Đáp án:

\(\begin{array}{l}
a,\\
{\left( {x - 5} \right)^2}\\
b,\\
{\left( {x - 10} \right)^2}\\
c,\\
5.\left( {2x - 1} \right)\\
d,\\
{\left( {{y^2} - 7} \right)^2}\\
e,\\
{\left( {x + 2y} \right)^2}\\
f,\\
{\left( {x + \frac{5}{2}} \right)^2}\\
g,\\
{\left( {x + 2} \right)^2}\\
h,\\
{\left( {x - 4} \right)^2}\\
i,\\
\left( {x - 1} \right)\left( {x + 1} \right)\\
j,\\
\left( {3x - 5{y^2}} \right)\left( {3x + 5{y^2}} \right)\\
k,\\
7.\left( {2x + 5} \right)\\
l,\\
{\left( {4x - 1} \right)^2}\\
m,\\
{\left( {2x + 3y} \right)^2}\\
n,\\
4.\left( {x - 2} \right)\left( {x + 2} \right)\\
o,\\
{\left( {x + 3} \right)^2}\\
p,\\
{\left( {4x - 1} \right)^2}
\end{array}\)

Giải thích các bước giải:

 Ta có:

\(\begin{array}{l}
a,\\
{x^2} - 10x + 25 = {x^2} - 2.x.5 + {5^2} = {\left( {x - 5} \right)^2}\\
b,\\
{x^2} - 20x + 100 = {x^2} - 2.x.10 + {10^2} = {\left( {x - 10} \right)^2}\\
c,\\
{\left( {x + 2} \right)^2} - {\left( {x - 3} \right)^2}\\
 = \left[ {\left( {x + 2} \right) - \left( {x - 3} \right)} \right].\left[ {\left( {x + 2} \right) + \left( {x - 3} \right)} \right]\\
 = \left( {x + 2 - x + 3} \right).\left( {x + 2 + x - 3} \right)\\
 = 5.\left( {2x - 1} \right)\\
d,\\
{y^4} - 14{y^2} + 49 = {\left( {{y^2}} \right)^2} - 2.{y^2}.7 + {7^2} = {\left( {{y^2} - 7} \right)^2}\\
e,\\
{x^2} + 4xy + 4{y^2} = {x^2} + 2.x.\left( {2y} \right) + {\left( {2y} \right)^2} = {\left( {x + 2y} \right)^2}\\
f,\\
{x^2} + 5x + \frac{{25}}{4} = {x^2} + 2.x.\frac{5}{2} + {\left( {\frac{5}{2}} \right)^2} = {\left( {x + \frac{5}{2}} \right)^2}\\
g,\\
{x^2} + 4x + 4 = {x^2} + 2.x.2 + {2^2} = {\left( {x + 2} \right)^2}\\
h,\\
{x^2} - 8x + 16 = {x^2} - 2.x.4 + {4^2} = {\left( {x - 4} \right)^2}\\
i,\\
{x^2} - 1 = {x^2} - {1^2} = \left( {x - 1} \right)\left( {x + 1} \right)\\
j,\\
9{x^2} - 25{y^4} = {\left( {3x} \right)^2} - {\left( {5{y^2}} \right)^2} = \left( {3x - 5{y^2}} \right)\left( {3x + 5{y^2}} \right)\\
k,\\
\left( {{x^2} + 12x + 36} \right) - {\left( {x - 1} \right)^2}\\
 = \left( {{x^2} + 2.x.6 + {6^2}} \right) - {\left( {x - 1} \right)^2}\\
 = {\left( {x + 6} \right)^2} - {\left( {x - 1} \right)^2}\\
 = \left[ {\left( {x + 6} \right) - \left( {x - 1} \right)} \right].\left[ {\left( {x + 6} \right) + \left( {x - 1} \right)} \right]\\
 = \left( {x + 6 - x + 1} \right).\left( {x + 6 + x - 1} \right)\\
 = 7.\left( {2x + 5} \right)\\
l,\\
16{x^2} - 8x + 1 = {\left( {4x} \right)^2} - 2.4x.1 + {1^2} = {\left( {4x - 1} \right)^2}\\
m,\\
4{x^2} + 12xy + 9{y^2} = {\left( {2x} \right)^2} + 2.2x.3y + {\left( {3y} \right)^2} = {\left( {2x + 3y} \right)^2}\\
n,\\
4{x^2} - 16 = 4.\left( {{x^2} - 4} \right) = 4.\left( {{x^2} - {2^2}} \right) = 4.\left( {x - 2} \right)\left( {x + 2} \right)\\
o,\\
{x^2} + 6x + 9 = {x^2} + 2.x.3 + {3^2} = {\left( {x + 3} \right)^2}\\
p,\\
16{x^2} - 8x + 1 = {\left( {4x} \right)^2} - 2.4x.1 + {1^2} = {\left( {4x - 1} \right)^2}
\end{array}\)