\(|12,1.x+12,1.0,1|=|12,1.\left(x+0,1\right)|\)

=> 12,1.(x+0,1)=21,1 hoặc -21,1

=> x+0,1=\(\frac{211}{121}\)hoặc\(\frac{-211}{121}\)

=. x = \(\frac{1989}{1210}\)hoặc \(-1,844\)

| 12,1 .x + 12,1 . 0,1 | = 12,1

| 12,1 . ( x + 0,1 ) |     = 12,1

\(\Rightarrow\)| x + 0,1 |           = 12,1 :12,1

        | x + 0,1 |           = 1

        | x |                   = 1 - 0,1

        | x |                   = 0,9

         x                     = 0,9

Vậy x = 0,9

a) 12,1 x 5,5 + 12,1 x 4,5

= 12,1 x ( 5,5 + 4,5 )

= 12,1 x 10

= 121

b) 8,32 x 4 x 25

= 8,32 x ( 4 x 25 )

= 8,32 x 100

= 832

c) 0,81 x 8,4 + 2,6 x 0,81

= 0,81 x ( 8,4 + 2,6 )

= 0,81 x 11

= 8,91

d) 0,8 x 1,25 x 0,29

= 1 x 0,29

= 0,29

e) 16,5 x 47,8 + 47,8 x 3,5

= 47,8 x ( 16,5 + 3,5 )

= 47,8 x 20

= 956

g) 2,5 x 5 x 0,2

= 2,5 x ( 5 x 0,2 )

= 2,5 x 1

= 2,5

a) 12,1 x 5,5 + 12,1 x 4,5 

= 12,1 x ( 5,5 + 4,5 ) 

= 12,1 x 10 

= 121 

b) 8,32 x 4 x 25

= 8,32 x 100 

= 832 

c) 0,81 x 8,4 + 2,6 x 0,81

= 0,81 x ( 8,4 + 2,6 )

= 0,81 x 11

= 8,91

d) 0,8 x 1,25 x 0,29

= 1 x 0,29

= 0,29

e) 16,5 x 47,8 + 47,8 x 3,5 

= 47,8 x (16,5 + 3,5 ) 

= 47 ,8 x 20

= 956

g) 2,5 x 5 x 0,2

= 2,5 x 1

= 2,5

\(a.\)

\(\left|0,2x-3,1\right|=6,3\)

\(\Rightarrow\left[\begin{array}{nghiempt}0,2x-3,1=6,3\\0,2x-3,1=-6,3\end{array}\right.\)

\(\Rightarrow\left[\begin{array}{nghiempt}0,2x=6,3+3,1\\0,2=-6,3+3,1\end{array}\right.\)

\(\Rightarrow\left[\begin{array}{nghiempt}0,2x=9,4\\0,2x=-3,2\end{array}\right.\)

\(\Rightarrow\left[\begin{array}{nghiempt}x=9,4:0,2\\x=-3,2:0,2\end{array}\right.\)

\(\Rightarrow\left[\begin{array}{nghiempt}x=47\\x=-16\end{array}\right.\)

Vậy :    \(x\in\left\{-16;47\right\}\)

\(b.\)

\(\left|12,1x+12,1.0,1\right|=12,1\)

\(\Leftrightarrow\left|12,1.\left(x+0,1\right)\right|=12,1\)

\(\Rightarrow\left[\begin{array}{nghiempt}12,1\left(x+0,1\right)=12,1\\12,1\left(x+0,1\right)=-12,1\end{array}\right.\)

\(\Rightarrow\left[\begin{array}{nghiempt}x+0,1=12,1:12,1\\x+0,1=-12,1:12,1\end{array}\right.\)

\(\Rightarrow\left[\begin{array}{nghiempt}x+0,1=1\\x+0,1=-1\end{array}\right.\)

\(\Rightarrow\left[\begin{array}{nghiempt}x=1-0,1\\x=-1-0,1\end{array}\right.\)

\(\Rightarrow\left[\begin{array}{nghiempt}x=0,9\\x=-1,1\end{array}\right.\)

Vậy :    \(x\in\left\{-1,1;-,9\right\}\)

