Đáp án:
\(\begin{array}{l}
a,\,\,\,\,\dfrac{{{x^2}}}{9} + \dfrac{8}{3}xy + 16{y^2}\\
b,\,\,\,\,\dfrac{1}{{{x^2}}} - \dfrac{6}{{xy}} + \dfrac{9}{{{y^2}}}\\
c,\,\,\,\,\dfrac{{{x^2}}}{4} - \dfrac{{{y^2}}}{{36}}\\
d,\,\,\,\,{x^4} - \dfrac{4}{{25}}{y^2}\\
e,\,\,\,\,{y^2} - 4x{y^2} + 4{x^2}{y^2}\\
f,\,\,\,\,{x^2} - \dfrac{{32}}{5}xy + \dfrac{{256}}{{25}}{y^2}\\
g,\,\,\,\,\dfrac{1}{9}{x^2}{y^4} - {z^6}\\
h,\,\,\,\,{x^4} - \dfrac{8}{9}{x^2} + \dfrac{{16}}{{81}}
\end{array}\)
Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
a,\\
{\left( {\dfrac{x}{3} + 4y} \right)^2} = {\left( {\dfrac{x}{3}} \right)^2} + 2.\dfrac{x}{3}.4y + {\left( {4y} \right)^2} = \dfrac{{{x^2}}}{9} + \dfrac{8}{3}xy + 16{y^2}\\
b,\\
{\left( {\dfrac{1}{x} - \dfrac{3}{y}} \right)^2} = {\left( {\dfrac{1}{x}} \right)^2} - 2.\dfrac{1}{x}.\dfrac{3}{y} + {\left( {\dfrac{3}{y}} \right)^2} = \dfrac{1}{{{x^2}}} - \dfrac{6}{{xy}} + \dfrac{9}{{{y^2}}}\\
c,\\
\left( {\dfrac{x}{2} - \dfrac{y}{6}} \right)\left( {\dfrac{x}{2} + \dfrac{y}{6}} \right) = {\left( {\dfrac{x}{2}} \right)^2} - {\left( {\dfrac{y}{6}} \right)^2} = \dfrac{{{x^2}}}{4} - \dfrac{{{y^2}}}{{36}}\\
d,\\
\left( {{x^2} + \dfrac{2}{5}y} \right)\left( {{x^2} - \dfrac{2}{5}y} \right) = {\left( {{x^2}} \right)^2} - {\left( {\dfrac{2}{5}y} \right)^2} = {x^4} - \dfrac{4}{{25}}{y^2}\\
e,\\
{\left( {y - 2xy} \right)^2} = {y^2} - 2.y.2xy + {\left( {2xy} \right)^2} = {y^2} - 4x{y^2} + 4{x^2}{y^2}\\
f,\\
16.{\left( { - \dfrac{1}{4}x + \dfrac{4}{5}y} \right)^2}\\
= 16.\left[ {{{\left( { - \dfrac{1}{4}x} \right)}^2} + 2.\left( { - \dfrac{1}{4}x} \right).\dfrac{4}{5}y + {{\left( {\dfrac{4}{5}y} \right)}^2}} \right]\\
= 16.\left( {\dfrac{1}{{16}}{x^2} - \dfrac{2}{5}xy + \dfrac{{16}}{{25}}{y^2}} \right)\\
= {x^2} - \dfrac{{32}}{5}xy + \dfrac{{256}}{{25}}{y^2}\\
g,\\
\left( { - \dfrac{1}{3}x{y^2} + {z^3}} \right)\left( { - \dfrac{1}{3}x{y^2} - {z^3}} \right)\\
= {\left( { - \dfrac{1}{3}x{y^2}} \right)^2} - {\left( {{z^3}} \right)^2}\\
= \dfrac{1}{9}{x^2}{y^4} - {z^6}\\
h,\\
{\left( {x + \dfrac{2}{3}} \right)^2}.{\left( {x - \dfrac{2}{3}} \right)^2}\\
= {\left[ {\left( {x + \dfrac{2}{3}} \right).\left( {x - \dfrac{2}{3}} \right)} \right]^2}\\
= {\left[ {{x^2} - {{\left( {\dfrac{2}{3}} \right)}^2}} \right]^2}\\
= {\left( {{x^2} - \dfrac{4}{9}} \right)^2}\\
= {\left( {{x^2}} \right)^2} - 2.{x^2}.\dfrac{4}{9} + {\left( {\dfrac{4}{9}} \right)^2}\\
= {x^4} - \dfrac{8}{9}{x^2} + \dfrac{{16}}{{81}}
\end{array}\)
