Đáp án:
a.$x\ne \pm2$
B.$x=1$
b.$x>1$
Giải thích các bước giải:
a.Để $C$ xác định
$\to\begin{cases}x^2-4\ne 0\\x-2\ne 0\\x+2\ne 0\end{cases}$
$\to\begin{cases}x^2\ne 4\\x\ne 2\\x\ne -2\end{cases}$
$\to\begin{cases}x\ne \pm2\\x\ne 2\\x\ne -2\end{cases}$
$\to x\ne \pm2$
b.Ta có:
$C=\dfrac{x^3}{x^2-4}-\dfrac{x}{x-2}-\dfrac2{x+2}$
$\to C=\dfrac{x^3}{\left(x-2\right)\left(x+2\right)}-\dfrac{x\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}-\dfrac{2\left(x-2\right)}{\left(x+2\right)\left(x-2\right)}$
$\to C=\dfrac{x^3-x\left(x+2\right)-2\left(x-2\right)}{\left(x-2\right)\left(x+2\right)}$
$\to C=\dfrac{\left(x-1\right)\left(x+2\right)\left(x-2\right)}{\left(x-2\right)\left(x+2\right)}$
$\to C=x-1$
Để $C=0\to x-1=0\to x=1$
b.Để $C>0\to x-1>0\to x>1$
