\(\left(x+2\right)\left(3-4x\right)=x^2+4x+4\)

\(\Leftrightarrow\left(x+2\right)\left(3-4x\right)=\left(x+2\right)^2\)

\(\Leftrightarrow\left(x+2\right)\left(3-4x\right)-\left(x+2\right)^2=0\)

\(\Leftrightarrow\left(x+2\right)\left(3-4x-x-2\right)=0\)

\(\Leftrightarrow\left(x+2\right)\left(-5x-5\right)=0\Leftrightarrow x=-2;-1\)

Answer:

\(5x+53=2xy+8y^2\)

\(\Rightarrow2\left(5x+53\right)=2\left(2xy+8y^2\right)\)

\(\Rightarrow10x+106=4xy+16y^2\)

\(\Rightarrow10x-4xy=16y^2-106\)

\(\Rightarrow x=\frac{16y^2-106}{10-4y}\)

\(\Rightarrow x=\frac{\left(16y^2-100\right)-6}{10-4y}\)

\(\Rightarrow x=\frac{-\left(10-4y\right)\left(4y+10\right)}{10-4y}-\frac{6}{10-4y}\)

\(\Rightarrow x=-4y-10-\frac{6}{10-4y}\)

Để cho x và y thuộc Z thì 6 chia hết cho 10 - 4y

\(\Rightarrow10-4y\inƯ\left(6\right)=\left\{\pm1;\pm2;\pm3;\pm6\right\}\)

Trường hợp: \(\orbr{\begin{cases}10-4y=1\\10-4y=-1\end{cases}}\Rightarrow\orbr{\begin{cases}4y=9\left(l\right)\\4y=11\left(l\right)\end{cases}}\)

Trường hợp: \(\orbr{\begin{cases}10-4y=2\\10-4y=-2\end{cases}}\Rightarrow\orbr{\begin{cases}4y=8\\4y=12\end{cases}}\Rightarrow\orbr{\begin{cases}y=2\Rightarrow x=-21\\y=3\Rightarrow x=-19\end{cases}}\)

Trường hợp: \(\orbr{\begin{cases}10-4y=3\\10-4y=-3\end{cases}}\Rightarrow\orbr{\begin{cases}4y=7\left(l\right)\\4y=13\left(l\right)\end{cases}}\)

Trường hợp: \(\orbr{\begin{cases}10-4y=6\\10-4y=-6\end{cases}}\Rightarrow\orbr{\begin{cases}4y=4\\4y=16\end{cases}}\Rightarrow\orbr{\begin{cases}y=1\Rightarrow x=-15\\y=4\Rightarrow x=-25\end{cases}}\)

a, 

\(Fe_xO_y+2yHCl\rightarrow xFeCl_{\frac{2y}{x}}+yH_2O\)

b, Theo pt : 

  \(n_{Fe_xO_y}=n_{HCl}\times\frac{1}{2y}=\frac{0,075}{y} \left(mol\right)\)

         \(\Rightarrow56x+16y=4\div\frac{0,075}{y}=\frac{160}{3}y\)

           \(\Rightarrow\frac{x}{y}=\frac{2}{3}\)

                      Vậy CT là Fe2O3

Cách 2

     Theo pt :

       \(n_{O\left(oxit\right)}=\frac{1}{2}n_{HCl}=0,075 \left(mol\right)\)

             \(\Rightarrow m_O=1,2\left(g\right)\)

             \(\Rightarrow m_{Fe}=4-1,2=2,8\left(g\right)\)

             \(\Rightarrow n_{Fe}=0,05 \left(mol\right)\)

  CÓ 

     x : y = 0,05 : 0,075 = 2 : 3

             Vậy CT là Fe2O3