$\frac{x}{x+1}$ + $\frac{x}{x-3}$ = $\frac{4x}{(x+1)(x-3)}$ 21/11/2021 Bởi Arianna $\frac{x}{x+1}$ + $\frac{x}{x-3}$ = $\frac{4x}{(x+1)(x-3)}$
Giải thích các bước giải: $\frac{x}{x+1}$ + $\frac{x}{x-3}$ = $\frac{4x}{(x+1)(x-3)}$ ( ĐK: x $\neq$ 3;-1) ⇔ $\frac{x²-3x}{(x+1)(x-3)}$ + $\frac{x²+x}{(x+1)(x−3)}$ = $\frac{4x}{(x+1)(x−3)}$ ⇒ x²-3x+x²+x=4x ⇔ 2x² -6x=0 ⇔ 2x(x-3)=0 ⇔ \(\left[ \begin{array}{l}2x=0\\x-3=0\end{array} \right.\) ⇔ \(\left[ \begin{array}{l}x=0\\x=3( loại)\end{array} \right.\) Vậy S={0} Bình luận
$\frac{x}{x+1}$ +$\frac{x}{x-3}$ =$\frac{4x}{(x+1)(x-3)}$ (=)$\frac{x}{x+1}$ +$\frac{x}{x-3}$ -$\frac{4x}{(x+1)(x-3)}$=0 (=)$\frac{x(x-3)+x(x+1)-4x}{(x-3)(x+1)}$ =0 (=)x²-3x+x²+x-4x=0 (=)2x²-6x=0 (=)2x(x-3)=0 (=)\(\left[ \begin{array}{l}2x=0\\x-3=0\end{array} \right.\) (=)\(\left[ \begin{array}{l}x=0(TMĐK)\\x=3(KTMĐK)\end{array} \right.\) S={0} Bình luận
Giải thích các bước giải:
$\frac{x}{x+1}$ + $\frac{x}{x-3}$ = $\frac{4x}{(x+1)(x-3)}$ ( ĐK: x $\neq$ 3;-1)
⇔ $\frac{x²-3x}{(x+1)(x-3)}$ + $\frac{x²+x}{(x+1)(x−3)}$ = $\frac{4x}{(x+1)(x−3)}$
⇒ x²-3x+x²+x=4x
⇔ 2x² -6x=0
⇔ 2x(x-3)=0
⇔ \(\left[ \begin{array}{l}2x=0\\x-3=0\end{array} \right.\)
⇔ \(\left[ \begin{array}{l}x=0\\x=3( loại)\end{array} \right.\)
Vậy S={0}
$\frac{x}{x+1}$ +$\frac{x}{x-3}$ =$\frac{4x}{(x+1)(x-3)}$
(=)$\frac{x}{x+1}$ +$\frac{x}{x-3}$ -$\frac{4x}{(x+1)(x-3)}$=0
(=)$\frac{x(x-3)+x(x+1)-4x}{(x-3)(x+1)}$ =0
(=)x²-3x+x²+x-4x=0
(=)2x²-6x=0
(=)2x(x-3)=0
(=)\(\left[ \begin{array}{l}2x=0\\x-3=0\end{array} \right.\) (=)\(\left[ \begin{array}{l}x=0(TMĐK)\\x=3(KTMĐK)\end{array} \right.\)
S={0}