Bài 3.4: Tìm x,biết: a,|2x-5=x+1 b,|3x-2|-1=x c,|3x-7|=2x+1 d,|2x-1|+1=x

Answer

`a, |2x - 5| = x + 1`

Ta có:

`|2x - 5| >= 0`

`=> x + 1 >= 0`

`=> x >= -1`

`=>` $\left[\begin{matrix} 2x - 5 = x + 1\\2x - 5 = -x - 1\end{matrix}\right.$

`=>` $\left[\begin{matrix} 2x - x = 1 + 5\\2x + x = -1 + 5\end{matrix}\right.$

`=>` $\left[\begin{matrix} x = 6\\3x = 4\end{matrix}\right.$

`=>` $\left[\begin{matrix} x = 6 \ (TM)\\x = \dfrac{4}{3} \ (TM)\end{matrix}\right.$

Vậy `x \in {6 ; 4/3}.`

`b, |3x - 2| - 1 = x`

`=> |3x - 2| = x + 1`

Ta có:

`|3x - 2| >= 0`

`=> x + 1 >= 0`

`=> x >= -1`

`=>` $\left[\begin{matrix} 3x - 2 = x + 1\\3x - 2 = -x - 1\end{matrix}\right.$

`=>` $\left[\begin{matrix} 3x - x = 1 + 2\\3x + x = -1 + 2\end{matrix}\right.$

`=>` $\left[\begin{matrix} 2x = 3\\4x = 1\end{matrix}\right.$

`=>` $\left[\begin{matrix} x = \dfrac{3}{2} \ (TM)\\x = \dfrac{1}{4} \ (TM)\end{matrix}\right.$

Vậy `x \in {3/2 ; 1/4}.`

`c, |3x - 7| = 2x + 1`

Ta có:

`|3x - 7| >= 0`

`=> 2x + 1 >= 0`

`=> 2x >= -1`

`=> x >= -1/2`

`=>` $\left[\begin{matrix} 3x - 7 = 2x + 1\\3x - 7 = -2x - 1\end{matrix}\right.$

`=>` $\left[\begin{matrix} 3x - 2x = 1 + 7\\3x + 2x = -1 + 7\end{matrix}\right.$

`=>` $\left[\begin{matrix} x = 8\\5x = 6\end{matrix}\right.$

`=>` $\left[\begin{matrix} x = 8 \ (TM)\\x = \dfrac{6}{5} \ (TM)\end{matrix}\right.$

Vậy `x \in {8 ; 6/5}.`

`d, |2x - 1| + 1 = x`

`=> |2x - 1| = x - 1`

Ta có:

`|2x - 1| >= 0`

`=> x - 1 >= 0`

`=> x >= 1`

`=>` $\left[\begin{matrix} 2x - 1 = x - 1\\2x - 1 = 1 - x\end{matrix}\right.$

`=>` $\left[\begin{matrix} 2x - x = 1 - 1\\2x + x = 1 - 1\end{matrix}\right.$

`=>` $\left[\begin{matrix} x = 0\\3x = 0\end{matrix}\right.$

`=>` $\left[\begin{matrix} x = 0\\x = 0\end{matrix}\right.$

`=> x = 0` (KTM)

Vậy $x \in {varnothing}$