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Đáp án:

\(\begin{array}{l}
a,\\
{\left( {1 - 5x} \right)^2}\\
b,\\
9{x^2} + 30xy + 25{y^2}\\
c,\\
9x - 6\sqrt x y + {y^2}\\
d,\\
{\left( {{x^2} + 9} \right)^2}\\
e,\\
9{x^2} - 24xy + 16{y^2}\\
f,\\
x + 4\sqrt x y + 4{y^2}\\
g,\\
{\left( {x - 2{y^2}} \right)^2}\\
h,\\
4{x^2} - 12xy + 9{y^2}\\
i,\\
\left( {3\sqrt x  - \sqrt y } \right).\left( {3\sqrt x  + \sqrt y } \right)\\
m,\\
\dfrac{{{x^2}}}{{{y^2}}} - \dfrac{4}{9}\\
n,\\
\dfrac{{{y^2}}}{9} - \dfrac{{{x^2}}}{4}\\
p,\\
4x - \dfrac{4}{9}
\end{array}\)

Giải thích các bước giải:

 Ta có:

\(\begin{array}{l}
a,\\
1 - 10x + 25{x^2} = {1^2} - 2.1.5x + {\left( {5x} \right)^2} = {\left( {1 - 5x} \right)^2}\\
b,\\
{\left( {3x + 5y} \right)^2} = {\left( {3x} \right)^2} + 2.3x.5y + {\left( {5y} \right)^2} = 9{x^2} + 30xy + 25{y^2}\\
c,\\
{\left( {3\sqrt x  - y} \right)^2} = {\left( {3\sqrt x } \right)^2} - 2.3\sqrt x .y + {y^2} = 9x - 6\sqrt x y + {y^2}\\
d,\\
{x^4} + 81 + 18{x^2} = {x^4} + 18{x^2} + 81\\
 = {\left( {{x^2}} \right)^2} + 2.{x^2}.9 + {9^2} = {\left( {{x^2} + 9} \right)^2}\\
e,\\
{\left( {3x - 4y} \right)^2} = {\left( {3x} \right)^2} - 2.3x.4y + {\left( {4y} \right)^2} = 9{x^2} - 24xy + 16{y^2}\\
f,\\
{\left( {\sqrt x  + 2y} \right)^2} = {\sqrt x ^2} + 2.\sqrt x .2y + {\left( {2y} \right)^2} = x + 4\sqrt x y + 4{y^2}\\
g,\\
{x^2} + 4{y^4} - 4x{y^2} = {x^2} - 4x{y^2} + 4{y^4}\\
 = {x^2} - 2.x.2{y^2} + {\left( {2{y^2}} \right)^2} = {\left( {x - 2{y^2}} \right)^2}\\
h,\\
{\left( {2x - 3y} \right)^2} = {\left( {2x} \right)^2} - 2.2x.3y + {\left( {3y} \right)^2} = 4{x^2} - 12xy + 9{y^2}\\
i,\\
{\left( {3\sqrt x } \right)^2} - {\left( {\sqrt y } \right)^2} = \left( {3\sqrt x  - \sqrt y } \right).\left( {3\sqrt x  + \sqrt y } \right)\\
m,\\
\left( {\dfrac{x}{y} - \dfrac{2}{3}} \right)\left( {\dfrac{x}{y} + \dfrac{2}{3}} \right) = {\left( {\dfrac{x}{y}} \right)^2} - {\left( {\dfrac{2}{3}} \right)^2} = \dfrac{{{x^2}}}{{{y^2}}} - \dfrac{4}{9}\\
n,\\
\left( {\dfrac{x}{2} + \dfrac{y}{3}} \right)\left( {\dfrac{y}{3} - \dfrac{x}{2}} \right) = \left( {\dfrac{y}{3} + \dfrac{x}{2}} \right)\left( {\dfrac{y}{3} - \dfrac{x}{2}} \right)\\
 = {\left( {\dfrac{y}{3}} \right)^2} - {\left( {\dfrac{x}{2}} \right)^2} = \dfrac{{{y^2}}}{9} - \dfrac{{{x^2}}}{4}\\
p,\\
\left( {2\sqrt x  - \dfrac{2}{3}} \right)\left( {\dfrac{2}{3} + 2\sqrt x } \right) = \left( {2\sqrt x  - \dfrac{2}{3}} \right)\left( {2\sqrt x  + \dfrac{2}{3}} \right)\\
 = {\left( {2\sqrt x } \right)^2} - {\left( {\dfrac{2}{3}} \right)^2} = 4x - \dfrac{4}{9}
\end{array}\)