Giải thích các bước giải:
ĐKXĐ: \({x^4} - 10{x^2} + 9 \ne 0 \Leftrightarrow \left( {{x^2} - 1} \right)\left( {{x^2} - 9} \right) \ne 0 \Leftrightarrow \left\{ \begin{array}{l}
x \ne \pm 1\\
x \ne \pm 3
\end{array} \right.\)
Ta có:
\(\begin{array}{l}
A = \frac{{{x^4} - 5{x^2} + 4}}{{{x^4} - 10{x^2} + 9}}\\
= \frac{{\left( {{x^4} - {x^2}} \right) - \left( {4{x^2} - 4} \right)}}{{\left( {{x^4} - {x^2}} \right) - \left( {9{x^2} - 9} \right)}}\\
= \frac{{{x^2}\left( {{x^2} - 1} \right) - 4\left( {{x^2} - 1} \right)}}{{{x^2}\left( {{x^2} - 1} \right) - 9\left( {{x^2} - 1} \right)}}\\
= \frac{{\left( {{x^2} - 1} \right)\left( {{x^2} - 4} \right)}}{{\left( {{x^2} - 1} \right)\left( {{x^2} - 9} \right)}} = \frac{{{x^2} - 4}}{{{x^2} - 9}}
\end{array}\)
b,
Ta có:
\[A = 0 \Leftrightarrow \frac{{{x^2} - 4}}{{{x^2} - 9}} = 0 \Leftrightarrow {x^2} - 4 = 0 \Leftrightarrow x = \pm 2\]
c,
\[\begin{array}{l}
\left| {2x - 1} \right| = 7 \Leftrightarrow \left[ \begin{array}{l}
2x - 1 = 7\\
2x - 1 = - 7
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
x = 4\\
x = - 3\left( L \right)
\end{array} \right.\\
x = 4 \Rightarrow A = \frac{{{4^2} - 4}}{{{4^2} - 9}} = \frac{{12}}{7}
\end{array}\]
