Bài 1.3 : Tìm x, biết : a, 2|3x – 1|+ 1= 5 b, |x/2 – 1| = 3 c, |-x + 2/5| + 1/2 = 3,5 d, |x – 1/3| = 2/1/5

Đáp án:

 `a) x=1` hoặc `x=-1/3`

`b) x=8` hoặc `x=-4`

`c) x=-13/5` hoặc `x=17/5`

`d) x=38/15` hoặc `x=-28/15`

Giải thích các bước giải:

`a) 2|3x-1|+1=5`

`⇔2|3x-1|=4`

`⇔|3x-1|=2`

`⇔`\(\left[ \begin{array}{l}3x-1=2\\3x-1=-2\end{array} \right.\) 

`⇔`\(\left[ \begin{array}{l}3x=3\\3x=-1\end{array} \right.\) 

`⇔`\(\left[ \begin{array}{l}x=1\\x=-1/3\end{array} \right.\) 

vậy `x=1` hoặc `x=-1/3`

`b) |x/2-1|=3`

`⇔`\(\left[ \begin{array}{l}x/2-1=3\\x/2-1=-3\end{array} \right.\) 

`⇔`\(\left[ \begin{array}{l}x/2=4\\x/2=-2\end{array} \right.\) 

`⇔`\(\left[ \begin{array}{l}x=8\\x=-4\end{array} \right.\) 

vậy `x=8` hoặc `x=-4`

`c) |2/5-x|+1/2=3,5`

`⇔|2/5-x|=7/2-1/2`

`⇔|2/5-x|=3`

`⇔`\(\left[ \begin{array}{l}2/5-x=3\\2/5-x=-3\end{array} \right.\) 

`⇔`\(\left[ \begin{array}{l}x=2/5-3\\x=2/5+3\end{array} \right.\) 

`⇔`\(\left[ \begin{array}{l}x=-13/5\\x=17/5\end{array} \right.\) 

vậy `x=-13/5` hoặc `x=17/5`

`d) |x-1/3|=2“1/5`

`⇔|x-1/3|=11/5`

`⇔`\(\left[ \begin{array}{l}x-1/3=11/5\\x-1/3=-11/5\end{array} \right.\) 

`⇔`\(\left[ \begin{array}{l}x=11/5+1/3\\x=-11/5+1/3\end{array} \right.\) 

`⇔`\(\left[ \begin{array}{l}x=38/15\\x=-28/15\end{array} \right.\) 

vậy `x=38/15` hoặc `x=-28/15`