a) Ta có: \(A = B\)
\( \Leftrightarrow \left( {x - 3} \right)\left( {x + 4} \right) - 2\left( {3x - 2} \right) \) \(= {\left( {x - 4} \right)^2}\)
\( \Leftrightarrow {x^2} + 4x - 3x - 12 - 6x + 4 \) \(= {x^2} - 8x + 16 \)
\( \Leftrightarrow {x^2} - {x^2} + 4x - 3x - 6x + 8x \) \(= 16 + 12 - 4\)
\( \Leftrightarrow 3x = 24 \Leftrightarrow x = 8 \)
Vậy với \(x = 8\) thì \(A = B\).
b) Ta có : \(A = B\)
\( \Leftrightarrow \left( {x + 2} \right)\left( {x - 2} \right) + 3{x^2} \) \(= {\left( {2x + 1} \right)^2} + 2x\)
\( \Leftrightarrow {x^2} - 4 + 3{x^2} \) \(= 4{x^2} + 4x + 1 + 2x \)
\( \Leftrightarrow {x^2} + 3{x^2} - 4{x^2} - 4x - 2x \) \( = 1 + 4 \)
\( \displaystyle \Leftrightarrow - 6x = 5 \Leftrightarrow x = - {5 \over 6} \)
Vậy với \( \displaystyle x = - {5 \over 6} \) thì \(A = B\).
c) Ta có: \(A = B\)
\( \Leftrightarrow \left( {x - 1} \right)\left( {{x^2} + x + 1} \right) - 2x \) \(= x\left( {x - 1} \right)\left( {x + 1} \right)\)
\(\eqalign{ & \Leftrightarrow {x^3} - 1 - 2x = x\left( {{x^2} - 1} \right) \cr & \Leftrightarrow {x^3} - 1 - 2x = {x^3} - x \cr & \Leftrightarrow {x^3} - {x^3} - 2x + x = 1 \cr & \Leftrightarrow - x = 1 \Leftrightarrow x = - 1 \cr} \)
Vậy với \(x = -1\) thì \(A = B\).
d) Ta có : \(A = B\)
\( \Leftrightarrow {\left( {x + 1} \right)^3} - {\left( {x - 2} \right)^3} \) \(= \left( {3x - 1} \right)\left( {3x + 1} \right)\)
\( \Leftrightarrow {x^3} + 3{x^2} + 3x + 1 - {x^3} + 6{x^2} \) \( - 12x + 8 = 9{x^2} - 1 \)
\( \Leftrightarrow {x^3} - {x^3} + 3{x^2} + 6{x^2} - 9{x^2} + 3x \) \( - 12x = - 1 - 1 - 8 \)
\(\displaystyle \Leftrightarrow - 9x = - 10 \Leftrightarrow x = {{10} \over 9} \)
Vậy với \(\displaystyle x = {{10} \over 9}\) thì \(A = B\).
