Cho $T =  - 9{x^2} + 6x - 5$. Chọn khẳng định đúng.

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

$\dfrac{{{x^2}}}{2} - 2xy + 2{y^2}$

${\left( {5x - 2y} \right)^2}$

${\left( {2x - 5y} \right)^2}$

${\left( {25x - 4y} \right)^2}$

${\left( {5x + 2y} \right)^2}$

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

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

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

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

$342$

$243$

$324$

$- 324$