Question

tìm ba số x ; y; z biết  :      (x-\(\dfrac{1}{5}\))²⁰⁰⁴   +      (y +0,4)¹⁰⁰   +   (z-3)⁶⁷⁸    = 0

 

Answer 1

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

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

Dấu "=" xảy ra \(\Leftrightarrow\left\{{}\begin{matrix}x-\dfrac{1}{5}=0\\y+0,4=0\\z-3=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{1}{5}\\y=-0,4\\z=3\end{matrix}\right.\)

Vậy \(\left(x,y,z\right)=\left(\dfrac{1}{5};-0,4;3\right)\)

Answer 2

Vì \(\left(x-\dfrac{1}{5}\right)^{2004}\ge0\forall x\)

\(\left(y+0,4\right)^{100}\ge\forall y\)

\(\left(z-3\right)^{678}\ge0\forall z\)

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

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

Dấu ''='' xảy ra khi \(x=\dfrac{1}{5};y=-0,4;z=3\)