Question

Giúp mik bài này với ạ:

\(\dfrac{x+1}{x-2}\) = \(\dfrac{2}{x^2-4}\)

Answer 1

ĐKXĐ: \(x\ne\pm2\)

\(\dfrac{x+1}{x-2}=\dfrac{2}{x^2-4}\)

\(\Rightarrow\dfrac{\left(x+1\right)\left(x+2\right)}{x^2-4}=\dfrac{2}{x^2-4}\)

\(\Rightarrow\left(x+1\right)\left(x+2\right)=2\)

\(\Leftrightarrow x^2+3x+2=2\)

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

\(\Leftrightarrow x\left(x+3\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-3\end{matrix}\right.\) (thỏa mãn)

Answer 2

đkxđ: \(x ≠2; x ≠-2\)

\(\dfrac{x+1}{x-2}=\dfrac{2}{x^2-4}\)

\(⇔\dfrac{(x+1)(x+2)}{x^2-4}=\dfrac{2}{x^2-4}\)

\(⇔(x+1)(x+2)=2\)

\(⇔x^2+3x=0\)

\(⇔x(x+3)=0\)

\(⇔\left[\begin{array}{} x=0\\ x+3=0 \end{array} \right.\)

\(⇔\left[\begin{array}{} x=0\\ x=-3 \end{array} \right.\)