\(A\text{) }x^2-4x=0\)
\(\Leftrightarrow x\left(x-4\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x-4=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=4\end{matrix}\right.\)
Vậy...
\(B\text{) }\frac{5}{x-2}-\frac{3}{x+2}=\frac{3x-1}{x^2-4}\text{ }\left(ĐKXĐ:x\ne\pm2\right)\)
\(\Leftrightarrow5x+10-3x+6=3x-1\)
\(\Leftrightarrow x=17\)
Vậy...
a) \(x^2-4=0\)
\(\Leftrightarrow\left(x-2\right)\left(x+2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-2=0\\x+2=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=2\\x=-2\end{matrix}\right.\)
b)\(đk\left\{{}\begin{matrix}x-2\ne0\\x+2\ne0\\x^2-4\ne0\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x\ne2\\x\ne-2\end{matrix}\right.\)
\(\frac{5}{x-2}-\frac{3}{x+2}=\frac{3x-1}{x^2-4}\)
\(\Leftrightarrow\frac{5\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}-\frac{3\left(x-2\right)}{\left(x+2\right)\left(x-2\right)}=\frac{3x-1}{\left(x+2\right)\left(x-2\right)}\)
\(\Leftrightarrow5x+10-3x+6=3x-1\)
\(\Leftrightarrow-x=-17\)
\(\Leftrightarrow x=17\)(t/m)
