Đáp án:
b) x=0
Giải thích các bước giải:
\(\begin{array}{l}
a)DK:x \ne \pm 2\\
\dfrac{{x\left( {x + 2} \right) + 3\left( {x - 2} \right)}}{{{{\left( {x + 2} \right)}^2}\left( {x - 2} \right)}} = 0\\
\to {x^2} + 2x + 3x - 6 = 0\\
\to {x^2} + 5x - 6 = 0\\
\to {x^2} - x + 6x - 6 = 0\\
\to x\left( {x - 1} \right) + 6\left( {x - 1} \right) = 0\\
\to \left( {x - 1} \right)\left( {x + 6} \right) = 0\\
\to \left[ \begin{array}{l}
x = 1\\
x = - 6
\end{array} \right.\\
b)\dfrac{{1 - {x^2} - x - 1 + {x^3} + {x^2} + x}}{{{x^2} + x + 1}} = 0\\
\to {x^3} = 0\\
\to x = 0\\
c)DK:x \ne \pm 2\\
\left[ {\dfrac{{2\left( {x + 2} \right) - 2\left( {x - 2} \right)}}{{\left( {x + 2} \right)\left( {x - 2} \right)}}} \right].\dfrac{{{{\left( {x + 2} \right)}^2}}}{8} = 0\\
\to \dfrac{{2x + 4 - 2x + 4}}{{\left( {x + 2} \right)\left( {x - 2} \right)}}.\dfrac{{{{\left( {x + 2} \right)}^2}}}{8} = 0\\
\to \dfrac{8}{{\left( {x + 2} \right)\left( {x - 2} \right)}}.\dfrac{{{{\left( {x + 2} \right)}^2}}}{8} = 0\\
\to x + 2 = 0\\
\to x = - 2\left( l \right)\\
\to x \in \emptyset
\end{array}\)
