Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
a,\\
\left( {x - 3} \right)\left( {x + 3} \right) = {\left( {x + 3} \right)^2} - 15\\
\Leftrightarrow {x^2} - 9 = {x^2} + 6x + 9 - 15\\
\Leftrightarrow 6x = - 3\\
\Leftrightarrow x = - \dfrac{1}{2}\\
b,\\
\dfrac{2}{{x + 1}} - \dfrac{3}{x} = \dfrac{4}{{{x^2} + x}}\,\,\,\,\,\,\,\,\,\,\,\,\,\left( \begin{array}{l}
x \ne 0\\
x \ne - 1
\end{array} \right)\\
\Leftrightarrow \dfrac{{2x - 3.\left( {x + 1} \right)}}{{x\left( {x + 1} \right)}} = \dfrac{4}{{x\left( {x + 1} \right)}}\\
\Leftrightarrow \dfrac{{ - x - 3}}{{x\left( {x + 1} \right)}} = \dfrac{4}{{x\left( {x + 1} \right)}}\\
\Leftrightarrow - x - 3 = 4\\
\Leftrightarrow x = - 7\\
2,\\
{\left( {2x + 1} \right)^2} - 4x\left( {x + 1} \right) \ge 15\\
\Leftrightarrow \left( {4{x^2} + 4x + 1} \right) - \left( {4{x^2} + 4x} \right) \ge 15\\
\Leftrightarrow 1 \ge 15\,\,\,\,\left( {vn} \right)
\end{array}\)
