Lời giải chi tiết:

Hướng dẫn giải chi tiết

\(\begin{array}{l}a)\;A = \left( {{x^2} - 3x + 9} \right)\left( {x + 3} \right) - \left( {54 + {x^3}} \right)\\A = \left( {{x^2} - 3x + {3^2}} \right)\left( {x + 3} \right) - \left( {54 + {x^3}} \right)\\A = {x^3} + {3^3} - 54 - {x^3}\\A = 27 - 54 = - 27\end{array}\)

\(\begin{array}{l}b)\;B = {\left( {x - 1} \right)^2} + 2\left( {x - 1} \right)\left( {x + 1} \right) + {\left( {x + 1} \right)^2} + 5\\B = {\left( {\left( {x - 1} \right) + \left( {x + 1} \right)} \right)^2} + 5\\B = {\left( {x - 1 + x + 1} \right)^2} + 5\\B = {\left( {2x} \right)^2} + 5 = 4{x^2} + 5\end{array}\)

\(\begin{array}{l}c)\;C = 9{x^2} - 2xy + \dfrac{1}{9}{y^2} - 2\left( {3x + \dfrac{1}{3}y} \right)\left( {3x - \dfrac{1}{3}y} \right) + {\left( {3x + \dfrac{1}{3}y} \right)^2}\\C = {\left( {3x} \right)^2} - 2.3x.\dfrac{1}{3}y + {\left( {\dfrac{1}{3}y} \right)^2} - 2\left( {3x + \dfrac{1}{3}y} \right)\left( {3x - \dfrac{1}{3}y} \right) + {\left( {3x + \dfrac{1}{3}y} \right)^2}\\C = {\left( {3x - \dfrac{1}{3}y} \right)^2} - 2\left( {3x + \dfrac{1}{3}y} \right)\left( {3x - \dfrac{1}{3}y} \right) + {\left( {3x + \dfrac{1}{3}y} \right)^2}\\C = {\left( {\left( {3x - \dfrac{1}{3}y} \right) - \left( {3x + \dfrac{1}{3}y} \right)} \right)^2}\\C = {\left( {3x - \dfrac{1}{3}y - 3x - \dfrac{1}{3}y} \right)^2}\\C = {\left( { - \dfrac{2}{3}y} \right)^2} = \dfrac{4}{9}{y^2}\end{array}\)