Toán Lớp 8: $\frac{x^2-6x+9}{x^-9}$:$\frac{x+1}{2x+6}$
a) Xác định điều kiện của biểu thức A
b)Rút gọn A
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Toán Lớp 8: $\frac{x^2-6x+9}{x^-9}$:$\frac{x+1}{2x+6}$
a) Xác định điều kiện của biểu thức A
b)Rút gọn A
Comments ( 1 )
a)x \ne \left\{ { – 3; – 1;3} \right\}\\
b)\dfrac{{2x – 6}}{{x + 1}}
\end{array}\)
a)DK:x \ne \left\{ { – 3; – 1;3} \right\}\\
b)A = \dfrac{{{x^2} – 6x + 9}}{{{x^2} – 9}}:\dfrac{{x + 1}}{{2x + 6}}\\
= \dfrac{{{{\left( {x – 3} \right)}^2}}}{{\left( {x – 3} \right)\left( {x + 3} \right)}}.\dfrac{{2\left( {x + 3} \right)}}{{x + 1}}\\
= \dfrac{{2\left( {x – 3} \right)}}{{x + 1}} = \dfrac{{2x – 6}}{{x + 1}}
\end{array}\)
d)A = \dfrac{{2x – 6}}{{x + 1}} = \dfrac{{2\left( {x + 1} \right) – 8}}{{x + 1}}\\
= 2 – \dfrac{8}{{x + 1}}\\
A \in Z \to \dfrac{8}{{x + 1}} \in Z\\
\to x + 1 \in U\left( 8 \right)\\
\to \left[ \begin{array}{l}
x + 1 = 8\\
x + 1 = – 8\\
x + 1 = 4\\
x + 1 = – 4\\
x + 1 = 2\\
x + 1 = – 2\\
x + 1 = 1\\
x + 1 = – 1
\end{array} \right. \to \left[ \begin{array}{l}
x = 7\\
x = – 9\\
x = 3\left( l \right)\\
x = – 5\\
x = 1\\
x = – 3\left( l \right)\\
x = 0\\
x = – 2
\end{array} \right.
\end{array}\)