Ta có:

\(2x+33=\left(-11\right)-x\)

\(\Leftrightarrow2x+33=\left(-11\right)+\left(-x\right)\)

\(\Leftrightarrow2x+33=-\left(11+x\right)\)

\(\Leftrightarrow2x=-\left(11+x\right)-33\)

\(\Leftrightarrow2x=-\left(11+x\right)+\left(-33\right)\)

\(\Leftrightarrow2x=-\left(11+33+x\right)\)

\(\Leftrightarrow2x=-\left(44+x\right)\)

\(\Leftrightarrow2x=2\cdot\frac{-\left(44+x\right)}{2}\)

\(\Leftrightarrow2x=2\cdot\left[\left(-44\right):2\right]+\left[\left(-x\right):2\right]\)

\(\Leftrightarrow2x=2\cdot\left(-22\right)+\frac{-x}{2}\)

\(\Leftrightarrow x=\left(-22\right)+\frac{-x}{2}\)

\(\Leftrightarrow\left(-22\right)+\frac{-x}{2}=x\)

\(\Leftrightarrow\frac{-x}{2}=x-22\)

\(\Leftrightarrow\frac{-x}{2}=-22\)

\(\Leftrightarrow-x=\left(-22\right):2\)

\(\Leftrightarrow-x=-11\)

\(\Leftrightarrow x=11\)

\(\left(2x+1\right)^2-9=0\)

\(\Leftrightarrow\left(2x+1\right)^2=9\)

\(\Leftrightarrow\left(2x+1\right)^2=3^2\)

\(\Leftrightarrow\left(2x+1\right)=3\)

\(\Leftrightarrow2x=3-1\)

\(\Leftrightarrow2x=2\)

\(\Leftrightarrow x=1\)

Vậy x = 1

Ta có:

\(\left(2x+1\right)^2-9=0\)

\(\Leftrightarrow\left(2x+1\right)^2=0+9\)

\(\Leftrightarrow\left(2x+1\right)^2=9\)

\(\Leftrightarrow\hept{\begin{cases}\left(2x+1\right)^2=3^2\\\left(2x+1\right)^2=-3^2\end{cases}}\)

\(\Leftrightarrow\hept{\begin{cases}2x+1=3\\2x+1=-3\end{cases}}\)

\(\Leftrightarrow\hept{\begin{cases}2x=3-1\\2x=\left(-3\right)-1\end{cases}}\)

\(\Leftrightarrow\hept{\begin{cases}2x=2\\2x=-4\end{cases}}\)

\(\Leftrightarrow\hept{\begin{cases}x=2:2\\x=\left(-4\right):2\end{cases}}\)

\(\Leftrightarrow\hept{\begin{cases}x=1\\x=-2\end{cases}}\)