Question

Đã trả lời bởi chuyên gia

Answer 1

\[\begin{array}{l}
Dat:A = \frac{{{x^3} - 7x + 6}}{{{x^3} + 5{x^2} - 2x - 24}}\\
= \frac{{{x^3} - 4x - 3x + 6}}{{{x^3} - 2{x^2} + 7{x^2} - 14x + 12x - 24}}\\
= \frac{{x\left( {{x^2} - 4} \right) - 3\left( {x - 2} \right)}}{{{x^2}\left( {x - 2} \right) + 7x\left( {x - 2} \right) + 12\left( {x - 2} \right)}}\\
= \frac{{x\left( {x - 2} \right)\left( {x + 2} \right) - 3\left( {x - 2} \right)}}{{\left( {x - 2} \right)\left( {{x^2} + 7x + 12} \right)}}\\
= \frac{{\left( {x - 2} \right)\left( {{x^2} - x + 3x - 3} \right)}}{{\left( {x - 2} \right)\left( {{x^2} + 3x + 4x + 12} \right)}}\\
= \frac{{\left( {x - 2} \right)\left[ {x\left( {x - 1} \right) + 3\left( {x - 1} \right)} \right]}}{{\left( {x - 2} \right)\left[ {x\left( {x + 3} \right) + 4\left( {x + 3} \right)} \right]}}\\
= \frac{{\left( {x - 2} \right)\left( {x - 1} \right)\left( {x + 3} \right)}}{{\left( {x - 2} \right)\left( {x + 3} \right)\left( {x + 4} \right)}}\\
dk:\left( {x - 2} \right)\left( {x + 3} \right)\left( {x + 4} \right) \ne 0 \Leftrightarrow \left\{ \begin{array}{l}
x - 2 \ne 0\\
x + 3 \ne 0\\
x + 4 \ne 0
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
x \ne 2\\
x \ne - 3\\
x \ne - 4
\end{array} \right.\\
A = \frac{{x - 1}}{{x + 4}}
\end{array}\]

Vậy với \[\left\{ \begin{array}{l}
x \ne 2\\
x \ne - 3\\
x \ne - 4
\end{array} \right.\] thì \[A = \frac{{x - 1}}{{x + 4}}\]

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Answer 2

\[\begin{array}{l}
Dat:A = \frac{{{x^3} - 7x + 6}}{{{x^3} + 5{x^2} - 2x - 24}}\\
= \frac{{{x^3} - 4x - 3x + 6}}{{{x^3} - 2{x^2} + 7{x^2} - 14x + 12x - 24}}\\
= \frac{{x\left( {{x^2} - 4} \right) - 3\left( {x - 2} \right)}}{{{x^2}\left( {x - 2} \right) + 7x\left( {x - 2} \right) + 12\left( {x - 2} \right)}}\\
= \frac{{x\left( {x - 2} \right)\left( {x + 2} \right) - 3\left( {x - 2} \right)}}{{\left( {x - 2} \right)\left( {{x^2} + 7x + 12} \right)}}\\
= \frac{{\left( {x - 2} \right)\left( {{x^2} - x + 3x - 3} \right)}}{{\left( {x - 2} \right)\left( {{x^2} + 3x + 4x + 12} \right)}}\\
= \frac{{\left( {x - 2} \right)\left[ {x\left( {x - 1} \right) + 3\left( {x - 1} \right)} \right]}}{{\left( {x - 2} \right)\left[ {x\left( {x + 3} \right) + 4\left( {x + 3} \right)} \right]}}\\
= \frac{{\left( {x - 2} \right)\left( {x - 1} \right)\left( {x + 3} \right)}}{{\left( {x - 2} \right)\left( {x + 3} \right)\left( {x + 4} \right)}}\\
dk:\left( {x - 2} \right)\left( {x + 3} \right)\left( {x + 4} \right) \ne 0 \Leftrightarrow \left\{ \begin{array}{l}
x - 2 \ne 0\\
x + 3 \ne 0\\
x + 4 \ne 0
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
x \ne 2\\
x \ne - 3\\
x \ne - 4
\end{array} \right.\\
A = \frac{{x - 1}}{{x + 4}}
\end{array}\]

Vậy với \[\left\{ \begin{array}{l}
x \ne 2\\
x \ne - 3\\
x \ne - 4
\end{array} \right.\] thì \[A = \frac{{x - 1}}{{x + 4}}\]

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