Question

(x-1/5)^2004 +(y+0,4)^100 + (z-3)^678 = 0

Answer 1

\(\left(x-\frac{1}{5}\right)^{2004}+\left(y+0,4\right)^{100}+\left(z-3\right)^6=0.\)

\(Nx:\left(x-\frac{1}{5}\right)^{2004}\ge0;\left(y+0,4\right)^{100}\ge0;\left(z-3\right)^{678}\ge0\)

\(\Rightarrow VT=0\Leftrightarrow\hept{\begin{cases}\left(x-\frac{1}{5}\right)^{2004}=0\\\left(y+0,4\right)^{100}=0\\\left(z-3\right)^{678}=0\end{cases}}\)

\(\left(x-\frac{1}{5}\right)^{2004}\Leftrightarrow x-\frac{1}{5}=0\Leftrightarrow x=\frac{1}{5}\)

\(\left(y+0,4\right)^{100}=0\Leftrightarrow y+0,4=0\Leftrightarrow y=-0,4\)

\(\left(z-3\right)^{678}=0\Leftrightarrow z-3=0\Leftrightarrow z=3\)

Vậy \(x=\frac{1}{5};y=-0,4;z=3\)