a) ĐKXĐ: \(x\ne\pm1\)
b) \(A=\dfrac{x^3-1}{x^2-1}\cdot\left(\dfrac{1}{x-1}-\dfrac{x+1}{x^2+x+1}\right)\left(dkxd:x\ne\pm1\right)\)
\(=\dfrac{\left(x-1\right)\left(x^2+x+1\right)}{\left(x-1\right)\left(x+1\right)}\cdot\left[\dfrac{x^2+x+1}{\left(x-1\right)\left(x^2+x+1\right)}-\dfrac{\left(x-1\right)\left(x+1\right)}{\left(x-1\right)\left(x^2+x+1\right)}\right]\)
\(=\dfrac{x^2+x+1}{x+1}\cdot\dfrac{x^2+x+1-\left(x^2-1\right)}{\left(x-1\right)\left(x^2+x+1\right)}\)
\(=\dfrac{x^2+x+1-x^2+1}{\left(x-1\right)\left(x+1\right)}\)
\(=\dfrac{x+2}{x^2-1}\)
c) Có: \(\left|x+3\right|=1\Leftrightarrow\left[{}\begin{matrix}x+3=1\\x+3=-1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-2\\x=-4\end{matrix}\right.\left(tmdk\right)\)
+) Với \(x=-2\), thay vào \(A\), ta được:
\(A=\dfrac{-2+2}{\left(-2\right)^2-1}=0\)
+) Với \(x=-4\), thay vào \(A\), ta được:
\(A=\dfrac{-4+2}{\left(-4\right)^2-1}=-\dfrac{2}{15}\)
\(\text{#}Toru\)
