Toán Lớp 8: Cho A= ( x/x+3 – 1/1+3 + x^2-1/9 -x^2 ) :2/ x+3
a) tìm ĐKXĐ , rút gọn A
b) tìm x thuộc z để nhận gtri nguyên
giải hộ e vs ạ
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Toán Lớp 8: Cho A= ( x/x+3 – 1/1+3 + x^2-1/9 -x^2 ) :2/ x+3
a) tìm ĐKXĐ , rút gọn A
b) tìm x thuộc z để nhận gtri nguyên
giải hộ e vs ạ
Comments ( 2 )
x = 8\\
x = – 2\\
x = 4\\
x = 2
\end{array} \right.\)
a)DK:x \ne \pm 3\\
A = \left( {\dfrac{x}{{x – 3}} – \dfrac{1}{{x + 3}} + \dfrac{{{x^2} – 1}}{{9 – {x^2}}}} \right):\dfrac{2}{{x + 3}}\\
= \left[ {\dfrac{{x\left( {x + 3} \right) – x + 3 – {x^2} + 1}}{{\left( {x – 3} \right)\left( {x + 3} \right)}}} \right].\dfrac{{x + 3}}{2}\\
= \dfrac{{{x^2} + 3x – x – {x^2} + 4}}{{\left( {x – 3} \right)\left( {x + 3} \right)}}.\dfrac{{x + 3}}{2}\\
= \dfrac{{2x + 4}}{{\left( {x – 3} \right)\left( {x + 3} \right)}}.\dfrac{{x + 3}}{2}\\
= \dfrac{{x + 2}}{{x – 3}}\\
b)A = \dfrac{{x + 2}}{{x – 3}} = \dfrac{{x – 3 + 5}}{{x – 3}}\\
= 1 + \dfrac{5}{{x – 3}}\\
A \in Z \to \dfrac{5}{{x – 3}} \in Z\\
\to x – 3 \in U\left( 5 \right)\\
\to \left[ \begin{array}{l}
x – 3 = 5\\
x – 3 = – 5\\
x – 3 = 1\\
x – 3 = – 1
\end{array} \right. \to \left[ \begin{array}{l}
x = 8\\
x = – 2\\
x = 4\\
x = 2
\end{array} \right.
\end{array}\)