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

Do \(\left|0,2.x-3,1\right|\ge0;\left|0,2.x+3.1\right|\ge0\)

=> \(\left|0,2.x-3,1\right|+\left|0,2.x+3,1\right|\ge0\)

Mà theo đề bài: |0,2.x - 3,1| + |0,2.x + 3,1| = 0

=> \(\begin{cases}\left|0,2.x-3,1\right|=0\\\left|0,2.x+3,1\right|=0\end{cases}\)=> \(\begin{cases}0,2.x-3,1=0\\0,2.x+3,1=0\end{cases}\), vô lí

Vậy không tồn tại giá trị của x thỏa mãn đề bài

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 ^_^

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) \(\left|0,2x-3,1\right|=6,3\)

\(\Leftrightarrow\begin{cases}x\ge15,5\\0,2x-3,1=6,3\end{cases}\)  hoặc \(\begin{cases}x< 15,5\\0,2x-3,1=-6,3\end{cases}\)

\(\Leftrightarrow\begin{cases}x\ge15,5\\x=47\left(tm\right)\end{cases}\) hoạc \(\begin{cases}x< 15,5\\x=-47\left(tm\right)\end{cases}\)

Vậy x={47;-47}

b) \(\left|12,1x-3,1\right|=12,1\)

\(\Leftrightarrow\begin{cases}x\ge\frac{31}{121}\\12,1x-3,1=12,1\end{cases}\)  hoạc \(\begin{cases}x< \frac{31}{121}\\12,1x-3,1=-12,1\end{cases}\)

\(\Leftrightarrow\begin{cases}x\ge\frac{31}{121}\\x=\frac{152}{121}\left(tm\right)\end{cases}\)  hoặc \(\begin{cases}x< \frac{31}{121}\\x=-\frac{90}{121}\left(tm\right)\end{cases}\)

Vaayj x={152/121;-90/121}

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

\(\Leftrightarrow\begin{cases}0,2x-3,1=0\\0,2x+3,1=0\end{cases}\)\(\Leftrightarrow\begin{cases}x=15,5\\x=-15,5\end{cases}\)(vô lý)

Vậy k có x thỏa mãn

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

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 ^_^

\(a.\)

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

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

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

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

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

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

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

a) Ta có: /0.2x - 3.1/=6.3

→0.2x-3.1=6.3 hoặc 0.2x-3.1=-6.3

→0.2x=6.3+3.1=9.4 hoặc 0.2x=(-6.3)+3.1=-3.2

→x=9.4:0.2=47 hoặc x=-3.2:0.2=-16

Vậy x=47 hoặc x=-16

toán lớp 7 k phải lớp 10 đâu nha ấn nhầm

bo chiu