Toán Lớp 8: M=2.(1-9x^2)/3x^2+6x:2-6x/3x
a)rút gọn M
b) tìm x đẻ M là giá trị nguyên
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Toán Lớp 8: M=2.(1-9x^2)/3x^2+6x:2-6x/3x
a)rút gọn M
b) tìm x đẻ M là giá trị nguyên
Comments ( 1 )
a)\dfrac{{1 + 3x}}{{x + 2}}\\
b)\left[ \begin{array}{l}
x = 3\\
x = – 7\\
x = – 1\\
x = – 3
\end{array} \right.
\end{array}\)
a)DK:x \ne \left\{ { – 2;0;\dfrac{1}{3}} \right\}\\
M = \dfrac{{2\left( {1 – 9{x^2}} \right)}}{{3{x^2} + 6x}}:\dfrac{{2 – 6x}}{{3x}}\\
= \dfrac{{2\left( {1 – 3x} \right)\left( {1 + 3x} \right)}}{{3x\left( {x + 2} \right)}}.\dfrac{{3x}}{{2\left( {1 – 3x} \right)}}\\
= \dfrac{{1 + 3x}}{{x + 2}}\\
b)M = \dfrac{{3x + 1}}{{x + 2}} = \dfrac{{3\left( {x + 2} \right) – 5}}{{x + 2}}\\
= 3 – \dfrac{5}{{x + 2}}\\
M \in Z \to \dfrac{5}{{x + 2}} \in Z\\
\to x + 2 \in U\left( 5 \right)\\
\to \left[ \begin{array}{l}
x + 2 = 5\\
x + 2 = – 5\\
x + 2 = 1\\
x + 2 = – 1
\end{array} \right.\\
\to \left[ \begin{array}{l}
x = 3\\
x = – 7\\
x = – 1\\
x = – 3
\end{array} \right.
\end{array}\)