Ta có : \(42+\frac{3}{7}\left|3x-1\right|=12\)
=> \(\left|3x-1\right|=-70\)
TH1 : \(3x-1\ge0\left(x\ge\frac{1}{3}\right)\)
=> \(3x-1=-70\)
=> \(3x=-69\)
=> \(x=-23\) ( KTM )
TH2 : \(3x-1< 0\left(x< \frac{1}{3}\right)\)
=> \(1-3x=-70\)
=> \(-3x=-71\)
=> \(x=\frac{71}{3}\) ( KTM )
Vậy phương trình trên vô nghiệm .
a, |x - 1 | = | 2x + 3 |
b, |x - 1| + 3x = 1
c, |5x - 3 | - x = 7
d, |y| + |y - 2 | = 2
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tìm x biết
a,4/3x-1/+/x/-2/x-5/+7/x-3/=12
Tìm x biết :
a) 5/6 + (2/3x- -7/12) = 1/2
b) x/15 + 1,2 = 1/3
c) 7/1 - 5/18 = -3/2
d) 3 + | x-1,34 | = 5
e) 5-|x+2,56|=3
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\(a.A=[\frac{1,5+1-0,75}{2,5+\frac{5}{3}-1,25}+\frac{0,375-0,3+\frac{3}{11}+\frac{3}{12}}{-0,625+0,5-\frac{5}{11}-\frac{5}{12}}]+\frac{1890}{2005}+115\)
b.B=\(\left[\frac{1\frac{11}{31}\cdot4\frac{3}{7}-\left(15-6\cdot\frac{1}{3}\cdot\frac{2}{19}\right)}{4\frac{5}{6}+\frac{1}{6}\left(42-5\frac{1}{3}\right)}\cdot\left(-1\frac{19}{93}\right)\right]\cdot\frac{31}{50}\)
