tìm x a) 1 1/5 – 8 /5x = -3 2/3 b) | 2x – 1 | – 6/7 = 1/7 16/08/2021 Bởi Rylee tìm x a) 1 1/5 – 8 /5x = -3 2/3 b) | 2x – 1 | – 6/7 = 1/7
`1 1/5 – 8 /5x = -3 2/3` `⇔6/5 – 8 /5x =-(11)/3` `⇔8 /5x=(73)/(15)` `⇔x=(73)/(14)` `| 2x – 1 | – 6/7 = 1/7` `⇔| 2x – 1 | =1` `⇔`\(\left[ \begin{array}{l}2x-1=1\\2x-1=-1\end{array} \right.\) `⇔`\(\left[ \begin{array}{l}x=1\\x=0\end{array} \right.\) Bình luận
Đáp án: $a/$ `1 1/5 – 8/5x = -3 2/3` `⇔ 6/5 – 8/5x = (-11)/3` `⇔ 8/5x = 6/5 – (-11)/3` `⇔ 8/5x = 73/15` `⇔ x = 73/15 : 8/5` `⇔ x = 73/24` `text{Vậy x=}` `73/24` $b/$ `|2x – 1| – 6/7 = 1/7` `⇔ |2x – 1| = 1/7 + 6/7` `⇔ |2x – 1| = 1` `⇔ 2x – 1 = 1` `text{hoặc}` `2x – 1= -1` `⇔ 2x = 2` `text{hoặc}` `2x = 0` `⇔ x = 1` `text{hoặc}` `x = 0` `text{Vậy x = 1, x = 0}` Bình luận
`1 1/5 – 8 /5x = -3 2/3`
`⇔6/5 – 8 /5x =-(11)/3`
`⇔8 /5x=(73)/(15)`
`⇔x=(73)/(14)`
`| 2x – 1 | – 6/7 = 1/7`
`⇔| 2x – 1 | =1`
`⇔`\(\left[ \begin{array}{l}2x-1=1\\2x-1=-1\end{array} \right.\)
`⇔`\(\left[ \begin{array}{l}x=1\\x=0\end{array} \right.\)
Đáp án:
$a/$ `1 1/5 – 8/5x = -3 2/3`
`⇔ 6/5 – 8/5x = (-11)/3`
`⇔ 8/5x = 6/5 – (-11)/3`
`⇔ 8/5x = 73/15`
`⇔ x = 73/15 : 8/5`
`⇔ x = 73/24`
`text{Vậy x=}` `73/24`
$b/$ `|2x – 1| – 6/7 = 1/7`
`⇔ |2x – 1| = 1/7 + 6/7`
`⇔ |2x – 1| = 1`
`⇔ 2x – 1 = 1` `text{hoặc}` `2x – 1= -1`
`⇔ 2x = 2` `text{hoặc}` `2x = 0`
`⇔ x = 1` `text{hoặc}` `x = 0`
`text{Vậy x = 1, x = 0}`