Đáp án:
\(A = \dfrac{9}{4}\)
Giải thích các bước giải:
\(DK:x \ne \left\{ { - 2;3} \right\}\)
\(\begin{array}{l}
A = - \dfrac{{3\left( {x + 1} \right)}}{{{x^2} - x - 6}}\\
= - \dfrac{{3\left( {x + 1} \right)}}{{{x^2} + 2x - 3x - 6}}\\
= - \dfrac{{3\left( {x + 1} \right)}}{{x\left( {x + 2} \right) - 3\left( {x + 2} \right)}}\\
= - \dfrac{{3\left( {x + 1} \right)}}{{\left( {x + 2} \right)\left( {x - 3} \right)}}
\end{array}\)
Có:
\(\begin{array}{l}
{x^2} - 4 = 0\\
\to \left( {x - 2} \right)\left( {x + 2} \right) = 0\\
\to \left[ \begin{array}{l}
x = 2\\
x = - 2\left( l \right)
\end{array} \right.\\
Thay:x = 2\\
\to A = - \dfrac{{3\left( {2 + 1} \right)}}{{\left( {2 + 2} \right)\left( {2 - 3} \right)}} = \dfrac{9}{4}
\end{array}\)
