Question

rút gọn \(\dfrac{1}{x}\)-\(\dfrac{x}{2x+1}\)+\(\dfrac{2x^2-3x-1}{x\left(2x+1\right)}\)

Answer 1

=\(\dfrac{2x+1}{x\left(2x+1\right)}\)-\(\dfrac{x^2}{x\left(2x+1\right)}\)+\(\dfrac{2x^2-3x-1}{x\left(2x+1\right)}\)

= 2x+1 - x² + 2x²-3x-1

= -x + x²

Answer 2

ĐKXĐ : x ≠ 0 ; x ≠ \(-\dfrac{1}{2}\)

\(\Rightarrow2x+1-x^2=2x^2-3x-1\)

\(\Leftrightarrow-3x^2+5x+2=0\)

\(\Leftrightarrow-3x^2-x+6x+2=0\)

\(\Leftrightarrow-x\left(3x+1\right)+2\left(3x+1\right)=0\)

\(\Leftrightarrow\left(3x+1\right)\left(-x+2\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{1}{3}\\x=2\end{matrix}\right.\) (N)