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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 ^_^
a/ \(\left|12,1x+12,1.0,1\right|=12,1\)
\(\Leftrightarrow\left|12,1.\left(x+0,1\right)\right|=12,1\)
\(\Leftrightarrow\left[{}\begin{matrix}12,1.\left(x+0,1\right)=12,1\\12,1.\left(x+0,1\right)=-12,1\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x+0,1=1\\x+0,1=-1\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0,9\\x=-1,1\end{matrix}\right.\)
Vậy ................
b/ \(\left|0,2x-3,1\right|+\left|0,2x+3,1\right|=0\)
Mà \(\left\{{}\begin{matrix}\left|0,2x-3,1\right|\ge0\\\left|0,2x+3,1\right|\ge0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\left|0,2x-3,1\right|=0\\\left|0,2x+3,1\right|=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}0,2x-3,1=0\\0,2x+3,1=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}0,2x=3,1\\0,2x=-3,1\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=15,5\\x=-15,5\end{matrix}\right.\)
Vậy ..
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/ \(\left|12,1x+1,1.01\right|=12,1\)
\(\Leftrightarrow\left|12,1x+0,11\right|=12,1\)
\(\Leftrightarrow\left[{}\begin{matrix}12,1x+0,11=12,1\\12,1x+0,11=-12,1\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}12,1x=11,99\\12,1x=-12,21\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{11,99}{12,1}\\x=\dfrac{-12,21}{12,1}\end{matrix}\right.\)
\(a.\)
\(\left|0,2x-3,1\right|=6,3\)
\(\Rightarrow\left[\begin{array}{nghiempt}0,2x-3,1=6,3\\0,2x-3,1=-6,3\end{array}\right.\)
\(\Rightarrow\left[\begin{array}{nghiempt}0,2x=6,3+3,1\\0,2=-6,3+3,1\end{array}\right.\)
\(\Rightarrow\left[\begin{array}{nghiempt}0,2x=9,4\\0,2x=-3,2\end{array}\right.\)
\(\Rightarrow\left[\begin{array}{nghiempt}x=9,4:0,2\\x=-3,2:0,2\end{array}\right.\)
\(\Rightarrow\left[\begin{array}{nghiempt}x=47\\x=-16\end{array}\right.\)
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 .............................
|0,2 . x - 3,1| + |0,2 . x + 3,1| = 0
Ta có: \(\hept{\begin{cases}\left|0,2.x-3,1\right|\ge0\\\left|0,2.x+3,1\right|\ge0\end{cases}}\)
Mà theo đề bài: |0,2 . x - 3,1| + |0,2 . x + 3,1| = 0
\(\Rightarrow\hept{\begin{cases}0,2.x-3,1=0\\0,2.x+3,1=0\end{cases}}\Rightarrow\hept{\begin{cases}0,2.x=3,1\\0,2.x=-3,1\end{cases}}\Rightarrow\hept{\begin{cases}x=15,5\\x=-15,5\end{cases}}\)
Vay x = 15,5 ; -15,5
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 ^_^