## giúp mình bài này với x/x+1 + x/x-3 = 3x/(x-2)(x-3) tìm ĐKXĐ, MC

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giúp mình bài này với x/x+1 + x/x-3 = 3x/(x-2)(x-3) tìm ĐKXĐ, MC

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1 giờ 2021-10-08T14:06:46+00:00 2 Answers 0 views 0

1. ĐKXĐ: $\begin{cases}x+1\ne 0\\x-3\ne 0\\x-2\ne 0\end{cases}↔\begin{cases}x\ne -1\\x\ne 3\\x\ne 2\end{cases}$

$\\$

$\dfrac{x}{x+1}+\dfrac{x}{x-3}=\dfrac{3x}{(x-2)(x-3)}$

$↔x(x-2)(x-3)+x(x+1)(x-2)=3x.(x+1)$

$↔x(x²-5x+6)+x(x²-x-2)=3x²+3x$

$↔x³-5x²+6x+x³-x²-2x-3x²-3x=0$

$↔2x³-9x²+x=0$

$↔x(2x²-9x+1)=0$

$↔2x.(x²-2.\dfrac{9}{4}.x+\dfrac{81}{4}-\dfrac{79}{4})=0$

$↔2x.[(x-\dfrac{9}{4})-\dfrac{79}{4}]=0$

$↔2x.(x+\dfrac{-9+2\sqrt{79}}{4})(x-\dfrac{9+2\sqrt{79}}{4})=0$

$↔x=0\quad or\quad x+\dfrac{-9+2\sqrt{79}}{4}=0\quad or\quad x-\dfrac{9+2\sqrt{79}}{4}=0$

$↔x=0(TM)\quad or\quad x=\dfrac{9-2\sqrt{79}}{4}(TM)\quad or\quad x=\dfrac{9+2\sqrt{79}}{4}(TM)$

2. ĐKXĐ :

$\begin{cases}\\ x+1 \ne0 \\\\ x-3 \ne0\\\\ x-2 \ne0 \\\\\end{cases}$
$\to$
$\begin{cases}\\ x \ne-1 \\\\ x\ne3\\\\ x \ne2 \\\\\end{cases}$
Vậy ĐKXĐ : $x \ne -1 ; x \ne 3 ; x \ne2$
MSC của ba phân thức là $(x+1)(x-3)(x-2)$