a) \(\left|1-x\right|+\left|y-\frac{2}{3}\right|+\left|x+z\right|=0\)

\(\Leftrightarrow\hept{\begin{cases}1-x=0\\y-\frac{2}{3}=0\\x+z=0\end{cases}\Leftrightarrow\hept{\begin{cases}x=1-0=1\\y=0+\frac{2}{3}=\frac{2}{3}\\z=0-1=-1\end{cases}}}\)

Vậy \(x=1,y=\frac{2}{3},z=-1\)

b) \(\left|\frac{1}{4}-x\right|+\left|x+y+z\right|+\left|\frac{2}{3}+y\right|=0\)

\(\Leftrightarrow\hept{\begin{cases}\frac{1}{4}-x=0\\x+y+z=0\\\frac{2}{3}+y=0\end{cases}\Leftrightarrow\hept{\begin{cases}x=\frac{1}{4}-0=\frac{1}{4}\\x+y+z=0\\y=0+\frac{2}{3}=\frac{2}{3}\end{cases}\Leftrightarrow}\hept{\begin{cases}x=\frac{1}{4}\\z=0-\frac{1}{4}-\frac{2}{3}=\frac{-11}{12}\\y=\frac{2}{3}\end{cases}}}\)

Vậy \(x=\frac{1}{4},y=\frac{-11}{12},z=\frac{2}{3}\)

a)x=78

b)x=20

c)x=10

d)x=25

e)x=50

( x - 78 ) . 2015 = 0

x - 78=0:2015

x - 78=0

x=0+78

x=78

( y - 14 ) : 2 = 3

y - 14=3.2

y - 14=6

y=6+14

y=20

y - 14 : 2 = 3

y - 7 = 3

y=3+7

y=10

124 + 118 - x = 217

118 - x = 217-124

118 - x = 93

x=118-93

x=25

x - 32 : 16 = 48

x - 2 = 48

x=48+2

x=50

\(\left(y:2\right)+14=32\\ y:2=32-14\\ y:2=18\\ y=18\times2\\ y=36\)

Ta có: |15/32 - x| ≥ 0; |4/25 - y| ≥ 0; |z - 14/31| ≥ 0

=> |15/32 - x| +|4/25 - y|+ |z - 14/31| ≥ 0

Mà |15/32 - x| +|4/25 - y|+ |z - 14/31| < 0

=> x, y, z ∈ \(\varnothing\)