a) \(\dfrac{x+1}{32}=\dfrac{2}{x+1}\)

\(\Leftrightarrow\dfrac{x+1}{32}=\dfrac{2}{x+1}\left(đk:xe1\right)\)

\(\Leftrightarrow\left(x+1\right)\left(x+1\right)=64\)

\(\Leftrightarrow\left(x+1\right)^2-64=0\)

\(\Leftrightarrow x^2+2x+1-64=0\)

\(\Leftrightarrow x^2+6x-63=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{-2+16}{2}\\x=\dfrac{-2-16}{2}\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=7\\x=-9\end{matrix}\right.\left(đk:xe-1\right)\)

\(\Leftrightarrow\left[{}\begin{matrix}x=7\\x=-9\end{matrix}\right.\)

Vậy \(x_1=-9;x_2=7\)

b) \(\dfrac{x+1}{5}=\dfrac{7}{x-1}\)

\(\Leftrightarrow\dfrac{x+1}{5}=\dfrac{7}{x-1}\left(đk:xe1\right)\)

\(\Leftrightarrow\left(x+1\right)\left(x-1\right)=35\)

\(\Leftrightarrow\left(x+1\right)\left(x-1\right)-35=0\)

\(\Leftrightarrow x^2-1-35=0\)

\(\Leftrightarrow x^2-36=0\)

\(\Leftrightarrow x^2=36\)

\(\Leftrightarrow\left[{}\begin{matrix}x=6\\x=-6\end{matrix}\right.\left(đk:xe1\right)\)

\(\Leftrightarrow\left[{}\begin{matrix}x=6\\x=-6\end{matrix}\right.\)

Vậy \(x_1=-6;x_2=6\)

c) \(\left|4,5-2x\right|:1\dfrac{7}{4}=\dfrac{11}{14}\)

\(\Leftrightarrow\left|4,5-2x\right|:\dfrac{11}{4}=\dfrac{11}{4}\)

\(\Leftrightarrow\left|4,5-2x\right|\cdot\dfrac{4}{11}=\dfrac{11}{14}\)

\(\Leftrightarrow\dfrac{4}{11}\cdot\left|4,5-2x\right|=\dfrac{11}{14}\)

\(\Leftrightarrow\left|4,5-2x\right|=\dfrac{121}{56}\)

\(\Leftrightarrow\left[{}\begin{matrix}4,5-2x=\dfrac{121}{56}\\4,5-2x=-\dfrac{121}{56}\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{131}{112}\\x=\dfrac{373}{112}\end{matrix}\right.\)

Vậy \(x_1=\dfrac{131}{112};x_2=\dfrac{373}{112}\)