Toán Lớp 9: P= (x+3/x^2-1 – 3/x+1) :(1-2/x-1) d) rút gọn biểu thức P e) tìm x để P<0 f) tìm x là số nguyên để Q=x.p nhận giá trị nguyên
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Toán Lớp 9: P= (x+3/x^2-1 – 3/x+1) :(1-2/x-1) d) rút gọn biểu thức P e) tìm x để P<0 f) tìm x là số nguyên để Q=x.p nhận giá trị nguyên
Comments ( 1 )
d) – \dfrac{2}{{x + 1}}\\
e)x > – 1\\
f)\left[ \begin{array}{l}
x = – 3\\
x = 0\\
x = – 2
\end{array} \right.
\end{array}\)
d)DK:x \ne \left\{ { – 1;1;3} \right\}\\
P = \dfrac{{x + 3 – 3\left( {x – 1} \right)}}{{\left( {x – 1} \right)\left( {x + 1} \right)}}:\dfrac{{x – 1 – 2}}{{x – 1}}\\
= \dfrac{{ – 2x + 6}}{{\left( {x – 1} \right)\left( {x + 1} \right)}}:\dfrac{{x – 3}}{{x – 1}}\\
= \dfrac{{ – 2\left( {x – 3} \right)}}{{\left( {x – 1} \right)\left( {x + 1} \right)}}.\dfrac{{x – 1}}{{x – 3}}\\
= – \dfrac{2}{{x + 1}}\\
e)P < 0\\
\to – \dfrac{2}{{x + 1}} < 0\\
\to x + 1 > 0\\
\to x > – 1\\
f)Q = x.P = x. – \dfrac{2}{{x + 1}}\\
= – \dfrac{{2x}}{{x + 1}} = – \dfrac{{2\left( {x + 1} \right) – 2}}{{x + 1}}\\
= – 2 + \dfrac{2}{{x + 1}}\\
Q \in Z\\
\to \dfrac{2}{{x + 1}} \in Z\\
\to x + 1 \in U\left( 2 \right)\\
\to \left[ \begin{array}{l}
x + 1 = 2\\
x + 1 = – 2\\
x + 1 = 1\\
x + 1 = – 1
\end{array} \right. \to \left[ \begin{array}{l}
x = 1\left( l \right)\\
x = – 3\\
x = 0\\
x = – 2
\end{array} \right.
\end{array}\)