Question

Tìm x;y;z​​​\(\in Q\)a,\(\left|x+\frac{19}{5}\right|+\left|y+\frac{1890}{1975}\right|+\left|z-2004\right|=0\)

Answer 1

Vì \(\left|x+\frac{19}{5}\right|\ge0\) với \(\forall x\)

\(\left|y+\frac{1890}{1975}\right|\ge0\) với \(\forall y\)

\(\left|z-2004\right|\ge0\)với \(\forall z\)

\(\Rightarrow\left|x+\frac{19}{5}\right|+\left|y+\frac{1890}{1975}\right|+\left|z-2004\right|\ge0\)

Dấu "=" xảy ra \(\Leftrightarrow\hept{\begin{cases}\left|x+\frac{19}{5}\right|=0\\\left|y+\frac{1890}{1975}\right|=0\\\left|z-2004\right|=0\end{cases}}\Rightarrow\hept{\begin{cases}x=-\frac{19}{5}\\y=-\frac{1890}{1975}\\z=2004\end{cases}}\)