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\(\left(x-\frac{1}{5}\right)^{2004}+\left(y+0,4\right)^{100}+\left(z-3\right)^{678}=0\)
Do \(\left(x-\frac{1}{5}\right)^{2004};\left(y+0,4\right)^{100};\left(z-3\right)^{678}\ge0\forall x,y,z\)
\(\Rightarrow\left(x-\frac{1}{5}\right)^{2004}+\left(y+0,4\right)^{100}+\left(z-3\right)^{678}=0\)
\(\Leftrightarrow\hept{\begin{cases}x-\frac{1}{5}=0\\y+0,4=0\\z-3=0\end{cases}}\Leftrightarrow\hept{\begin{cases}x=0,2\\y=-0,4\\z=3\end{cases}}\)
....
\(3x=y\)=> \(\frac{x}{1}=\frac{y}{3}\)
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
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:
\(\frac{x}{2}=\frac{y}{-5}=\frac{x-y}{2-\left(-5\right)}=\frac{10}{7}\)
\(\Rightarrow\left\{{}\begin{matrix}x=\frac{20}{7}\\y=\frac{-50}{7}\end{matrix}\right.\)
Ta có:
\(\left\{{}\begin{matrix}\left(x-\frac{1}{5}\right)^{2004}\ge0\\\left(y+0,4\right)^{100}\ge0\\\left(z-3\right)^{678}\ge0\\\left(x-\frac{1}{5}\right)^{2004}+\left(y+0,4\right)^{100}+\left(z-3\right)^{678}=0\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}\left(x-\frac{1}{5}\right)^{2004}=0\\\left(y+0,4\right)^{100}=0\\\left(z-3\right)^{678}=0\\\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}\left(x-\frac{1}{5}\right)=0\\\left(y+0,4\right)=0\\\left(z-3\right)=0\\\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}x=\frac{1}{5}\\y=-0,4\\z=3\end{matrix}\right.\)