a) \(x^2+6x+9=49\)
\(\Leftrightarrow\left(x+3\right)^2=49\)
\(\Leftrightarrow\left[{}\begin{matrix}x+3=7\\x+3=-7\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=4\\x=-10\end{matrix}\right.\)
c) \(\dfrac{1}{x}+\dfrac{2}{x-2}=0\left(đk:x\ne0,x\ne2\right)\)
\(\Leftrightarrow\dfrac{x-2+2x}{x\left(x-2\right)}=0\)
\(\Leftrightarrow3x-2=0\Leftrightarrow3x=2\Leftrightarrow x=\dfrac{2}{3}\left(tm\right)\)
d) \(\dfrac{x-1}{x+2}-\dfrac{x}{x-2}=\dfrac{5x-2}{4-x^2}\left(đk:x\ne2,x\ne-2\right)\)
\(\Leftrightarrow\dfrac{\left(x-1\right)\left(x-2\right)-x\left(x+2\right)}{x^2-4}=\dfrac{2-5x}{x^2-4}\)
\(\Leftrightarrow x^2-3x+2-x^2-2x=2-5x\)
\(\Leftrightarrow0=0\left(đúng\right)\)
Vậy \(x\in R,x\ne2,x\ne-2\)
e) \(\left|2x-5\right|=4\)
\(\Leftrightarrow\left[{}\begin{matrix}2x-5=4\\2x-5=-4\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{9}{2}\\x=\dfrac{1}{2}\end{matrix}\right.\)
f) \(\left|2x-6\right|=x+1\)
\(\Leftrightarrow\left[{}\begin{matrix}2x-6=x+1\\2x-6=-x-1\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=7\\x=\dfrac{5}{3}\end{matrix}\right.\)
