a: \(\Leftrightarrow\left\{{}\begin{matrix}3x-2>-4\\3x-2< 4\end{matrix}\right.\Leftrightarrow-\dfrac{2}{3}< x< 2\)

c: \(\Leftrightarrow\left[{}\begin{matrix}3x-1>5\\3x-1< -5\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x>2\\x< -\dfrac{4}{3}\end{matrix}\right.\)

d: \(\Leftrightarrow\left[{}\begin{matrix}3x+1>x-2\\3x+1< -x+2\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x>-3\\4x< 1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x>-\dfrac{3}{2}\\x< \dfrac{1}{4}\end{matrix}\right.\)

um mk chiu

a) |x+1|+|x+2+|x+3|=4x

<=> x+1+x+2+x+3=4x

<=> 3x+6=4x

<=> 6=4x-3x

<=> x=6

\(\left|3x-1\right|=\left|2x+5\right|\)

\(\Rightarrow\orbr{\begin{cases}3x-1=2x+5\\3x-1+2x+5=0\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}3x-2x=5+1\\5x+4=0\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}x=6\\x=-\frac{4}{5}\end{cases}}\)

Ta có: \(\hept{\begin{cases}\left(x-1\right)^2\ge0\\\left|3y-1\right|\ge0\\\left|z+2\right|\ge0\end{cases}}\Rightarrow\left(x-1\right)^2+\left|3y-1\right|+\left|z+2\right|\ge0\)

Dấu "="\(\Leftrightarrow\hept{\begin{cases}\left(x-1\right)^2=0\\\left|3y-1\right|=0\\\left|z+2\right|=0\end{cases}}\)

\(\Leftrightarrow\hept{\begin{cases}x-1=0\\3y-1=0\\x+2=0\end{cases}}\)

\(\Leftrightarrow\hept{\begin{cases}x=1\\y=\frac{1}{3}\\z=-2\end{cases}}\)

Vậy x = 1, \(y=\frac{1}{3}\),z = -2

\(a,\left(x.\dfrac{1}{2}\right)^3=\dfrac{1}{27}=\left(\dfrac{1}{3}\right)^3\\ \Rightarrow x.\dfrac{1}{2}=\dfrac{1}{3}\\ \Rightarrow x=\dfrac{1}{3}:\dfrac{1}{2}=\dfrac{2}{3}\\ ---\\ b,\left(x+\dfrac{1}{2}\right)^2=\dfrac{4}{5}=\left(\dfrac{2}{\sqrt{5}}\right)^2=\left(-\dfrac{2}{\sqrt{5}}\right)^2 \\ \Rightarrow\left[{}\begin{matrix}x+\dfrac{1}{2}=\dfrac{2}{\sqrt{5}}\\x+\dfrac{1}{2}=-\dfrac{2}{\sqrt{5}}\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=\dfrac{2}{\sqrt{5}}-\dfrac{1}{2}\\x=-\dfrac{2}{\sqrt{5}}-\dfrac{1}{2}\end{matrix}\right.\\ Vậy:x=\pm\dfrac{2}{\sqrt{5}}-\dfrac{1}{2}\)

\(c,\left|3x-\dfrac{4}{5}\right|=\dfrac{11}{5}\\ \Rightarrow\left[{}\begin{matrix}3x-\dfrac{4}{5}=\dfrac{11}{5}\\3x-\dfrac{4}{5}=-\dfrac{11}{5}\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}3x=\dfrac{11}{5}+\dfrac{4}{5}=3\\3x=-\dfrac{11}{5}+\dfrac{4}{5}=-\dfrac{7}{5}\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=\dfrac{3}{3}=1\\x=-\dfrac{7}{5}:3=-\dfrac{7}{15}\end{matrix}\right.\\ ---\\ d,\left|2x-2\right|=0\\ \Leftrightarrow2x-2=0\\ \Leftrightarrow2x=2\\ \Leftrightarrow x=1\)

\(\left|x+\frac{1}{2}\right|+\left|y-\frac{3}{4}\right|+\left|z+1\right|=0\)

\(\Rightarrow\hept{\begin{cases}\left|x+\frac{1}{2}\right|=0\\\left|y-\frac{3}{4}\right|=0\\\left|z+1\right|=0\end{cases}}\)

\(\Rightarrow\hept{\begin{cases}x=0-\frac{1}{2}\\y=0+\frac{3}{4}\\z=0-1\end{cases}}\)

\(\Rightarrow\hept{\begin{cases}x=\frac{-1}{2}\\y=\frac{3}{4}\\z=-1\end{cases}}\)

a) 0

b) 3

c) -0,5

a là 0 

b là 3

c là -0,5

a) x=0

b) x=3

c) x= -2

tick mình đi

a) -2

b) \(\frac{-11}{3}\)

c) -0,5

A, bạn ơi mình biết làm hết r

B,cài này dễ lằm bạn nghĩ đi okeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeee