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

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

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

diễn giải giúp mình nha bạn

Vì (x-1/5)^2004 >= 0 với mọi x

    (y+0,4)^100 >= 0 với mọi y

    (z-3)^678 >= với mọi z

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

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

=> x-1/5 =0

     y+0,4 = 0

     z-3=0

=>x = 1/5; y =-0.4; z = 3

vậy .........................

Ta có:

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

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

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

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

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.

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

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

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

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

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

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

Bài 1:

a)Ta có:

\(\frac{4}{5}\left(\frac{7}{2}+\frac{1}{4}\right)^2=\frac{4}{5}\left(\frac{15}{4}\right)^2=\frac{4}{5}.\frac{15}{4}.\frac{15}{4}=\frac{45}{4}\)

b)Ta có:

\(\frac{5^4.20^4}{25^5.4^5}=\frac{\left(5.20\right)^4}{\left(25.4\right)^5}=\frac{100^4}{100^5}=\frac{1}{100}\)

Bài 2:

Ta có:

\(\frac{x}{2}=\frac{y}{-5}=\frac{x-y}{2-\left(-5\right)}=\frac{10}{7}\)

\(\Rightarrow\left\{{}\begin{matrix}x=\frac{20}{7}\\y=\frac{-50}{7}\end{matrix}\right.\)

Bài 3:

Ta có:

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

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

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

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