Câu 1:
1: \(=\dfrac{x+1}{3\left(x-2\right)}\cdot\dfrac{3\left(x-2\right)^2}{x}=\dfrac{\left(x+1\right)\left(x-2\right)}{x}\)
2: \(\Leftrightarrow x^3+8-x^3+3x=14\)
=>3x=6
hay x=2
câu 2:
a, ĐKXĐ:\(\left\{{}\begin{matrix}2x+4\ne0\\x^2-4\ne0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\ne-2\\x\ne\pm2\end{matrix}\right.\Leftrightarrow x\ne\pm2\)
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 \(\dfrac{x+2}{2\left(x-2\right)}=0\Leftrightarrow x+2=0\Leftrightarrow x=-2\left(ktm\right)\)
