Viết biểu thức \(25{x^2} - 20xy + 4{y^2}\)  dưới dạng bình phương của một hiệu

\(\left( {A - B} \right)\left( {A + B} \right) = {A^2} + 2AB + {B^2}\)

\(\left( {A + B} \right)\left( {A - B} \right) = {A^2} - {B^2}\)         

\(\left( {A + B} \right)\left( {A - B} \right) = {A^2} - 2AB + {B^2}\)

\(\left( {A + B} \right)\left( {A - B} \right) = {A^2} + {B^2}\)

\({\left( {x + 2y} \right)^2} = {x^2} + 4xy + 4{y^2}\).

\({\left( {x - 2y} \right)^2} = {x^2} - 4xy + 4{y^2}\).

\({\left( {x - 2y} \right)^2} = {x^2} - 4{y^2}\).

\(\left( {x - 2y} \right)\left( {x + 2y} \right) = {x^2} - 4{y^2}\).

\(\left( {\dfrac{x}{9} - \dfrac{y}{{64}}} \right)\left( {\dfrac{x}{9} + \dfrac{y}{{64}}} \right)\)      

\(\left( {\dfrac{x}{3} - \dfrac{y}{4}} \right)\left( {\dfrac{x}{3} + \dfrac{y}{4}} \right)\)

\(\left( {\dfrac{x}{9} - \dfrac{y}{8}} \right)\left( {\dfrac{x}{9} + \dfrac{y}{8}} \right)\)    

\(\left( {\dfrac{x}{3} - \dfrac{y}{8}} \right)\left( {\dfrac{x}{3} + \dfrac{y}{8}} \right)\)

\(\dfrac{{{x^2}}}{4} - xy + 4{y^2}\)

\(\dfrac{{{x^2}}}{4} - 2xy + 4{y^2}\)

\(\dfrac{{{x^2}}}{4} - 2xy + 2{y^2}\)

\(\dfrac{{{x^2}}}{2} - 2xy + 2{y^2}\)

\(4 - {\left( {a + b} \right)^2} = \left( {2 + a + b} \right)\left( {2 - a + b} \right)\).

\(4 - {\left( {a + b} \right)^2} = \left( {4 + a + b} \right)\left( {4 - a - b} \right)\).

\(4 - {\left( {a + b} \right)^2} = \left( {2 + a - b} \right)\left( {2 - a + b} \right)\).

\(4 - {\left( {a + b} \right)^2} = \left( {2 + a + b} \right)\left( {2 - a - b} \right)\).

\( 342\)

\(243\)

\(324\)

\( - 324\)