Toán Lớp 8: Cho biểu thức P=(1/x-1 + x/x^3-1 * x^2+x+1/x+1) :2x+1/x^2+2x+1
1) Rút gọn P; 2) Tính P biết I x – 2 I = 3: 3) Tìm x ∈ Z, để P ∈ Z
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Toán Lớp 8: Cho biểu thức P=(1/x-1 + x/x^3-1 * x^2+x+1/x+1) :2x+1/x^2+2x+1
1) Rút gọn P; 2) Tính P biết I x – 2 I = 3: 3) Tìm x ∈ Z, để P ∈ Z
Giúp mình với!
Comments ( 1 )
1)Dkxd:x\# 1;x\# – \dfrac{1}{2};x\# – 1\\
P = \left( {\dfrac{1}{{x – 1}} + \dfrac{x}{{{x^3} – 1}}.\dfrac{{{x^2} + x + 1}}{{x + 1}}} \right):\dfrac{{2x + 1}}{{{x^2} + 2x + 1}}\\
= \left( {\dfrac{1}{{x – 1}} + \dfrac{x}{{\left( {x – 1} \right)\left( {{x^2} + x + 1} \right)}}.\dfrac{{{x^2} + x + 1}}{{x + 1}}} \right)\\
.\dfrac{{{{\left( {x + 1} \right)}^2}}}{{2x + 1}}\\
= \left( {\dfrac{1}{{x – 1}} + \dfrac{x}{{\left( {x – 1} \right)\left( {x + 1} \right)}}} \right).\dfrac{{{{\left( {x + 1} \right)}^2}}}{{2x + 1}}\\
= \dfrac{{x + 1 + x}}{{\left( {x – 1} \right)\left( {x + 1} \right)}}.\dfrac{{{{\left( {x + 1} \right)}^2}}}{{2x + 1}}\\
= \dfrac{{x + 1}}{{x – 1}}\\
2)\left| {x – 2} \right| = 3\\
\Leftrightarrow \left[ \begin{array}{l}
x – 2 = 3 \Leftrightarrow x = 5\left( {tm} \right)\\
x – 2 = – 3 \Leftrightarrow x = – 1\left( {ktm} \right)
\end{array} \right.\\
Khi:x = 5\\
\Leftrightarrow P = \dfrac{{5 + 1}}{{5 – 1}} = \dfrac{6}{4} = \dfrac{3}{2}\\
3)P = \dfrac{{x + 1}}{{x – 1}} = \dfrac{{x – 1 + 2}}{{x – 1}} = 1 + \dfrac{2}{{x – 1}}\\
P \in Z\\
\Leftrightarrow \dfrac{2}{{x – 1}} \in Z\\
\Leftrightarrow \left( {x – 1} \right) \in \left\{ { – 2; – 1;1;2} \right\}\\
\Leftrightarrow x \in \left\{ { – 1;0;2;3} \right\}\\
Do:x\# – 1\\
Vay\,x \in \left\{ {0;2;3} \right\}
\end{array}$