a) \(\left|\left|x-1\right|-1\right|=2\Rightarrow\orbr{\begin{cases}\left|x-1\right|-1=2\\\left|x-1\right|-1=-2\end{cases}}\Rightarrow\orbr{\begin{cases}\left|x-1\right|=3\\\left|x-1\right|=-1\left(l\right)\end{cases}}\)

TH1: x - 1 = 3

         x      = 4

TH2: x - 1 = - 3

        x       = - 2 

b) Tương tự câu a.

c) \(\left|\left|2x-3\right|-x+1\right|=42-8\)

\(\left|\left|2x-3\right|-x+1\right|=34\)

TH1: \(\left|2x-3\right|-x+1=34\)

\(\left|2x-3\right|-x=33\)

Với \(x\ge\frac{3}{2}\), ta có \(2x-3-x=33\Rightarrow x=36\)  (tm)

Với \(x< \frac{3}{2}\), ta có \(3-2x-x+1=34\Rightarrow-3x=30\Rightarrow x=-10\left(tm\right)\)

TH2: \(\left|2x-3\right|-x+1=-34\)

\(\left|2x-3\right|-x=-35\)

Với \(x\ge\frac{3}{2}\), ta có \(2x-3-x=-35\Rightarrow x=-32\)  (l)

Với \(x< \frac{3}{2}\), ta có \(3-2x-x+1=-34\Rightarrow-3x=38\Rightarrow x=\frac{38}{3}\left(l\right)\)

d) Tương tự câu c.

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

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

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

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

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, !x-1!=0

\(\Rightarrow x-1=0\)

\(\Rightarrow x=0+1\)

\(\Rightarrow x=1\)

Vậy x=1

b,-11.!3x-1!=-22

\(\Rightarrow!3x-1!=-22:\left(-11\right)\)

\(\Rightarrow!3x-1!=2\)

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

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

\(\Rightarrow\orbr{\begin{cases}3x=3\\3x=-1\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}x=3:3\\x=-1:3\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}x=1\\x=\frac{-1}{3}\end{cases}}\)

Vậy \(\orbr{\begin{cases}x=1\\x=-\frac{1}{3}\end{cases}}\)

c, !x!<2

\(\Rightarrow x\in\left\{-1;0;1\right\}\)

Vậy \(x\in\left\{-1;0;1\right\}\)

Hok tốt nhé!!

A ) 

| x - 1 | = 0 

  x        = 0 + 1 

  x        = 1 

B ) 

-11 . | 3x - 1 | = -22 

         | 3x - 1 | = -22 : ( -11 ) 

         | 3x - 1 | = 2 

           3x        = 2 + 1 

           3x        = 3 

             x        = 3 : 3 

             x        = 1 

Ta có:\(\hept{\begin{cases}\left|x-2\right|\ge0\\\left|1,5-y\right|\ge0\\\left|3-z\right|\ge0\end{cases}\Rightarrow\left|x-2\right|+\left|1,5-y\right|+\left|3-z\right|\ge0}\)

Để \(\left|x-2\right|+\left|1,5-y\right|+\left|3-z\right|=0\) thì \(\hept{\begin{cases}\left|x-2\right|=0\\\left|1,5-y\right|=0\\\left|3-z\right|=0\end{cases}\Rightarrow\hept{\begin{cases}x=2\\y=1,5\\z=3\end{cases}}}\)

Vì |x-2| ; |1,5-y| ; |3-z| đều >= 0 nên VT >= 0 

=> VT= 0 <=> x-2=0;1,5-y=0;3-z=0

<=> x=2;y=1,5;z=3

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

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

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

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

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