Question

Rút gọn:

\(A=\dfrac{x}{x-2}+\dfrac{1}{x+2}-\dfrac{x^2-2}{x^2-4}\)

Answer 1

Lời giải:
ĐKXĐ: $x\neq \pm 2$

\(A=\frac{x(x+2)}{(x-2)(x+2)}+\frac{x-2}{(x+2)(x-2)}-\frac{x^2-2}{(x-2)(x+2)}\\ =\frac{x(x+2)+x-2-(x^2-2)}{(x-2)(x+2)}\\ =\frac{3x}{(x-2)(x+2)}=\frac{3x}{x^2-4}\)

Answer 2

\(A=\dfrac{x}{x-2}+\dfrac{1}{x+2}-\dfrac{x^2-2}{x^2-4}\left(1\right)\)

\(ĐKXĐ\Leftrightarrow\left\{{}\begin{matrix}x-2\ne0\\x+2\ne0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\ne2\\x\ne-2\end{matrix}\right.\)

\(\left(1\right)=\dfrac{x\left(x+2\right)}{\left(x+2\right)\left(x-2\right)}+\dfrac{x-2}{\left(x+2\right)\left(x-2\right)}-\dfrac{x^2-2}{\left(x+2\right)\left(x-2\right)}\)

\(=x\left(x+2\right)+x-2-x^2-2\)

\(=x^2+2x+x-2-x^2-2\)

\(=4x-4\)