\(\left(x-\frac{1}{5}\right)^{2004}+\left(y+0,4\right)^{100}+\left(z-3\right)^6=0.\)

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

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

\(\left(x-\frac{1}{5}\right)^{2004}\Leftrightarrow x-\frac{1}{5}=0\Leftrightarrow x=\frac{1}{5}\)

\(\left(y+0,4\right)^{100}=0\Leftrightarrow y+0,4=0\Leftrightarrow y=-0,4\)

\(\left(z-3\right)^{678}=0\Leftrightarrow z-3=0\Leftrightarrow z=3\)

Vậy \(x=\frac{1}{5};y=-0,4;z=3\)

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

\(\left(y+\dfrac{2}{5}\right)^{100}\ge0\forall y\)

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

Do đó: \(\left(x-\dfrac{1}{5}\right)^{2004}+\left(y+\dfrac{2}{5}\right)^{100}+\left(z-3\right)^{678}\ge0\forall x,y,z\)

Dấu '=' xảy ra khi \(\left(x,y,z\right)=\left(\dfrac{1}{5};\dfrac{-2}{5};3\right)\)

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

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 \(x=\dfrac{1}{5};y=-0,4;z=3\)

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

Ta có:

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

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

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

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

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

Vậy \(\left(x;y;z\right)\in\left\{\frac{1}{5};-0,4;3\right\}.\)

Chúc bạn học tốt!

vì (x-1/5)²⁰⁰⁴≥0 với mọi x

(y+0,4)¹⁰⁰≥0 với mọi y

(z-3)⁶⁷⁸≥0 với mọi z

=>(x-1/5)²⁰⁰⁴+(y+0.4)¹⁰⁰+(x-3)⁶⁷⁸≥0 với mọi x,y,z

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

a) \(\frac{4}{5}\)✖\(\left(\frac{7}{2}+\frac{1}{4}\right)^2\)

= \(\frac{4}{5}\)✖ \((\frac{15}{4})^2\)

= \(\frac{4}{5}\)✖ \(\frac{225}{16}\)

= \(\frac{1}{1}\times\frac{45}{4}\)

= \(\frac{45}{4}\)