c: =>x+1+2(x-1)=5
=>x+1+2x-2=5
=>3x-1=5
=>3x=6
hay x=2(nhận)
d: \(\left|x-7\right|=2x+3\)
\(\Leftrightarrow\left\{{}\begin{matrix}x>=-\dfrac{3}{2}\\\left(2x+3-x+7\right)\left(2x+3+x-7\right)=0\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}x>=-\dfrac{3}{2}\\\left(x+10\right)\left(3x-4\right)=0\end{matrix}\right.\Leftrightarrow x=\dfrac{4}{3}\)
c)\(\dfrac{1}{x-1}+\dfrac{2}{x+1}=\dfrac{5}{x^2-1}\)
\(\Leftrightarrow\cdot\dfrac{x+1+2\left(x-1\right)}{\left(x-1\right)\left(x+1\right)}=\dfrac{5}{x^2-1}\)
\(\Leftrightarrow x+1+2x-2=5\)
\(\Leftrightarrow3x-1=5\Leftrightarrow3x=6\Leftrightarrow x=2\left(tm\right)\)
b)\(\left|x-7\right|=2x+3\)
\(\left[{}\begin{matrix}x-7=2x+3\\x-7=-2x-3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}-x-10=0\\3x-4=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}-x=10\\3x=-4\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-10\\x=\dfrac{4}{3}\end{matrix}\right.\)