\(c.\)

\(\left|0,2x-3,1\right|+\left|0,2x=3,1\right|=0\)

\(\Rightarrow\left[\begin{array}{nghiempt}0,2x-3,1=0\\0,2+3,1=0\end{array}\right.\)

\(\Rightarrow\left[\begin{array}{nghiempt}0,2x=0+3,1\\0,2x=0-3,1\end{array}\right.\)

\(\Rightarrow\left[\begin{array}{nghiempt}0,2x=3,1\\0,2x=-3,1\end{array}\right.\)

\(\Rightarrow\left[\begin{array}{nghiempt}x=3,1:0,2\\x=-3,1:0,2\end{array}\right.\)

\(\Rightarrow\left[\begin{array}{nghiempt}x=15,5\\x=-15,5\end{array}\right.\)

Vậy :    \(x\in\left\{-15,5;15,5\right\}\)

a ) \(\left|0,2.x-3,1\right|=6,3\)

\(\Leftrightarrow\left[\begin{array}{nghiempt}0,2x-3,1=6,3\\0,2x-3,1=-6,3\end{array}\right.\)

\(\Leftrightarrow\left[\begin{array}{nghiempt}0,2x=9,4\\0,2x=-3,2\end{array}\right.\)

\(\Leftrightarrow\left[\begin{array}{nghiempt}x=47\\x=-16\end{array}\right.\)

Vậy ........

b ) \(\left|12,1.x+12,1.0,1\right|=12,1\)

\(\Leftrightarrow\left|12,1.\left(x+0,1\right)\right|=12,1\)

\(\Leftrightarrow\left[\begin{array}{nghiempt}12,1.\left(x+0,1\right)=12,1\\12,1.\left(x+0,1\right)=-12,1\end{array}\right.\)

\(\Leftrightarrow\left[\begin{array}{nghiempt}\left(x+0,1\right)=1\\\left(x+0,1\right)=-1\end{array}\right.\)

\(\Leftrightarrow\left[\begin{array}{nghiempt}x=0,9\\x=-1,1\end{array}\right.\)

Vậy ...........................

c ) \(\left|0,2.x-3,1\right|+\left|0,2.x+3,1\right|=0\)

=> \(\left[\begin{array}{nghiempt}0,2x-3,1=0\\0,2x+3,1=0\end{array}\right.\)

\(\Rightarrow\left[\begin{array}{nghiempt}0,2x=3,1\\0,2x=-3,1\end{array}\right.\)

\(\Rightarrow\left[\begin{array}{nghiempt}x=15,5\\x=-15,5\end{array}\right.\)

Vậy .............................

x=0,9 nha

Bạn làm cả cách làm ra hộ vs

|12,1.x+12,1.0,1|=12,1

=> TH1: 12,1.x+12,1.0,1=12,1

       (=) 12,1.(x+0,1) = 12,1

       (=)     x+0,1      = 12,1: 12,1

       (=)     x+0,1      = 1

       (=)     x             = 1-0,1

       (=)     x             = 0,9

TH2:  12,1.x+12,1.0,1= -12,1

    (=) 12,1.(x+0,1) = -12,1

    (=)    x+0,1       = -1

    (=)      x            = -1-0,1

    (=)      x            = -1,1

Vậy, x = 0,9 hoặc x = -1,1

\(12,1+12,1\cdot101+99\cdot12,1-101\cdot12,1\)

\(=12,1\cdot\left(1+101+99-101\right)\)

\(=12,1\cdot\left(1+99\right)\)

\(=12,1\cdot100\)

\(=1210\)

`12,1 + 12,1 x 101 + 99 x 12,1 - 101 x 12,1`

`= 1 x 12,1 + 12,1 x 101 + 99 x 12,1 - 101 x 12,1`

`= 12,1 x (1 + 101 + 99 - 101)`

`= 12,1 x 100`

`= 1 210`

Có hai trường hợp:

Nếu phần trị tuyệt đối bằng -12,1 thì x = -1,1

Nếu phần trị tuyệt đối bằng 12,1 thì x= 0,9

Cảm ơn nha