`|x+ 3/4 |+ | y - 1/5| + | x+ y +z | = 0`
Vì :`|x + 3/4| ; |y - 1/5| ; |x+y+z|` $≥$ $0$ $∀$ $x;y;z$
`⇒ |x+3/4 |= |y - 1/5 | = |x+y+z| = 0`
$⇒$ $\left \{ {{x + \dfrac{3}{4} = 0} \atop {y - \dfrac{1}{5}=0}} \atop {x + y+ z =0} \right.$
$⇒$ $\left \{ {{x= - \dfrac{3}{4} } \atop {y = \dfrac{1}{5}}} \atop {\dfrac{-3}{4} + \dfrac{1}{5} + z=0} \right.$
$⇔$ $\left \{ {{x= - \dfrac{3}{4} } \atop {y = \dfrac{1}{5}}} \atop {\dfrac{-11}{20} + z=0} \right.$
$⇔$ $\left \{ {{x= - \dfrac{3}{4} } \atop {y = \dfrac{1}{5}}} \atop {z = \dfrac{11}{20}} \right.$
Vậy `(x;y;z)=({-3}/4;1/5;{11}/{20})`
