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

b) |x+1|+|x+2|+|x+3|+|x+4|=5x

<=> x+1+x+2+x+3+x+4=5x

<=> 4x+10=5x

<=> 10=5x-4x

<=> x=10

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

\(\left|3x-1\right|=\left|\dfrac{-1}{3}x+2\right|\)

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

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

<=> \(\left[{}\begin{matrix}\dfrac{10}{3}x=3\\\dfrac{-8}{3}x=1\end{matrix}\right.\)

<=> \(\left[{}\begin{matrix}x=\dfrac{9}{10}\\x=\dfrac{-3}{8}\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}3x-1=-\dfrac{1}{3}x+2\left(x\ge\dfrac{1}{3}\right)\\3x-1=\dfrac{1}{3}x-2\left(x< \dfrac{1}{3}\right)\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}\dfrac{10}{3}x=3\\\dfrac{8}{3}x=-1\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{9}{10}\left(tm\right)\\x=-\dfrac{3}{8}\left(tm\right)\end{matrix}\right.\)

a) 0

b) 3

c) -0,5

a là 0 

b là 3

c là -0,5

a)0

b)3

c)-0,5

tick mjnh nhé

a) x=0

b) x=3

c) x= -2

tick mình đi

a) -2

b) \(\frac{-11}{3}\)

c) -0,5

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

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

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

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

Ta có:

\(B-2011=\left|x-1\right|+\left|x-2\right|+\left|x-3\right|\)

\(\ge x-1+0+3-x=2\)

\(\Rightarrow B-2011\ge2\)\(\Rightarrow B\ge2013\)

Dấu = khi \(\begin{cases}x-1\ge0\\x-2=0\\3-x\ge0\end{cases}\)\(\Leftrightarrow\begin{cases}x\ge1\\x=2\\x\le3\end{cases}\)\(\Leftrightarrow x=2\)

Vậy MinB=2013 khi x=2

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

\(\Leftrightarrow\left\{{}\begin{matrix}x-2=x\\x-2=-x\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x-x=2\left(loại\right)\\x+x=2\end{matrix}\right.\)

\(\Leftrightarrow2x=2\)

\(\Leftrightarrow x=1\left(tm\right)\)

Vậy ......................

b) \(\left|x+2\right|=x\)

\(\Leftrightarrow\left\{{}\begin{matrix}x+2=x\\x+2=-x\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x-x=-2\left(loại\right)\\x+x=-2\end{matrix}\right.\)

\(\Leftrightarrow2x=-2\)

\(\Leftrightarrow x=-1\left(tm\right)\)

Vậy ...............

c) Ta có ;

\(\left|x-3,4\right|+\left|2,6-x\right|=0\)

Mà :

\(\left\{{}\begin{matrix}\left|x-3,4\right|\ge0\\\left|2,6-x\right|\ge0\end{matrix}\right.\)

\(\Leftrightarrow\left|x-3,4\right|+\left|2,6-x\right|\ge\left|x-3,4+2,6-x\right|=\left|-0,8\right|=0,8>0\)

\(\Leftrightarrow\) ko tồn tại \(x\)