Question

Tìm \(x\)\(\in\)\(Q\):

\(a\)\()\) \(|0,5.x-2|-|x+\dfrac{2}{3}|=0\)

\(b)\) \(2x-|x+1|=\dfrac{-1}{2}\)

 

Answer 1

a)

TH1: \(x< \dfrac{-2}{3}\)

<=> \(\left\{{}\begin{matrix}\left|0,5x-2\right|=2-0,5x\\\left|x+\dfrac{2}{3}\right|=-x-\dfrac{2}{3}\end{matrix}\right.\)

PT <=> \(2-0,5x+x+\dfrac{2}{3}=0< =>x=\dfrac{-16}{3}\left(c\right)\)

TH2: \(\dfrac{-2}{3}\le x< 4\)

<=> \(\left\{{}\begin{matrix}\left|0,5x-2\right|=2-0,5x\\\left|x+\dfrac{2}{3}\right|=x+\dfrac{2}{3}\end{matrix}\right.\)

PT <=> \(2-0,5x-x-\dfrac{2}{3}=0< =>x=\dfrac{8}{9}\left(c\right)\)

TH3: \(x\ge4\)

<=> \(\left\{{}\begin{matrix}\left|0,5x-2\right|=0,5x-2\\\left|x+\dfrac{2}{3}\right|=x+\dfrac{2}{3}\end{matrix}\right.\)

PT <=> \(0,5x-2-x-\dfrac{2}{3}=0< =>x=\dfrac{-16}{3}\left(l\right)\)

KL: x \(\left\{\dfrac{-16}{3};\dfrac{8}{9}\right\}\)

b) TH1: \(x\ge-1< =>\left|x+1\right|=x+1\)

PT <=> 2x - x -1 = \(\dfrac{-1}{2}\)

<=> x = \(\dfrac{1}{2}\) (c)

TH2: x < -1 <=> \(\left|x+1\right|=-x-1\)

PT <=> 2x + x + 1 = \(\dfrac{-1}{2}\)

<=> x = \(\dfrac{-1}{2}\) (l)

KL: x \(\in\left\{\dfrac{1}{2}\right\}\)