mình biết làm đấy nhưng không biết ghi vào đây như thế nào!

làm sao để mở dấu ngoặc vuông vậy

a) |0,2x - 3,1| = 6,3

\(0,2x-3,1=\pm6,3\)

Th1:

0,2x - 3,1 = 6,3

0,2x = 6,3 + 3,1

0,2x = 9,4

x = 9,4 : 0,2

x = 47

Th2:

0,2x - 3,1 = - 6,3

0,2x = - 6,3 + 3,1

0,2x = - 3,2

x = - 3,2 : 0,2

x = - 16

Vậy x = 47 hoặc x = - 16

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

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

\(12,1\left(x+0,1\right)=\pm12,1\)

Th1:

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

x + 0,1 = 1

x = 1 - 0,1

x = 0,9

Th2:

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

c) |0,2x - 3,1| + |0,2.x + 3,1| = 0

|0,2x - 3,1| + |0,2x + 3,1| \(\ge\) |0,2x - 3,1 + 0,2x + 3,1| = 0,4x

mà |0,2x - 3,1| + |0,2.x + 3,1| = 0

=> x = 0

x + x : 0,2 = 1,35
x * 1 + x * 5 = 1,35
x * ( 1 + 5 ) = 1,35
x * 6 = 1,35
x = 1,35 : 6
x = 0,225

hok tốt nha ^_^

x + x : 0,2 = 1,35
x * 1 + x * 5 = 1,35
x * ( 1 + 5 ) = 1,35
x * 6 = 1,35
x = 1,35 : 6
x = 0,225

hok tốt nha ^_^

bài 2)

a)|12,1x+12,1.0,1|=12,1

<=> |12,1(x+0,1)|=12,1

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

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

b) |0,2x-3,1|+|0,2+3,1|=0

<=>|0,2x-3,1|+3,3=0

ta có |A|>=0

=> pt trên vô nghiệm

bài 1) a) 6,5+1,2+3,5-5,2+6,5-4,8

=(6,5+3,5+6,5)-(5,2-1,2+4,8)

=15,5-8,8=24,6

tt nha bạn

a, | 0,2 . x - 3,1 | = 6,3

\(\Rightarrow0,2.x-3,1=\pm6,3\)

+) \(0,2.x-3,1=6,3\Rightarrow x=47\)

+) \(0,2.x-3,1=-6,3\Rightarrow x=16\)

Vậy x = 47 hoặc x = 16

b, | 12,1 . x - 3,1 | = 12,1

\(\Rightarrow12,1.x-3,1=\pm12,1\)

+) \(12,1.x-3,1=12,1\Rightarrow x=\frac{151}{121}\)

+) \(12,1.x-3,1=-12,1\Rightarrow x=\frac{90}{121}\)

Vậy \(x=\frac{151}{121}\) hoặc \(x=\frac{90}{121}\)

c, | 0,2 . x - 3,1 | + | 0,2 . x + 3,1 | = 0

\(\Rightarrow0,2.x-3,1=0\) hoặc \(0,2.x+3,1=0\)

+) \(0,2.x-3,1=0\Rightarrow x=15,5\)

+) \(0,2.x+3,1=0\Rightarrow x=-15,5\)

Vậy x = 15,5 hoặc x = -15,5

c) ( sửa lại )

| 0,2 . x - 3,1 | + | 0,2 . x + 3,1 | = 0

\(\Rightarrow\) 0,2 . x - 3,1 = 0 hoặc 0,2 . x + 3,1 = 0

+) \(0,2.x-3,1=0\Rightarrow x=15,5\)

+) \(0,2.x+3,1=0\Rightarrow x=-15,5\)

Vì x không thể bằng 15,5 và -15,5 ( 15,5 khác -15,5 ) nên x không có giá trị thỏa mãn đề bài.

Vậy x không có giá trị thỏa mãn đề bài

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

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

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

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

Vậy ...

b/ \(\left|12,1x+1,1.01\right|=12,1\)

\(\Leftrightarrow\left|12,1x+0,11\right|=12,1\)

\(\Leftrightarrow\left[{}\begin{matrix}12,1x+0,11=12,1\\12,1x+0,11=-12,1\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}12,1x=11,99\\12,1x=-12,21\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{11,99}{12,1}\\x=\dfrac{-12,21}{12,1}\end{matrix}\right.\)

dễ mà bài này mk bk làm ak

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

\(\Leftrightarrow\left\{{}\begin{matrix}\left|0,2x-3,1\right|=0\\\left|0,2x+3,1\right|=0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}0,2x-3,1=0\\0,2x+3,1=0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}0,2x=3,1\\0,2x=-3,1\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x=15,5\\x=-15,5\end{matrix}\right.\) (vô lí)

Vậy ko tìm dc x thỏa mãn theo yêu cầu

\(a.\)

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

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

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

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

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

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

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 .............................