Toán Giải bất phương trình:(5x-3/x+2)-4=6/x+2 04/07/2021 By Skylar Giải bất phương trình:(5x-3/x+2)-4=6/x+2
$\dfrac{5x-3}{x+2}-4=\dfrac{6}{x+2}$ $↔\dfrac{5x-3}{x+2}=\dfrac{6}{x+2}+4$ $↔\dfrac{5x-3}{x+2}=\dfrac{14+4x}{x+2}$ $↔\dfrac{5x-3}{x+2}-\dfrac{14+4x}{x+2}=0$ $↔\dfrac{5x-3-(14+4x)}{x+2}=0$ $↔\dfrac{x-17}{x+2}=0$ $↔\dfrac{x}{x+2}-\dfrac{17}{x+2}=0$ $↔\dfrac{x}{x+2}=\dfrac{17}{x+2}$ $↔x(x+2)=17(x+2)$ $↔x=17$ Trả lời
Đáp án: `(5x-3)/(x+2)-4=6/(x+2)` `⇔(5x-3)/(x+2)-6/(x+2)=4` `⇔(5x-3-6)/(x+2)=4` `⇔(5x-9)/(x+2)=4` `⇔4(x+2)=5x-9` `⇔4x+8=5x-9` `⇔5x-4x=8+9` `⇔x=17` vậy `x=17` Trả lời
$\dfrac{5x-3}{x+2}-4=\dfrac{6}{x+2}$
$↔\dfrac{5x-3}{x+2}=\dfrac{6}{x+2}+4$
$↔\dfrac{5x-3}{x+2}=\dfrac{14+4x}{x+2}$
$↔\dfrac{5x-3}{x+2}-\dfrac{14+4x}{x+2}=0$
$↔\dfrac{5x-3-(14+4x)}{x+2}=0$
$↔\dfrac{x-17}{x+2}=0$
$↔\dfrac{x}{x+2}-\dfrac{17}{x+2}=0$
$↔\dfrac{x}{x+2}=\dfrac{17}{x+2}$
$↔x(x+2)=17(x+2)$
$↔x=17$
Đáp án:
`(5x-3)/(x+2)-4=6/(x+2)`
`⇔(5x-3)/(x+2)-6/(x+2)=4`
`⇔(5x-3-6)/(x+2)=4`
`⇔(5x-9)/(x+2)=4`
`⇔4(x+2)=5x-9`
`⇔4x+8=5x-9`
`⇔5x-4x=8+9`
`⇔x=17`
vậy `x=17`