Tìm x, biết |2x-1|=2x-1

Bài 1:

a, \(\left|2x-1\right|=2x-1\)

+) Xét \(x\ge\dfrac{1}{2}\) ta có:
\(2x-1=2x-1\)

\(\Rightarrow x\) tùy ý với \(x\ge\dfrac{1}{2}\)

+) Xét \(x< \dfrac{1}{2}\) ta có:

\(1-2x=2x-1\)

\(\Rightarrow4x=2\)

\(\Rightarrow x=\dfrac{1}{2}\) ( không t/m )

Vậy...

b, \(\left|0,5-3x\right|=3x-0,5\)

+) Xét \(x\ge\dfrac{1}{6}\) ta có:

\(0,5-3x=3x-0,5\)

\(\Rightarrow6x=1\)

\(\Rightarrow x=\dfrac{1}{6}\) ( t/m )
+) Xét \(x< \dfrac{1}{6}\) ta có:

\(3x-0,5=3x-0,5\)

\(\Rightarrow x\) tùy ý với \(x< \dfrac{1}{6}\)

Vậy \(x\le\dfrac{1}{6}\)

c, \(\left|5x+1\right|-10x=0,5\)

+) Xét \(x\ge\dfrac{-1}{5}\) ta có:

\(5x+1-10x=0,5\)

\(\Rightarrow-5x=-0,5\)

\(\Rightarrow x=\dfrac{1}{10}\) ( t/m )

+) Xét \(x< \dfrac{-1}{5}\) ta có:

\(-5x-1-10x=0,5\)

\(\Rightarrow-15x=1,5\)

\(\Rightarrow x=\dfrac{-1}{10}\) ( không t/m )

Vậy \(x=\dfrac{1}{10}\)

Bài 2:

a, Ta có: \(-\left|x-3,5\right|\le0\)

\(\Rightarrow A=0,5-\left|x-3,5\right|\le3,5\)

Dấu " = " xảy ra khi \(-\left|x-3,5\right|=0\Rightarrow x=3,5\)

Vậy \(MIN_A=0,5\) khi x = 3,5

b, Ta có: \(-\left|1,4-x\right|\le0\)

\(\Rightarrow B=-\left|1,4-x\right|-2\le-2\)

Dấu " = " xảy ra khi \(-\left|1,4-x\right|=0\Rightarrow x=1,4\)

Vậy \(MIN_B=-2\) khi \(x=1,4\)