Tìm x biết: e) trị tuyệt đối của( x + 19 5) + trị tuyệt đối của( y + 1890 /1975 ) + | z − 2004 | = 0 ;

e) \(\left| {x + \frac{{19}}{5}} \right| + \left| {y + \frac{{1890}}{{1975}}} \right| + \left| {z - 2004} \right| = 0\)

Vì \[\left| {x + \frac{{19}}{5}} \right| \ge 0\]; \[\left| {y + \frac{{1890}}{{1975}}} \right| \ge 0\]; \[\left| {z - 2004} \right| \ge 0\]

Khi đó để \(\left| {x + \frac{{19}}{5}} \right| + \left| {y + \frac{{1890}}{{1975}}} \right| + \left| {z - 2004} \right| = 0\) thì\[\left\{ \begin{array}{l}\left| {x + \frac{{19}}{5}} \right| = 0\\\left| {y + \frac{{1890}}{{1975}}} \right| = 0\\\left| {z - 2004} \right| = 0\end{array} \right.\]

Suy ra\[\left\{ \begin{array}{l}x + \frac{{19}}{5} = 0\\x + \frac{{1890}}{{1975}} = 0\\z - 2004 = 0\end{array} \right.\], do đó \[\left\{ \begin{array}{l}x = - \frac{{19}}{5}\\y = - \frac{{1890}}{{1975}} = - \frac{{378}}{{395}}\\z = 2004\end{array} \right.\].

Vậy \[\left( {x;y;z} \right) = \left( { - \frac{{19}}{5}; - \frac{{378}}{{395}};2004} \right)\]