Rút gọn P=[2x/(2x^2-5x+3)-5/(2x-3)]:[3+2/(1-x)] 01/11/2021 Bởi Rylee Rút gọn P=[2x/(2x^2-5x+3)-5/(2x-3)]:[3+2/(1-x)]
Điều kiện xác định:$\begin{cases}2x^2-5x+3\ne0\\2x-3\ne0\\1-x\ne0\end{cases}↔\begin{cases}x\ne \dfrac32\\x\ne1\end{cases}$ `P=((2x)/(2x^2-5x+3)-5/(2x-3)):(3+2/(1-x))` `=[(2x)/((x-1)(2x-3))-(5(x-1))/((x-1)(2x-3))]:[(3(x-1))/(x-1)-2/(x-1)]` `=(2x-5x+5)/((x-1)(2x-3)):(3x-3-2)/(x-1)` `=(-3x+5)/((x-1)(2x-3)).(x-1)/(3x-5)` `=-1/(2x-3)` Bình luận
$\text{ĐKXĐ:}$ •$2x^2-5x+3 \ne 0$ $\to 2x^2-2x-3x+3\ne0$ $\to 2x(x-1)-3(x-1)\ne0$ $\to (x-1)(2x-3)\ne$ $\to x-1\ne0 \to x\ne1$ $2x-3\ne0 \to x\ne \dfrac{3}{2}$ •$2x-3\ne0 \to x\ne \dfrac{3}{2}$ •$1-x\ne \to x\ne 1$ Bình luận
Điều kiện xác định:
$\begin{cases}2x^2-5x+3\ne0\\2x-3\ne0\\1-x\ne0\end{cases}↔\begin{cases}x\ne \dfrac32\\x\ne1\end{cases}$
`P=((2x)/(2x^2-5x+3)-5/(2x-3)):(3+2/(1-x))`
`=[(2x)/((x-1)(2x-3))-(5(x-1))/((x-1)(2x-3))]:[(3(x-1))/(x-1)-2/(x-1)]`
`=(2x-5x+5)/((x-1)(2x-3)):(3x-3-2)/(x-1)`
`=(-3x+5)/((x-1)(2x-3)).(x-1)/(3x-5)`
`=-1/(2x-3)`
$\text{ĐKXĐ:}$
•$2x^2-5x+3 \ne 0$
$\to 2x^2-2x-3x+3\ne0$
$\to 2x(x-1)-3(x-1)\ne0$
$\to (x-1)(2x-3)\ne$
$\to x-1\ne0 \to x\ne1$
$2x-3\ne0 \to x\ne \dfrac{3}{2}$
•$2x-3\ne0 \to x\ne \dfrac{3}{2}$
•$1-x\ne \to x\ne 1$