Toán giúp mình bài này với x/x+1 + x/x-3 = 3x/(x-2)(x-3) tìm ĐKXĐ, MC 08/10/2021 By aihong giúp mình bài này với x/x+1 + x/x-3 = 3x/(x-2)(x-3) tìm ĐKXĐ, MC
ĐKXĐ: $\begin{cases}x+1\ne 0\\x-3\ne 0\\x-2\ne 0\end{cases}↔\begin{cases}x\ne -1\\x\ne 3\\x\ne 2\end{cases}$ $\\$ $\dfrac{x}{x+1}+\dfrac{x}{x-3}=\dfrac{3x}{(x-2)(x-3)}$ $↔x(x-2)(x-3)+x(x+1)(x-2)=3x.(x+1)$ $↔x(x²-5x+6)+x(x²-x-2)=3x²+3x$ $↔x³-5x²+6x+x³-x²-2x-3x²-3x=0$ $↔2x³-9x²+x=0$ $↔x(2x²-9x+1)=0$ $↔2x.(x²-2.\dfrac{9}{4}.x+\dfrac{81}{4}-\dfrac{79}{4})=0$ $↔2x.[(x-\dfrac{9}{4})-\dfrac{79}{4}]=0$ $↔2x.(x+\dfrac{-9+2\sqrt{79}}{4})(x-\dfrac{9+2\sqrt{79}}{4})=0$ $↔x=0\quad or\quad x+\dfrac{-9+2\sqrt{79}}{4}=0\quad or\quad x-\dfrac{9+2\sqrt{79}}{4}=0$ $↔x=0(TM)\quad or\quad x=\dfrac{9-2\sqrt{79}}{4}(TM)\quad or\quad x=\dfrac{9+2\sqrt{79}}{4}(TM)$ Trả lời
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 $(x+1)(x-3)(x-2)$ Trả lời
ĐKXĐ: $\begin{cases}x+1\ne 0\\x-3\ne 0\\x-2\ne 0\end{cases}↔\begin{cases}x\ne -1\\x\ne 3\\x\ne 2\end{cases}$
$\\$
$\dfrac{x}{x+1}+\dfrac{x}{x-3}=\dfrac{3x}{(x-2)(x-3)}$
$↔x(x-2)(x-3)+x(x+1)(x-2)=3x.(x+1)$
$↔x(x²-5x+6)+x(x²-x-2)=3x²+3x$
$↔x³-5x²+6x+x³-x²-2x-3x²-3x=0$
$↔2x³-9x²+x=0$
$↔x(2x²-9x+1)=0$
$↔2x.(x²-2.\dfrac{9}{4}.x+\dfrac{81}{4}-\dfrac{79}{4})=0$
$↔2x.[(x-\dfrac{9}{4})-\dfrac{79}{4}]=0$
$↔2x.(x+\dfrac{-9+2\sqrt{79}}{4})(x-\dfrac{9+2\sqrt{79}}{4})=0$
$↔x=0\quad or\quad x+\dfrac{-9+2\sqrt{79}}{4}=0\quad or\quad x-\dfrac{9+2\sqrt{79}}{4}=0$
$↔x=0(TM)\quad or\quad x=\dfrac{9-2\sqrt{79}}{4}(TM)\quad or\quad x=\dfrac{9+2\sqrt{79}}{4}(TM)$
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 $(x+1)(x-3)(x-2)$