Resolve as seguintes equações

Casos notáveis:
$${\left( {A + B} \right)^2} = {A^2} + 2AB + {B^2}$$

$$\left( {A + B} \right)\left( {A – B} \right) = {A^2} – {B^2}$$

Lei do anulamento do produto:
$$\begin{array}{*{20}{c}}
{A \times B = 0}& \Leftrightarrow &{\begin{array}{*{20}{c}}
{A = 0}& \vee &{B = 0}
\end{array}}
\end{array}$$

1. Resolvendo a equação, temos:
   $$\begin{array}{*{20}{l}}
   {{x^2} – 2x + 1 = 0}& \Leftrightarrow &{{{\left( {x – 1} \right)}^2} = 0} \\
   {}& \Leftrightarrow &{x – 1 = 0} \\
   {}& \Leftrightarrow &{x = 1}
   \end{array}$$
2. Resolvendo a equação, temos:
   $$\begin{array}{*{20}{l}}
   {9{x^2} + 12x + 4 = 0}& \Leftrightarrow &{{{\left( {3x + 2} \right)}^2} = 0} \\
   {}& \Leftrightarrow &{3x + 2 = 0} \\
   {}& \Leftrightarrow &{x =  – \frac{2}{3}}
   \end{array}$$
3. Resolvendo a equação, temos:
   $$\begin{array}{*{20}{l}}
   {4{x^2} – 20x + 25 = 0}& \Leftrightarrow &{{{\left( {2x – 5} \right)}^2} = 0} \\
   {}& \Leftrightarrow &{2x – 5 = 0} \\
   {}& \Leftrightarrow &{x = \frac{5}{2}}
   \end{array}$$
4. Resolvendo a equação, temos:
   $$\begin{array}{*{20}{l}}
   {{x^2} – 8x = 4}& \Leftrightarrow &{{x^2} – 8x – 4 = 0} \\
   {}& \Leftrightarrow &{{{\left( {x – 4} \right)}^2} – 16 – 4 = 0} \\
   {}& \Leftrightarrow &{{{\left( {x – 4} \right)}^2} = 20} \\
   {}& \Leftrightarrow &{\begin{array}{*{20}{l}}
   {x – 4 =  – \sqrt {20} }& \vee &{x – 4 = \sqrt {20} }
   \end{array}} \\
   {}& \Leftrightarrow &{\begin{array}{*{20}{l}}
   {x = 4 – \sqrt {20} }& \vee &{x = 4 + \sqrt {20} }
   \end{array}}
   \end{array}$$
   Alternativa:
   $$\begin{array}{*{20}{l}}
   {{x^2} – 8x = 4}& \Leftrightarrow &{{x^2} – 8x – 4 = 0} \\
   {}& \Leftrightarrow &{{{\left( {x – 4} \right)}^2} – 16 – 4 = 0} \\
   {}& \Leftrightarrow &{{{\left( {x – 4} \right)}^2} – 20 = 0} \\
   {}& \Leftrightarrow &{{{\left( {x – 4} \right)}^2} – {{\left( {\sqrt {20} } \right)}^2} = 0} \\
   {}& \Leftrightarrow &{\left[ {\left( {x – 4} \right) + \sqrt {20} } \right]\left[ {\left( {x – 4} \right) – \sqrt {20} } \right] = 0} \\
   {}& \Leftrightarrow &{\begin{array}{*{20}{l}}
   {x – 4 + \sqrt {20}  = 0}& \vee &{x – 4 – \sqrt {20} }
   \end{array}} \\
   {}& \Leftrightarrow &{\begin{array}{*{20}{l}}
   {x = 4 – \sqrt {20} }& \vee &{x = 4 + \sqrt {20} }
   \end{array}}
   \end{array}$$