Đáp án:
B1:
\(\dfrac{{{x^2} - 1}}{x}\)
Giải thích các bước giải:
\(\begin{array}{l}
B1:\\
\dfrac{{{x^3} - 4{x^2} - x + 4}}{{{x^2} - 4x}} = \dfrac{{{x^2}\left( {x - 4} \right) - \left( {x - 4} \right)}}{{x\left( {x - 4} \right)}}\\
= \dfrac{{\left( {{x^2} - 1} \right)\left( {x - 4} \right)}}{{x\left( {x - 4} \right)}}\\
= \dfrac{{{x^2} - 1}}{x}\\
B2:\\
a)DK:{x^3} - 1 \ne 0 \to x - 1 \ne 0 \to x \ne 1\\
\dfrac{{{x^2} - 9x + 8}}{{{x^3} - 1}} = 0\\
\to {x^2} - 9x + 8 = 0\\
\to \left( {x - 8} \right)\left( {x - 1} \right) = 0\\
\to \left[ \begin{array}{l}
x = 8\left( {TM} \right)\\
x = 1\left( l \right)
\end{array} \right.\\
b)DK:{x^2} - 1 \ne 0 \to x \ne \pm 1\\
\dfrac{{3{x^2} + x - 2}}{{{x^2} - 1}} = 0\\
\to 3{x^2} + x - 2 = 0\\
\to \left( {3x - 2} \right)\left( {x + 1} \right) = 0\\
\to \left[ \begin{array}{l}
x = \dfrac{2}{3}\left( {TM} \right)\\
x = - 1\left( l \right)
\end{array} \right.\\
c)DK:{x^2} + 2x - 3 \ne 0 \to \left( {x - 1} \right)\left( {x + 3} \right) \ne 0\\
\to x \ne \left\{ { - 3;1} \right\}\\
\dfrac{{{x^3} + 2{x^2} - x - 2}}{{{x^2} + 2x - 3}} = 0\\
\to {x^2}\left( {x + 2} \right) - \left( {x + 2} \right) = 0\\
\to \left( {x + 2} \right)\left( {{x^2} - 1} \right) = 0\\
\to \left[ \begin{array}{l}
x = - 2\\
x - 1 = 0\\
x + 1 = 0
\end{array} \right.\\
\to \left[ \begin{array}{l}
x = - 2\left( {TM} \right)\\
x = 1\left( l \right)\\
x = - 1\left( {TM} \right)
\end{array} \right.
\end{array}\)
