\(\left|2x-3\right|=1-5x\)
\(\Leftrightarrow\left[{}\begin{matrix}2x-3=1-5x\\2x-3=-\left(1-5x\right)\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}2x+5x=1+3\\2x-3=-1+5x\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}7x=4\\2x-5x=-1+3\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{4}{7}\\-3x=2\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{4}{7}\\x=-\dfrac{2}{3}\end{matrix}\right.\)
Vậy \(S=\left\{\dfrac{4}{7};-\dfrac{2}{3}\right\}\)
\(\left|2x-3\right|=1-5x\)
ta có \(\left|2x-3\right|=\left\{{}\begin{matrix}2x-3\left(2x-3\ge0< =>x\ge\dfrac{3}{2}\right)\\3-2x\left(2x-3< 0< =>x< \dfrac{3}{2}\right)\end{matrix}\right.\)
với `x>=3/2`
`2x-3=1-5x`
`<=>2x+5x=1+3`
`<=>7x=4`
`<=>x=4/7(ktmđk)`
với `x<3/2`
`3-2x=1-5x`
`<=>-2x+5x=1-3`
`<=>3x=-2`
`<=>x=-2/3(tm)`
Vậy phương trinhg có tập nghiệm `S={-2/3}`
|2x - 3| = 1 - 5x (1)
|2x - 3| = 2x - 3 ⇔ 2x - 3 ≥ 0 ⇔ 2x ≥ 3 ⇔ x ≥ 3/2
|2x - 3| = -(2x - 3) = 3 - 2x ⇔ 2x - 3 < 0 ⇔ 2x < 3 ⇔ x < 3/2
*) Với x ≥ 3/2
(1) ⇔ 2x - 3 = 1 - 5x
⇔ 2x + 5x = 1 + 3
⇔ 7x = 4
⇔ x = 4/7 (loại)
*) Với x < 3/2
(1) ⇔ 3 - 2x = 1 - 5x
⇔ -2x + 5x = 1 - 3
⇔ 3x = -2
⇔ x = -2/3 (nhận)
Vậy S = {-2/3}
\(\left|2x-3\right|=1-5x\) (1)
TH1: 2x - 3 \(\ge\) 0 \(\Leftrightarrow\) x \(\ge\) \(\dfrac{3}{2}\)
(1) \(\Leftrightarrow\) 2x - 3 = 1 - 5x
\(\Leftrightarrow\) 2x - 3 - 1 + 5x = 0
\(\Leftrightarrow\) 7x - 4 = 0
\(\Leftrightarrow\) x = \(\dfrac{4}{7}\) (n)
TH2 : 2x - 3 < 0 \(\Leftrightarrow\) x < \(\dfrac{3}{2}\)
(1) \(\Leftrightarrow\) -(2x - 3) = 1 - 5x
\(\Leftrightarrow\) - 2x + 3 = 1 - 5x
\(\Leftrightarrow\) -2x + 3 - 1 + 5x = 0
\(\Leftrightarrow\) 3x + 2 = 0
\(\Leftrightarrow\) x = \(\dfrac{-2}{3}\) (n)
Vậy pt (1) có tập n0 S = \(\left\{\dfrac{4}{7},\dfrac{-2}{3}\right\}\)
