Question

(x - \(\frac{1}{5}\) )\(^{2004}\) + ( y + 0,4)\(^{100}\) +(z - 3)\(^{678}\) = 0
tìm x, y, z?

Answer 1

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

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

\(\Leftrightarrow\left\{{}\begin{matrix}\left(x-\frac{1}{5}\right)^{2004}=0\\\left(y+0,4\right)^{100}=0\\\left(z-3\right)^{678}=0\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}x-\frac{1}{5}=0\\y+0,4=0\\z-3=0\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}x=\frac{1}{5}\\x=-0,4=-\frac{2}{5}\\z=3\end{matrix}\right.\)

Vậy \(\left(x,y,z\right)\in\left\{\left(\frac{1}{5};-\frac{2}{5};3\right)\right\}\)

Answer 2

(x-\(\frac{1}{5}\))²⁰⁰⁴+(y+0,4)¹⁰⁰+(z-3)⁶⁷⁸=0

⇒\(\left\{{}\begin{matrix}x-\frac{1}{5}=0\\y+0,4=0\\z-3=0\end{matrix}\right.\)

⇒\(\left\{{}\begin{matrix}x=\frac{1}{5}\\y=-0,4\\z=3\end{matrix}\right.\)

Vậy...

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