\(\dfrac{1}{x+1}=1-\dfrac{x}{4}\)
\(\Leftrightarrow\dfrac{4}{4\left(x+1\right)}=\dfrac{4\left(x+1\right)-x\left(x+1\right)}{4\left(x+1\right)}\)
\(\Leftrightarrow4=4\left(x+1\right)-x\left(x+1\right)\)
\(\Leftrightarrow4=4x+4-x^2-x\)
\(\Leftrightarrow4-4=4x-x^2-x\)
\(\Leftrightarrow0=3x-x^2\)
\(\Leftrightarrow0=x\left(3-x\right)\)
\(\begin{cases} x=0\\ 3-x=0=>3=x \end{cases}\)
Vậy S={0;3}
\(\dfrac{1}{x+1}=1-\dfrac{x}{4}\)
\(\Leftrightarrow\dfrac{4}{4\left(x+1\right)}=\dfrac{4\left(x+1\right)-x\left(x+1\right)}{4\left(x+1\right)}\)
\(\Leftrightarrow4=4\left(x+1\right)-x\left(x+1\right)\)
\(\Leftrightarrow4=4x+4-x^2-x\)
\(\Leftrightarrow4-4=-x^2+3x\)
\(\Leftrightarrow-x^2+3x=0\)
\(\Leftrightarrow-x\left(x-3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x-3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=3\end{matrix}\right.\)
