Giải thích các bước giải:

Ta có:

\(\begin{array}{l}
a,\,\,\,3{x^2} - 75 = 0\\
 \Leftrightarrow 3.\left( {{x^2} - 25} \right) = 0\\
 \Leftrightarrow 3.\left( {x - 5} \right)\left( {x + 5} \right) = 0\\
 \Leftrightarrow \left[ \begin{array}{l}
x - 5 = 0\\
x + 5 = 0
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
x = 5\\
x =  - 5
\end{array} \right.\\
b,\,\,\,\,\,2{x^2} - 162 = 0\\
 \Leftrightarrow 2.\left( {{x^2} - 81} \right) = 0\\
 \Leftrightarrow 2.\left( {x - 9} \right)\left( {x + 9} \right) = 0\\
 \Leftrightarrow \left[ \begin{array}{l}
x - 9 = 0\\
x + 9 = 0
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
x = 9\\
x =  - 9
\end{array} \right.\\
c,\,\,\,\,\,{x^2} + 4x + 4 = 0\\
 \Leftrightarrow {x^2} + 2.x.2 + {2^2} = 0\\
 \Leftrightarrow {\left( {x + 2} \right)^2} = 0\\
 \Leftrightarrow x + 2 = 0\\
 \Leftrightarrow x =  - 2\\
d,\,\,\,\,{x^3} + 3{x^2} + 3x + 1 = 0\\
 \Leftrightarrow {x^3} + 3.{x^2}.1 + 3.x{.1^2} + {1^3} = 0\\
 \Leftrightarrow {\left( {x + 1} \right)^3} = 0\\
 \Leftrightarrow x + 1 = 0\\
 \Leftrightarrow x =  - 1\\
e,\,\,\,2{x^2} - 72 = 0\\
 \Leftrightarrow 2.\left( {{x^2} - 36} \right) = 0\\
 \Leftrightarrow 2.\left( {x - 6} \right)\left( {x + 6} \right) = 0\\
 \Leftrightarrow \left[ \begin{array}{l}
x - 6 = 0\\
x + 6 = 0
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
x = 6\\
x =  - 6
\end{array} \right.\\
g,\,\,\,\,\,{\left( {x + 3} \right)^2} = 0\\
 \Leftrightarrow x + 3 = 0\\
 \Leftrightarrow x =  - 3\\
h,\,\,\,\,{\left( {x - 1} \right)^2} - 81 = 0\\
 \Leftrightarrow {\left( {x - 1} \right)^2} - {9^2} = 0\\
 \Leftrightarrow \left[ {\left( {x - 1} \right) - 9} \right].\left[ {\left( {x - 1} \right) + 9} \right] = 0\\
 \Leftrightarrow \left( {x - 10} \right)\left( {x + 8} \right) = 0\\
 \Leftrightarrow \left[ \begin{array}{l}
x - 10 = 0\\
x + 8 = 0
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
x = 10\\
x =  - 8
\end{array} \right.\\
i,\,\,\,\,{x^2} - 14x + 49 = 0\\
 \Leftrightarrow {x^2} - 2.x.7 + {7^2} = 0\\
 \Leftrightarrow {\left( {x - 7} \right)^2} = 0\\
 \Leftrightarrow x - 7 = 0\\
 \Leftrightarrow x = 7\\
k,\,\,\,\,{x^3} - 3{x^2} + 3x - 1 = 0\\
 \Leftrightarrow {x^3} - 3.{x^2}.1 + 3.x{.1^2} - {1^3} = 0\\
 \Leftrightarrow {\left( {x - 1} \right)^3} = 0\\
 \Leftrightarrow x - 1 = 0\\
 \Leftrightarrow x = 1\\
l,\,\,\,\,2{x^2} - 98 = 0\\
 \Leftrightarrow 2.\left( {{x^2} - 49} \right) = 0\\
 \Leftrightarrow 2.\left( {x - 7} \right)\left( {x + 7} \right) = 0\\
 \Leftrightarrow \left[ \begin{array}{l}
x - 7 = 0\\
x + 7 = 0
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
x = 7\\
x =  - 7
\end{array} \right.\\
m,\,\,\,\,\,2{x^2} - 8 = 0\\
 \Leftrightarrow 2.\left( {{x^2} - 4} \right) = 0\\
 \Leftrightarrow 2.\left( {x - 2} \right)\left( {x + 2} \right) = 0\\
 \Leftrightarrow \left[ \begin{array}{l}
x - 2 = 0\\
x + 2 = 0
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
x = 2\\
x =  - 2
\end{array} \right.\\
p,\,\,\,\,{\left( {x + 2} \right)^2} = 25\\
 \Leftrightarrow \sqrt {{{\left( {x + 2} \right)}^2}}  = \sqrt {25} \\
 \Leftrightarrow \left| {x + 2} \right| = 5\\
 \Leftrightarrow \left[ \begin{array}{l}
x + 2 = 5\\
x + 2 =  - 5
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
x = 3\\
x =  - 7
\end{array} \right.\\
q,\,\,\,\,\,{x^2} + 6x + 9 = 0\\
 \Leftrightarrow {x^2} + 2.x.3 + {3^2} = 0\\
 \Leftrightarrow {\left( {x + 3} \right)^2} = 0\\
 \Leftrightarrow x + 3 = 0\\
 \Leftrightarrow x =  - 3
\end{array}\)

\(\begin{array}{l}
s,\\
{x^2} + 10x + 25 = 0\\
 \Leftrightarrow {x^2} + 2.x.5 + {5^2} = 0\\
 \Leftrightarrow {\left( {x + 5} \right)^2} = 0\\
 \Leftrightarrow x + 5 = 0\\
 \Leftrightarrow x =  - 5\\
t,\\
{x^3} + 6{x^2} + 12x + 8 = 0\\
 \Leftrightarrow {x^3} + 3.{x^2}.2 + 3.x{.2^2} + {2^3} = 0\\
 \Leftrightarrow {\left( {x + 2} \right)^3} = 0\\
 \Leftrightarrow x + 2 = 0\\
 \Leftrightarrow x =  - 2
\end{array}\)