a) |x-3|+5=-7

   |x+3|=-7-5

   |x-3|=-12( vô lí) 

Vậy không có giá trị nào của x thỏa mãn đề bài 

b)||x-7|-6|=3

suy ra |x-7|-6=3 hoặc |x-7|-6 = -3

          |x-7|=3+6        |x-7|=-3+6  

          |x-7|=9           |x-7|=3

suy ra x-7=9 hoặc x-7=-9 hoặc x-7=3 hoặc x-7=-3

TH1 ; x-7=9        TH2 : x-7=-9      TH3 : x-7=3        TH4 : x-7=-3 

          x=9+7                x= -9+7             x=3+7                x=-3+7 

         x=16                  x=-2                  x=10                  x=4

Vậy x thuộc {10;-2;10;4}

c) |x-1|=5-x 

suy ra x-1=5-x hoặc x-1=-(5-x)

         x+x=1+5        x-1=-5+x 

         2x=6             x-x=1-5|(vô lí) 

          x=6:2          

         x=3

Vậy x=3 

d) |1-x|+x+3=6

   |1-x|+x=6-3

   |1-x|+x=3

   |1-x|=3-x

suy ra 1-x=3-x hoặc 1-x=-(3-x) 

TH1 : 1-x=3-x                                 TH2 : 1-x=(-3-x)

        -x+x=-1+3 (vô lí )                   1-x=-3+x

                                                    -x-x=-1-3

                                                   -2x=-4

                                                    x= -4:(-2)

                                                   x=2

vậy x=2

a, l x-3 l +5 = 7                   

    l x-3 l =2 

=> x-3=2 hoac x-3=-2

     x=5   hoac x=1

B1:a )

(x-4).(y+3)=-3=-1.3=-3.1

ta có bảng sau:

a) -105 - 5.x = (-5)^2

=>-105 - 5.x = 25

=> 5.x = -105 - 25

=> 5.x = -130

=> x = -130: 5

=> x = -26

c) 400 -4.|5-x| = (-6)^2

=>400- 4.|5-x| = 36

=> 4.|5-x| = 400-36

=> 4.|5-x| = 364

=> |5-x| = 364:4

=> |5-x| = 91

=> \(\left[\begin{matrix}5-x=91\\5-x=-91\end{matrix}\right.\)

=> \(\left[\begin{matrix}x=5-91\\x=5-\left(-91\right)\end{matrix}\right.\)

^{=> \(\left[\begin{matrix}x=-86\\x=96\end{matrix}\right.\)}

d) 2.|x-1| -5 = -15

=>2.|x-1| = -15+5

=>2.|x-1| = 10

=> |x-1| = 10:2

=> |x-1| = 5

=> \(\left[\begin{matrix}x-1=5\\x-1=-5\end{matrix}\right.\)

=>

a) \(\frac{2}{3}+\frac{1}{3}:x=\frac{3}{5}\)

\(\frac{1}{3}:x=\frac{3}{5}-\frac{2}{3}=\frac{9}{15}-\frac{10}{15}=\frac{-1}{15}\)

\(x=\frac{-1}{15}.\frac{1}{3}\)

\(x=\frac{-1}{45}\)

Vậy x = \(\frac{-1}{45}\)

c) \(\left|2x-1\right|+1=4\)

\(\left|2x-1\right|=4-1=3\)

2x-1 = 3 ; -3

TH1: 2.x - 1 = 3

2.x = 3 + 1 = 4

x = 4 : 2 = 2

TH2: 2.x - 1 = -3

2.x = -3 + 1 = -2

x = -2 : 2 = -1

Vậy x \(\in\){ 2 ; -1 }

Ngại làm ấn máy ==

b)

\(\dfrac{1}{2}\left(\dfrac{4}{5}-\dfrac{3}{2}\right)+x=5\left(x-\dfrac{1}{3}\right)\)

=> \(-\dfrac{7}{20}+x=5x-\dfrac{5}{3}\)

=> \(\dfrac{79}{60}=4x\)

=> \(\dfrac{79}{240}=x\)

Vậy \(\dfrac{79}{240}=x\)

c)

\(2\left(x-5\right)-3\left(x+7\right)=14\)

=> \(2x-10-3x-21=14\)

=> \(-x-31=14\)

=> \(-x=45\)

=> x = -45

Vậy x là -45

d)

-8 . |x - 3| = -32

=> |x - 3| = 4

=> \(\left\{{}\begin{matrix}x-3=4\\x-3=-4\end{matrix}\right.\)

=> x = 7 và x = -1

Vậy x \(\in\left\{-1;7\right\}\)

a.|x-1|=5

=> hai TH

x-1=5 =>x=5+1=6

x-1=-5 => x= -5+1=-4

b. x² = 5²

=> x=5

a ) | x - 1 | = 5

d) 3. x + 5\(^{94}\) : 5\(^{92}\) = 3\(^2\) + 8

3x + \(5^2\) = 9 + 8

3x + 25 = 17

3x = 17 - 25

3x = -8

x = -8 : 3

x = \(\dfrac{-8}{3}\)

a: \(\left|x+\dfrac{4}{15}\right|-\left|-3.75\right|=-\left|-2.5\right|\)

\(\Leftrightarrow\left|x+\dfrac{4}{15}\right|=-2.5+3.75=1.25=\dfrac{5}{4}\)

\(\Leftrightarrow\left[{}\begin{matrix}x+\dfrac{4}{15}=\dfrac{5}{4}\\x+\dfrac{4}{15}=-\dfrac{5}{4}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{59}{60}\\x=-\dfrac{91}{60}\end{matrix}\right.\)

c: \(\left|x-y\right|+\left|y+\dfrac{9}{25}\right|=0\)

\(\Leftrightarrow\left\{{}\begin{matrix}x-y=0\\y+\dfrac{9}{25}=0\end{matrix}\right.\Leftrightarrow x=y=-\dfrac{9}{25}\)

d: Ta có: \(\left|x\left(x^2-\dfrac{5}{4}\right)\right|=x\)

\(\Leftrightarrow\left\{{}\begin{matrix}x>=0\\x\left(x^2-\dfrac{5}{4}\right)=x\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x>=0\\x\left(x^2-\dfrac{9}{4}\right)=0\end{matrix}\right.\Leftrightarrow x\in\left\{0;\dfrac{3}{2}\right\}\)