Đáp án:
c. x=4
d. \(\left[ \begin{array}{l}
x = 0\\
x = 3
\end{array} \right.\)
Giải thích các bước giải:
\(\begin{array}{l}
c.\frac{{{x^2} + 2x + 1 - {x^2} + 2x - 1 - 16}}{{\left( {x - 1} \right)\left( {x + 1} \right)}} = 0\\
\to {x^2} + 2x + 1 - {x^2} + 2x - 1 - 16 = 0\\
\to 4x - 16 = 0\\
\to x = 4\left( {TM} \right)\\
d.DK:x \ne \left\{ { - 3; - 2} \right\}\\
\frac{{{x^2} + 5x + 6 - x}}{{\left( {x + 2} \right)\left( {x + 3} \right)}} = \frac{{5x + 2x + 6}}{{\left( {x + 2} \right)\left( {x + 3} \right)}}\\
\to {x^2} + 5x + 6 - x - 7x - 6 = 0\\
\to {x^2} - 3x = 0\\
\to \left[ \begin{array}{l}
x = 0\\
x = 3
\end{array} \right.\left( {TM} \right)
\end{array}\)
