(\leftarrow\)0,25mol\(\rightarrow\)0,125mol

b. \(n_{NaOH}=\dfrac{10}{40}=0,25mol\\ m_{Cu\left(OH\right)_2}=0,125\cdot98=12,25g\)

c.\(C_{M_{Cu\left(OH\right)_2}}=\dfrac{0,125}{0,1}=1,25M\)