tìm x , y biết
\(|2x-27|^{2019}+|3y+10|^{2020}=0\)
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LM
1
T
23 tháng 11 2023
Ta có: \(\left\{{}\begin{matrix}\left(2x-3y\right)^{2018}\ge0\forall x,y\\\left(3y-4z\right)^{2020}\ge0\forall y,z\\\left|2x+3y-z-63\right|\ge0\forall x,y,z\end{matrix}\right.\)
\(\Rightarrow\left(2x-3y\right)^{2018}+\left(3y-4z\right)^{2020}+\left|2x+3y-z-63\right|\ge0\forall x,y,z\)
Mà: \(\left(2x-3y\right)^{2018}+\left(3y-4z\right)^{2020}+\left|2x+3y-z-63\right|=0\)
nên: \(\left\{{}\begin{matrix}2x-3y=0\\3y-4z=0\\2x+3y-z-63=0\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}2x=3y\\3y=4z\\z=2x+3y-63\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}2x=4z\\3y=4z\\z=4z+4z-63\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=4z:2\\y=4z:3\\z=8z-63\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}x=2z\\y=4z:3\\-7z=-63\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=2\cdot9=18\\y=4\cdot9:3=12\\z=9\end{matrix}\right.\)
Vậy \(x=18;y=12;z=9\).
$Toru$
YN
28 tháng 4 2022
Có:
\(\left|2x-27\right|^{2011}\ge0\forall x\)
\(\left(3y+10\right)^{2012}\ge0\forall y\)
\(\Rightarrow\left|2x-27\right|^{2011}+\left(3y+10\right)^{2012}\ge0\forall x;y\)
\(\Rightarrow\left|2x-27\right|^{2011}+\left(3y+10\right)^{2012}=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}2x-27=0\\3y+10=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{27}{2}\\y=-\dfrac{10}{3}\end{matrix}\right.\)
NT
1
DH
28 tháng 11 2019
Ta có : \(\left|2x-27\right|\ge0\forall x\Rightarrow\left|2x-27\right|^{2019}\ge0\forall x\)
\(\left(3x+10\right)^2\ge0\forall x\Rightarrow\left[\left(3x+10\right)^2\right]^{1009}=\left(3y+10\right)^{2018}\ge0\forall x\)
\(\Rightarrow\left|2x-27\right|^{2019}+\left(3y+10\right)^{2018}\ge0\forall x\)
Dấu " = " xảy ra \(\Leftrightarrow\left|2x-27\right|=\left(3y+10\right)=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=\frac{27}{2}\\y=-\frac{10}{3}\end{matrix}\right.\)
VG
0
BT
1
MH
23 tháng 12 2021
Vì \(\left\{{}\begin{matrix}\left|2x-27\right|^{2011}\text{≥0,∀x}\\\left(3y+10\right)^{2012}\text{≥0,∀y}\end{matrix}\right.\)
⇒ \(\left|2x-27\right|^{2011}+\left(3y+10\right)^{2012}\text{≥0,∀x},y\)
Dấu "=" ⇔ \(\left\{{}\begin{matrix}2x-27=0\\3y+10=0\end{matrix}\right.\)
⇒ \(\left\{{}\begin{matrix}x=\dfrac{27}{2}\\y=-\dfrac{10}{3}\end{matrix}\right.\)
Vậy ...
LN
3
XO
18 tháng 12 2020
Ta có \(\hept{\begin{cases}\left|2x-27\right|^{2011}\ge0\forall x\\\left(3y+10\right)^{2022}\ge0\forall y\end{cases}}\Rightarrow\left|2x-27\right|^{2011}+\left(3y+10\right)^{2022}\ge0\forall x;y\)
Dấu "=" xảy ra <=> \(\hept{\begin{cases}2x-27=0\\3y+10=0\end{cases}}\Rightarrow\hept{\begin{cases}x=\frac{27}{2}\\y=-\frac{10}{3}\end{cases}}\)
Vậy x = 27/2 ; y = -10/3 là giá trị cần tìm
18 tháng 12 2020
ta có |2x-27| > hoặc = 0=> |2x-27|^2011> hoặc = 0
(3y+10)^2012> hoặc 0 mà |2x-27|^2011+(3y+10)^2012=0
=>2x-27=0 hoặc 3y+10=0=>2x=27 hoặc 3y=-10
=>x=13,5 hoặc x=-10/3
vậy .............................
G
2
22 tháng 4 2019
Ta thấy \(\frac{2019}{3}.|x-3y|\ge0\forall x,y\)
\(|2x-2|\ge0\forall x\)
\(\Rightarrow\frac{7}{4}-\frac{2019}{3}.|x-3y|+|2x-2|+2020\ge\frac{1}{2}-0+2020\)
Hay \(C\ge\frac{4041}{2}\)
Dấu "=" xảy ra \(\Leftrightarrow\hept{\begin{cases}x-3y=0\\2x-2=0\end{cases}}\)
\(\Leftrightarrow\hept{\begin{cases}y=\frac{1}{3}\\x=1\end{cases}}\)
Vậy Min \(C=\frac{4041}{2}\)\(\Leftrightarrow\hept{\begin{cases}y=\frac{1}{3}\\x=1\end{cases}}\)
\(\left|2x-27\right|^{2019}+\left|3y+10\right|^{2020}=0\)
Ta có:
\(\left\{{}\begin{matrix}\left|2x-27\right|^{2019}\ge0\\\left|3y+10\right|^{2020}\ge0\end{matrix}\right.\forall x,y.\)
\(\Rightarrow\left|2x-27\right|^{2019}+\left|3y+10\right|^{2020}=0\)
\(\Rightarrow\left\{{}\begin{matrix}\left|2x-27\right|^{2019}=0\\\left|3y+10\right|^{2020}=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}\left|2x-27\right|=0\\\left|3y+10\right|=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}2x-27=0\\3y+10=0\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}2x=27\\3y=-10\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=27:2\\y=\left(-10\right):3\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=\frac{27}{2}\\y=-\frac{10}{3}\end{matrix}\right.\)
Vậy \(\left(x;y\right)\in\left\{\frac{27}{2};-\frac{10}{3}\right\}.\)
Chúc bạn học tốt!