Question

\(\dfrac{x+1}{x}+1=\dfrac{3x-1}{x+1}+\dfrac{1}{x\left(x+1\right)}\) giúp mik với

Answer 1

\(\dfrac{x+1}{x}+1=\dfrac{3x-1}{x+1}+\dfrac{1}{x\left(x+1\right)}\)

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

\(\Leftrightarrow2x^2+3x+1=3x^2-x+1\)

\(\Leftrightarrow2x^2+3x+1=0\)

\(\Leftrightarrow-x^2+4x=0\)

\(\Leftrightarrow x\left(-x+4\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\-x+4=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=4\end{matrix}\right.\)

Vậy \(x=4\)