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 \(x=\dfrac{1}{5};y=-0,4;z=3\)

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

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

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

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

Ta có :

(x−15)2014+(y+0,4)100+(z−3)678=0(x−15)2014+(y+0,4)100+(z−3)678=0

Mà ⎧⎩⎨⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪(x−15)2014≥0(y+0,4)100≥0(z−3)678≥0{(x−15)2014≥0(y+0,4)100≥0(z−3)678≥0

⇔(x−15)2014+(y+0,4)100+(z−3)678≥0⇔(x−15)2014+(y+0,4)100+(z−3)678≥0

Lại có : (x−15)2014+(y+0,4)100+(z−3)678=0(x−15)2014+(y+0,4)100+(z−3)678=0

⇔⎧⎩⎨⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪(x−15)2014=0(y+0,4)100=0(z−3)678=0⇔{(x−15)2014=0(y+0,4)100=0(z−3)678=0

⇔⎧⎩⎨⎪⎪⎪⎪⎪⎪x−15=0y+0,4=0z−3=0⇔{x−15=0y+0,4=0z−3=0

⇔⎧⎩⎨⎪⎪⎪⎪⎪⎪x=15y=−0,4z=3

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

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

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

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

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

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

Vậy \(x=\frac{1}{5};y=-0,4;z=3\)

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

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

= \(\frac{4}{5}\)✖ \(\frac{225}{16}\)

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

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

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 \(\left(x,y,z\right)=\left(\dfrac{1}{5};\dfrac{-2}{5};3\right)\)

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

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

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