Bài 1: Tìm x, biết: a) /2x – 3/ = 6 b) 2 . /3x + 1/ = 5 c) 7,5 – 3 /5 – 2x/ = -4,5 d) /3x -1/ = /x + 3/ 23/07/2021 Bởi Kaylee Bài 1: Tìm x, biết: a) /2x – 3/ = 6 b) 2 . /3x + 1/ = 5 c) 7,5 – 3 /5 – 2x/ = -4,5 d) /3x -1/ = /x + 3/
Bài 1 : a) `|2x-3|` = 6 ⇒ `2x` – `3` = `6` `2x` – `3` = `-6` ⇒ `2x` = `6` + `3` `2x` = `-6` + `3` ⇒ `2x` = `9` `2x` = `-3` ⇒ `x` = `9/2` `x` = `-3/2` b) `2.|3x+1|` = `5` ⇒ `|3x+1|` = `5` : `2` = `5/2` ⇒ `3x` + `1` = `5/2` `3x` + `1` = `-5/2` ⇒ `3x` = `5/2` – `1` = `3/2` `3x` = `-5/2` – `1` = `7/2` ⇒ `x` = `3/2` : `3` = `1/2` `x` = `-7/2` : `3` = `-7/6` c) `7,5` – `3|5-2x|` = `-4,5` ⇒ `3|5-2x|` = `7,5` – `(-4,5)` = `12` ⇒ `|5-2x|` = `12` : `3` = `4` ⇒ `5` – `2x` = `4` `5` – `2x` = `-4` ⇒ `2x` = `5` – `4` = `1` `2x` = `5` – `(-4)` = `9` ⇒ `x` = `1/2` `x` = `9/2` d) `|3x-1|` = `|x+3|` ⇒ `3x` – `1` = `x` + `3` `3x` – `1` = `-x` – `3` ⇒ `2x` = `4` `4x` = `-2` ⇒ `x` = `2` `x` = `-1/2` Bình luận
a) |2x – 3| = 6 <=> \(\left[ \begin{array}{l}2x-3=6\\2x – 3 = -6\end{array} \right.\) <=> \(\left[ \begin{array}{l}2x= 9\\2x = -3\end{array} \right.\) <=> \(\left[ \begin{array}{l}x=4,5\\x = -1,5\end{array} \right.\) Vậy… b) 2|3x+1| = 5 <=>|3x+1| =2,5 <=> \(\left[ \begin{array}{l}3x+1 = 2,5\\3x+1 = -2,5\end{array} \right.\) <=> \(\left[ \begin{array}{l}3x = 1,5\\3x = -3,5\end{array} \right.\) <=> \(\left[ \begin{array}{l}x=1,5\\x = -7/6\end{array} \right.\) Vậy… c) 7,5-3|5-2x| = -4,5 <=> 3|5-2x| = 12 <=>|5 – 2x| = 4 <=> \(\left[ \begin{array}{l}5-2x = 4\\5-2x = -4\end{array} \right.\) <=> \(\left[ \begin{array}{l}2x = 1\\2x = 9\end{array} \right.\) <=> \(\left[ \begin{array}{l}x = 0,5\\x = 4,5\end{array} \right.\) d) |3x-1| = |x+3| <=> \(\left[ \begin{array}{l}3x-1 = x+3\\3x – 1 = -x – 3\end{array} \right.\) <=> \(\left[ \begin{array}{l}2x = 4\\4x = -2\end{array} \right.\) <=> \(\left[ \begin{array}{l}x = 2\\x = -0,5\end{array} \right.\) Vậy… Bình luận
Bài 1 :
a) `|2x-3|` = 6
⇒ `2x` – `3` = `6`
`2x` – `3` = `-6`
⇒ `2x` = `6` + `3`
`2x` = `-6` + `3`
⇒ `2x` = `9`
`2x` = `-3`
⇒ `x` = `9/2`
`x` = `-3/2`
b) `2.|3x+1|` = `5`
⇒ `|3x+1|` = `5` : `2` = `5/2`
⇒ `3x` + `1` = `5/2`
`3x` + `1` = `-5/2`
⇒ `3x` = `5/2` – `1` = `3/2`
`3x` = `-5/2` – `1` = `7/2`
⇒ `x` = `3/2` : `3` = `1/2`
`x` = `-7/2` : `3` = `-7/6`
c) `7,5` – `3|5-2x|` = `-4,5`
⇒ `3|5-2x|` = `7,5` – `(-4,5)` = `12`
⇒ `|5-2x|` = `12` : `3` = `4`
⇒ `5` – `2x` = `4`
`5` – `2x` = `-4`
⇒ `2x` = `5` – `4` = `1`
`2x` = `5` – `(-4)` = `9`
⇒ `x` = `1/2`
`x` = `9/2`
d) `|3x-1|` = `|x+3|`
⇒ `3x` – `1` = `x` + `3`
`3x` – `1` = `-x` – `3`
⇒ `2x` = `4`
`4x` = `-2`
⇒ `x` = `2`
`x` = `-1/2`
a) |2x – 3| = 6
<=> \(\left[ \begin{array}{l}2x-3=6\\2x – 3 = -6\end{array} \right.\)
<=> \(\left[ \begin{array}{l}2x= 9\\2x = -3\end{array} \right.\)
<=> \(\left[ \begin{array}{l}x=4,5\\x = -1,5\end{array} \right.\)
Vậy…
b) 2|3x+1| = 5
<=>|3x+1| =2,5
<=> \(\left[ \begin{array}{l}3x+1 = 2,5\\3x+1 = -2,5\end{array} \right.\)
<=> \(\left[ \begin{array}{l}3x = 1,5\\3x = -3,5\end{array} \right.\)
<=> \(\left[ \begin{array}{l}x=1,5\\x = -7/6\end{array} \right.\)
Vậy…
c) 7,5-3|5-2x| = -4,5
<=> 3|5-2x| = 12
<=>|5 – 2x| = 4
<=> \(\left[ \begin{array}{l}5-2x = 4\\5-2x = -4\end{array} \right.\)
<=> \(\left[ \begin{array}{l}2x = 1\\2x = 9\end{array} \right.\)
<=> \(\left[ \begin{array}{l}x = 0,5\\x = 4,5\end{array} \right.\)
d) |3x-1| = |x+3|
<=> \(\left[ \begin{array}{l}3x-1 = x+3\\3x – 1 = -x – 3\end{array} \right.\)
<=> \(\left[ \begin{array}{l}2x = 4\\4x = -2\end{array} \right.\)
<=> \(\left[ \begin{array}{l}x = 2\\x = -0,5\end{array} \right.\)
Vậy…