a ) \(2x-\left(-17\right)=15\)

\(\Leftrightarrow2x+17=15\)

\(\Leftrightarrow2x=15-17\)

\(\Leftrightarrow2x=-2\)

\(\Leftrightarrow x=-2\div2\)

\(\Leftrightarrow x=-1\)

b ) \(-2x-8=72\)

\(\Leftrightarrow-2x=72+8\)

\(\Leftrightarrow-2x=80\)

\(\Leftrightarrow x=-40\)

a) 2x - ( -17 ) = 15

->  2x + 17 = 15

2x = 15 - 17

2x = -2

x = \(\frac{-2}{2}\)

x = -1

b) -2x - 8 = 72

-2x = 72 + 8

-2x = 80

x = 80 : ( -2 )

x = -40

c) 3 . \([x-1]\)= 27

x - 1 = \(\frac{27}{3}\)

x - 1 = 9

x = 9 + 1

x = 10

d) \([-2x+5]\)+ 8 = 21

-2x + 5 = 21 - 8

-2x + 5 = 13

-2x = 13 - 5

-2x = 8

x = \(\frac{8}{-2}\)

x = - 4

c ) \(3.\left|x-1\right|=27\)

\(\Leftrightarrow\left|x-1\right|=9\)

\(\Leftrightarrow\orbr{\begin{cases}x-1=9\\x-1=-9\end{cases}\Rightarrow}\orbr{\begin{cases}x=10\\x=-8\end{cases}}\)

Vậy \(x\in\left\{10;-8\right\}\)

d ) \(\left|-2x+5\right|+8=21\)

\(\Leftrightarrow\left|-2x+5\right|=13\)

\(\Leftrightarrow\orbr{\begin{cases}-2x+5=13\\-2x+5=-13\end{cases}\Rightarrow\orbr{\begin{cases}x=-4\\x=-9\end{cases}}}\)

Vậy \(x\in\left\{-4;-9\right\}\)

Bạn ơi, hình như câu 1 sai đề bài. Bạn xem lại giùm mk nhá!

đúng mà bạn

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.