\(A=\frac{a}{a-1}-\frac{a}{a+1}+a^2-1\left(đk:a\ne\pm1\right)\)
\(=\frac{a\left(a+1\right)}{a^2-1}-\frac{a\left(a-1\right)}{a^2-1}+a^2-1\)
\(=\frac{a^2+a-a^2+a}{a^2-1}+a^2-1\)
\(=\frac{2a}{a^2-1}+a^2-1\)
Bài làm:
a) đkxđ: \(\hept{\begin{cases}a-1\ne0\\a+1\ne0\\a^2-1\ne0\end{cases}}\Rightarrow\hept{\begin{cases}a\ne1\\a\ne-1\end{cases}}\)
b) Sửa đề:
\(A=\frac{a}{a-1}-\frac{a}{a+1}+\frac{2}{a^2-1}\)
\(A=\frac{a}{a-1}-\frac{a}{a+1}+\frac{2}{\left(a-1\right)\left(a+1\right)}\)
\(A=\frac{a\left(a+1\right)-a\left(a-1\right)+2}{\left(a-1\right)\left(a+1\right)}\)
\(A=\frac{a^2+a-a^2+a+2}{\left(a-1\right)\left(a+1\right)}\)
\(A=\frac{2a+2}{\left(a-1\right)\left(a+1\right)}=\frac{2\left(a+1\right)}{\left(a-1\right)\left(a+1\right)}\)
\(A=\frac{2}{a-1}\)
=> đpcm
c) \(A\inℤ\Rightarrow\frac{2}{a-1}\inℤ\Rightarrow\left(a-1\right)\inƯ\left(2\right)=\left\{\pm1;\pm2\right\}\)
\(\Rightarrow a\in\left\{-1;0;2;3\right\}\)
Mà \(a\ne-1\left(đkxd\right)\Rightarrow a\in\left\{0;2;3\right\}\)
d) Ta có: \(A\ge1\)
\(\Leftrightarrow\frac{2}{a-1}-1\ge0\)
\(\Leftrightarrow\frac{3-a}{a-1}\ge0\)
+ Nếu: \(\hept{\begin{cases}3-a\ge0\\a-1>0\end{cases}}\Rightarrow\hept{\begin{cases}3\ge a\\a>1\end{cases}}\Rightarrow1< a\le3\)
+ Nếu: \(\hept{\begin{cases}3-a\le0\\a-1< 0\end{cases}}\Rightarrow\hept{\begin{cases}a\ge3\\a< 1\end{cases}}\) (vô lý)
Vậy khi \(1< a\le3\) thì \(A\ge1\)
