(x-1/5)2004+(y+0,4)100+(z-3)678=0
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20 tháng 3 2020
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
NT
0
T
2 tháng 11 2018
\(\left(x-\frac{1}{5}\right)^{2004}+\left(y+0,4\right)^{100}+\left(z-3\right)^6=0.\)
\(Nx:\left(x-\frac{1}{5}\right)^{2004}\ge0;\left(y+0,4\right)^{100}\ge0;\left(z-3\right)^{678}\ge0\)
\(\Rightarrow VT=0\Leftrightarrow\hept{\begin{cases}\left(x-\frac{1}{5}\right)^{2004}=0\\\left(y+0,4\right)^{100}=0\\\left(z-3\right)^{678}=0\end{cases}}\)
\(\left(x-\frac{1}{5}\right)^{2004}\Leftrightarrow x-\frac{1}{5}=0\Leftrightarrow x=\frac{1}{5}\)
\(\left(y+0,4\right)^{100}=0\Leftrightarrow y+0,4=0\Leftrightarrow y=-0,4\)
\(\left(z-3\right)^{678}=0\Leftrightarrow z-3=0\Leftrightarrow z=3\)
Vậy \(x=\frac{1}{5};y=-0,4;z=3\)
PQ
1
13 tháng 7 2021
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 \(\left(x,y,z\right)=\left(\dfrac{1}{5};\dfrac{-2}{5};3\right)\)
LD
2
17 tháng 1 2022
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 \(\left(x,y,z\right)=\left(\dfrac{1}{5};-0,4;3\right)\)
17 tháng 1 2022
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 \(x=\dfrac{1}{5};y=-0,4;z=3\)
NT
1
VM
23 tháng 12 2019
\(\left(x-\frac{1}{5}\right)^{2004}+\left(y+0,4\right)^{100}+\left(z-3\right)^{678}=0\)
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!
AL
0
NT
1
27 tháng 5 2020
a) \(\frac{4}{5}\)✖\(\left(\frac{7}{2}+\frac{1}{4}\right)^2\)
= \(\frac{4}{5}\)✖ \((\frac{15}{4})^2\)
= \(\frac{4}{5}\)✖ \(\frac{225}{16}\)
= \(\frac{1}{1}\times\frac{45}{4}\)
= \(\frac{45}{4}\)
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 \(\left\{{}\begin{matrix}\left(x-\dfrac{1}{5}\right)^{2004}=0\\\left(y+0,4\right)^{100}=0\\\left(z-3\right)^{678}=0\end{matrix}\right.=>\left\{{}\begin{matrix}x=\dfrac{1}{5}\\y=-0,4\\z=3\end{matrix}\right.\)