x=1/5; y=-0.4; z=3

diễn giải giúp mình nha bạn

Vì (x-1/5)²⁰¹⁴ ; (y+0,4)¹⁰⁰; và (z-3)⁶⁷⁸ đều có mũ chẵn nên > 0

mà (x-1/5)²⁰⁰⁴+(y+0,4)¹⁰⁰+(z-3)⁶⁷⁸=0

=> x-1/5=0 và y+0,4=0 và z-3=0

=> x=1/5 và y=-0,4 và z=3.

Ta có:

\(\left\{{}\begin{matrix}\left(x-\frac{1}{5}\right)^{2004}\ge0\\\left(y+0,4\right)^{100}\ge0\\\left(z-3\right)^{678}\ge0\\\left(x-\frac{1}{5}\right)^{2004}+\left(y+0,4\right)^{100}+\left(z-3\right)^{678}=0\end{matrix}\right.\)

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

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

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

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

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

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

\(\Leftrightarrow\hept{\begin{cases}x-\frac{1}{5}=0\\y+0,4=0\\z-3=0\end{cases}}\Leftrightarrow\hept{\begin{cases}x=0,2\\y=-0,4\\z=3\end{cases}}\)

....

Tham khảo :

https://olm.vn/hoi-dap/detail/243970516929.html

\(3x=y\)=>  \(\frac{x}{1}=\frac{y}{3}\)

hay  \(\frac{x}{4}=\frac{y}{12}\)

\(5y=4z\)=>  \(\frac{y}{4}=\frac{z}{5}\)

hay  \(\frac{y}{12}=\frac{z}{15}\)

suy ra:   \(\frac{x}{4}=\frac{y}{12}=\frac{z}{15}\)

đến đây bạn ADTCDTSBN nhé

\(\left(x-\dfrac{1}{5}\right)^{2004}+\left(y+0.4\right)^{100}+\left(z-3\right)^{678}=0\)

\(\Rightarrow\left\{{}\begin{matrix}x-\dfrac{1}{5}=0\\y+0.4=0\\z-3=0\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}x=\dfrac{1}{5}\\y=-0.4\\z=3\end{matrix}\right.\)

Vì (x-1/5)²⁰⁰⁴ \(\ge\)0

(y+0,4)¹⁰⁰\(\ge\)0

(z-3)⁶⁷⁸\(\ge\)0

=>(x-1/5)²⁰⁰⁴+(y+0,4)¹⁰⁰+(z-3)⁶⁷⁸\(\ge0\)

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

Vậy x=1/5,y=-0,4,z=3

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

Nên để biểu thức bằng 0 \(\Leftrightarrow\hept{\begin{cases}x-\frac{1}{5}=0\\y+0,4=0\\z-3=0\end{cases}}\Leftrightarrow\hept{\begin{cases}x=\frac{1}{5}\\y=-0,4\\z=3\end{cases}}\)