Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
a,\\
\left( {x - 2} \right)\left( {x + 2} \right) - {\left( {x + 1} \right)^2} = 0\\
\Leftrightarrow \left( {{x^2} - {2^2}} \right) - \left( {{x^2} + 2x + 1} \right) = 0\\
\Leftrightarrow {x^2} - 4 - {x^2} - 2x - 1 = 0\\
\Leftrightarrow - 2x - 5 = 0\\
\Leftrightarrow - 2x = 5\\
\Leftrightarrow x = - \frac{5}{2}\\
b,\\
\left( {3x + 1} \right)\left( {x + 1} \right) - \left( {3x + 1} \right).\left( {3x - 1} \right) = 5\\
\Leftrightarrow \left( {3{x^2} + 4x + 1} \right) - \left( {9{x^2} - 1} \right) = 5\\
\Leftrightarrow - 4{x^2} + 4x + 2 = 5\\
\Leftrightarrow 4{x^2} - 4x + 3 = 0\\
\Leftrightarrow \left( {4{x^2} - 4x + 1} \right) + 2 = 0\\
\Leftrightarrow {\left( {2x - 1} \right)^2} + 2 = 0\,\,\,\,\,\,\,\,\,\,\,\left( {vn} \right)\\
2,\\
\left( {a + b} \right)\left( {{a^2} - ab + {b^2}} \right)\\
= {a^3} - {a^2}b + a{b^2} + {a^2}b - a{b^2} + {b^3}\\
= {a^3} + {b^3}\\
b,\\
\left( {a - b} \right)\left( {{a^2} + ab + {b^2}} \right)\\
= {a^3} + {a^2}b + a{b^2} - {a^2}b - a{b^2} + {b^3}\\
= {a^3} + {b^3}
\end{array}\)
