bo chiu

\(a.\)

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

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

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

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

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

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

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

| 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

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) |0,2x - 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 ^_^

giúp với

1) Tính nhanh

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

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

\(=10-10+7,7\)

\(=7,7\)

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