Đáp án:
Giải thích các bước giải:
`a) |2x - 1| < 3`
⇒ \(\left[ \begin{array}{l}2x - 1>3\\2x - 1<-3\end{array} \right.\)
\(\left[ \begin{array}{l}2x >4\\2x <-2\end{array} \right.\)
\(\left[ \begin{array}{l}x >2\\x <-1\end{array} \right.\)
`b) |2x - 1| > 2`
\(\left[ \begin{array}{l}2x - 1>2\\2x - 1<-2\end{array} \right.\)
\(\left[ \begin{array}{l}2x >3\\2x <-1\end{array} \right.\)
\(\left[ \begin{array}{l}x >\dfrac{3}{2}\\x <-\dfrac{1}{2}\end{array} \right.\)
bài 2
a,`|2x-1|+x=3`
`|2x - 1| = 3-x`
$\begin{cases} 3 - x ≥ 0\\\left[ \begin{array}{l}2x - 1=3-x\\2x - 1=x- 3\end{array} \right. \end{cases}$
$\begin{cases} x ≤ 3\\\left[ \begin{array}{l}3x = 4\\x = -2 (TM)\end{array} \right. \end{cases}$
$\begin{cases} x ≤ 3\\\left[ \begin{array}{l}x = \dfrac{4}{3} ( TM)\\x = -2 (TM)\end{array} \right. \end{cases}$
b) th1 `x ≥ \frac{1}{2}`
`⇒ 2x - 1 + x + 2 = x `
`3x + 1 = x `
`2x = -1 `
`x= \frac{-1}{2} `( L )
th2 `x ≤ -2`
`⇒ 1 - 2x - x - 2 = x `
`-3x - 1 = x `
`4x = -1 `
`x = \frac{-1}{4}` (L )
th3 `-2 < x < \frac{1}{2}`
`1-2x + x + 2 = x`
`-2x + x - x = -1-2`
`-2x = -3`
`x = \frac{3}{2}` ( L )
vậy ko có x thỏa mãn
`c) |x+1| + |x+2| + |x+3| = 4x`
`có |x+1| + |x+2| + |x+3| ≥ 0 ⇒ 4x ≥ 0`
`⇔ x ≥ 0 ⇒ x + 1 + x + 2 + x + 3 = 4x`
`3x + 6 = 4x `
`x = 6 ( TM )`
vậy `x = 6`
