$\frac{x}{2(x-3)}$ +$\frac{x}{2(x+1)}$=$\frac{2x}{(x-3)(x+1)}$ giúp vs ah!!! 08/10/2021 Bởi Natalia $\frac{x}{2(x-3)}$ +$\frac{x}{2(x+1)}$=$\frac{2x}{(x-3)(x+1)}$ giúp vs ah!!!
Đáp án: $\frac{x}{2(x-3)}$ + $\frac{x}{2(x+1)}$ = $\frac{2x}{((x-3)(x+1)}$ ĐKXĐ : x ≠ 3, x ≠ -1 ⇒ $\frac{x(x+1)}{2(x-3)(x+1)}$ + $\frac{x(x-3)}{2(x-3)(x+1)}$ = $\frac{4x}{2(x-3)(x+1)}$ ⇒ x ( x + 1 ) + x ( x – 3 ) = 4x ⇔ x² + x + x² – 3x – 4x = 0 ⇔ x ( x + 1 + x -3 – 4 ) = 0 ⇔ x ( 2x – 6 ) = 0 ⇔ \(\left[ \begin{array}{l}x=0\\2x-6 =0\end{array} \right.\) ⇔ \(\left[ \begin{array}{l}x=0\\x=3\end{array} \right.\) Bình luận
Đáp án:
$\frac{x}{2(x-3)}$ + $\frac{x}{2(x+1)}$ = $\frac{2x}{((x-3)(x+1)}$
ĐKXĐ : x ≠ 3, x ≠ -1
⇒ $\frac{x(x+1)}{2(x-3)(x+1)}$ + $\frac{x(x-3)}{2(x-3)(x+1)}$ = $\frac{4x}{2(x-3)(x+1)}$
⇒ x ( x + 1 ) + x ( x – 3 ) = 4x
⇔ x² + x + x² – 3x – 4x = 0
⇔ x ( x + 1 + x -3 – 4 ) = 0
⇔ x ( 2x – 6 ) = 0
⇔ \(\left[ \begin{array}{l}x=0\\2x-6 =0\end{array} \right.\)
⇔ \(\left[ \begin{array}{l}x=0\\x=3\end{array} \right.\)