câu 2:
a, ĐKXĐ:\(\left\{{}\begin{matrix}2x+4\ne0\\x^2-4\ne0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}2x\ne-4\\x^2\ne4\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\ne-2\\x\ne\pm2\end{matrix}\right.\Leftrightarrow x\ne\pm2\)
b, \(A=\dfrac{x}{2x+4}+\dfrac{3x+2}{x^2-4}\)
\(\Leftrightarrow A=\dfrac{x\left(x-2\right)}{2\left(x+2\right)\left(x-2\right)}+\dfrac{2\left(3x+2\right)}{2\left(x+2\right)\left(x-2\right)}\)
\(\Leftrightarrow A=\dfrac{x^2-2x}{2\left(x+2\right)\left(x-2\right)}+\dfrac{6x+4}{2\left(x+2\right)\left(x-2\right)}\)
\(\Leftrightarrow A=\dfrac{x^2-2x+6x+4}{2\left(x+2\right)\left(x-2\right)}\)
\(\Leftrightarrow A=\dfrac{x^2+4x+4}{2\left(x+2\right)\left(x-2\right)}\)
\(\Leftrightarrow A=\dfrac{\left(x+2\right)^2}{2\left(x+2\right)\left(x-2\right)}\)
\(\Leftrightarrow A=\dfrac{x+2}{2\left(x-2\right)}\)
c, Để A=0 thì \(\dfrac{x+2}{2\left(x-2\right)}=0\Leftrightarrow x+2=0\Leftrightarrow x=-2\left(ktm\right)\)
Vậy không có x thỏa mãn đề bài
a: ĐKXĐ: \(x\notin\left\{2;-2\right\}\)
b: \(A=\dfrac{x}{2\left(x+2\right)}+\dfrac{3x+2}{\left(x-2\right)\left(x+2\right)}=\dfrac{x^2-2x+6x+4}{2\left(x+2\right)\left(x-2\right)}=\dfrac{x^2+4x+4}{2\left(x+2\right)\left(x-2\right)}=\dfrac{x+2}{2x-4}\)
c: Để A=0 thì x+2=0
hay x=-2(loại)
