\(\text{a) }P=\dfrac{2a^2}{a^2-1}+\dfrac{a}{a+1}-\dfrac{a}{a-1}\\ \Rightarrow P=\dfrac{2a^2}{\left(a+1\right)\left(a-1\right)}+\dfrac{a}{a+1}-\dfrac{a}{a-1}\\ \Rightarrow P=\dfrac{2a^2}{\left(a+1\right)\left(a-1\right)}+\dfrac{a\left(a-1\right)}{\left(a+1\right)\left(a-1\right)}-\dfrac{a\left(a+1\right)}{\left(a-1\right)\left(a+1\right)}\\ \Rightarrow P=\dfrac{2a^2+a\left(a-1\right)-a\left(a+1\right)}{\left(a-1\right)\left(a+1\right)}\)
\(\Rightarrow\) Để \(P\) có nghĩa
\(\text{thì : }\Rightarrow\left(a-1\right)\left(a+1\right)\ne0\\ \Leftrightarrow\left\{{}\begin{matrix}a-1\ne0\\a+1\ne0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a\ne1\\a\ne-1\end{matrix}\right.\)
\(\text{ Ta có : }P=\dfrac{2a^2+a\left(a-1\right)-a\left(a+1\right)}{\left(a-1\right)\left(a+1\right)}\\ \Rightarrow P=\dfrac{a\left[2a+\left(a-1\right)-\left(a+1\right)\right]}{\left(a-1\right)\left(a+1\right)}\\ \Rightarrow P=\dfrac{a\left(2a+a-1-a-1\right)}{\left(a-1\right)\left(a+1\right)}\\ \Rightarrow P=\dfrac{a\left(2a-2\right)}{\left(a-1\right)\left(a+1\right)}\\ \Rightarrow P=\dfrac{2a\left(a-1\right)}{\left(a-1\right)\left(a+1\right)}\\ \Rightarrow P=\dfrac{2a}{a-1}\)
\(\text{c) Ta lại có : }P=\dfrac{2a}{a-1}\\ \Rightarrow P=\dfrac{2a-2+2}{a-1}\\\Rightarrow P=\dfrac{2\left(a-1\right)+2}{a-1}\\\Rightarrow P=\dfrac{2\left(a-1\right)}{a-1}+\dfrac{2}{a-1}\\\Rightarrow P=2+\dfrac{2}{a-1}\)
\(\Rightarrow\) Để \(P\in Z\)
\(\text{thì: }\Rightarrow\dfrac{2}{a-1}\in Z\\ \Rightarrow2⋮a-1\\ \Rightarrow a-1\inƯ_{\left(2\right)}\)
Mà \(Ư_{\left(2\right)}=\left\{\pm1;\pm2\right\}\)
Ta được bảng giá trị:
|
\(a-1\) |
\(-1\) |
\(-2\) |
\(1\) |
\(2\) |
|
\(a\) |
\(0\) |
\(-1\) |
\(2\) |
\(3\) |
Vậy a) Để \(P\) có nghĩa thì \(\left\{{}\begin{matrix}a\ne1\\a\ne-1\end{matrix}\right.\)
b) \(P=\dfrac{2a}{a-1}\)
c) để \(P\in Z\) thì \(a=\left\{0;-1;2;3\right\}\)
a) \(P\) có nghĩa \(\Leftrightarrow \left\{\begin{matrix} a+1\ne 0 \\ a-1\ne 0 \end{matrix} \right.\Leftrightarrow \left\{\begin{matric} a\ne -1 \\ a\ne 1 \end{matrix} \right.\)
b) \(P=\dfrac{2a^2}{(a-1)(a+1)}+\dfrac a{a+1}-\dfrac a{a-1}\)
\(=\dfrac{2a^2+a(a-1)-a(a+1)}{(a-1)(a+1)}\)
\(=\dfrac{2a^2+a^2-a-a^2-a}{(a-1)(a+1)}\)
\(=\dfrac{2a^2-2a}{(a-1)(a+1)}\)
\(=\dfrac{2a(a-1)}{(a-1)(a+1)}\)
\(=\dfrac{2a}{a+1}\)
c) Ta có: \(P=\dfrac{2a}{a+1}=\dfrac{2(a+1)-2}{a+1}=2-\dfrac 2{a+1}\)
\(P\in \mathbb{Z}\Leftrightarrow \dfrac 2{a+1}\in \mathbb{Z}\Leftrightarrow a+1\in Ư(2)=\left\{ \pm 1;\pm 2 \right\}\Leftrightarrow x\in \left\{ -3;-2;0;1 \right\}\)
Kết hợp với ĐKXĐ \(\Rightarrow a\in \left\{-3;-2;0 \right\}\)
a) P có nghĩa \(\Leftrightarrow\left\{{}\begin{matrix}a+1\ne0\\a-1\ne0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a\ne-1\\a\ne1\end{matrix}\right.\)
