Đáp án:
$\begin{array}{l}
a){x^2} - x + A = {B^2}\\
\Leftrightarrow {x^2} - 2.x.\dfrac{1}{2} + \dfrac{1}{4} - \dfrac{1}{4} + A = {B^2}\\
\Leftrightarrow {\left( {x - \dfrac{1}{2}} \right)^2} + A - \dfrac{1}{4} = {B^2}\\
\Leftrightarrow \left\{ \begin{array}{l}
A = \dfrac{1}{4}\\
B = x - \dfrac{1}{2}
\end{array} \right.\\
b)4{x^2} - A + 1 = {B^2}\\
\Leftrightarrow 4{x^2} - 4x + 1 + 4x - A = {B^2}\\
\Leftrightarrow {\left( {2x - 1} \right)^2} + 4x - A = {B^2}\\
\Leftrightarrow \left\{ \begin{array}{l}
A = 4x\\
B = 2x - 1
\end{array} \right.\\
c)\left( {5 - 3y} \right)\left( {5 + 3y} \right) = {A^2} - {B^2}\\
\Leftrightarrow {5^2} - {\left( {3y} \right)^2} = {A^2} - {B^2}\\
\Leftrightarrow \left\{ \begin{array}{l}
A = 5\\
B = 3y
\end{array} \right.\\
d)\left( {A - 6xy + 9{y^2}} \right) = {B^2}\\
\Leftrightarrow {x^2} - 6xy + 9{y^2} - {x^2} + A = {B^2}\\
\Leftrightarrow {\left( {x - 3y} \right)^2} + A - {x^2} = {B^2}\\
\Leftrightarrow \left\{ \begin{array}{l}
A = {x^2}\\
B = x - 3y
\end{array} \right.\\
e)2{x^2} + 8x + 10 = {A^2} + {B^2}\\
\Leftrightarrow 2.\left( {{x^2} + 4x + 4} \right) + 2 = {A^2} + {B^2}\\
\Leftrightarrow 2.{\left( {x + 2} \right)^2} + 2 = {A^2} + {B^2}\\
\Leftrightarrow \left\{ \begin{array}{l}
A = \sqrt 2 \left( {x + 2} \right)\\
B = \sqrt 2
\end{array} \right.\\
f)\left( {3x + 5} \right)\left( {x - 1} \right) = {A^2} - {B^2}\\
\Leftrightarrow 3{x^2} - 3x + 5x - 5 = {A^2} - {B^2}\\
\Leftrightarrow 3{x^2} + 2x - 5 = {A^2} - {B^2}\\
\Leftrightarrow {\left( {\sqrt 3 x} \right)^2} + 2.\sqrt 3 x.\dfrac{1}{{\sqrt 3 }} + \dfrac{1}{3} - \dfrac{1}{3} - 5 = {A^2} - {B^2}\\
\Leftrightarrow {\left( {\sqrt 3 x + \dfrac{1}{{\sqrt 3 }}} \right)^2} - \dfrac{{16}}{3} = {A^2} - {B^2}\\
\Leftrightarrow {\left( {\sqrt 3 x + \dfrac{1}{{\sqrt 3 }}} \right)^2} - {\left( {\dfrac{{4\sqrt 3 }}{3}} \right)^2}\\
\Leftrightarrow \left\{ \begin{array}{l}
A = \sqrt 3 x + \dfrac{1}{{\sqrt 3 }}\\
B = \dfrac{{4\sqrt 3 }}{3}
\end{array} \right.
\end{array}$
