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16 tháng 10 2019
\(\left(2x-5\right)^{2020}+\left(3y+4\right)^{2018}\le0\left(1\right)\)
Ta có: \(\hept{\begin{cases}\left(2x-5\right)^{2020}\ge0;\forall x,y\\\left(3y+4\right)^{2018}\ge0;\forall x,y\end{cases}}\)\(\Rightarrow\left(2x-5\right)^{2020}+\left(3y+4\right)^{2018}\ge0;\forall x,y\left(2\right)\)
Từ \(\left(1\right)\)và \(\left(2\right)\)\(\Rightarrow\left(2x-5\right)^{2020}+\left(3y+4\right)^{2018}=0\)
\(\Leftrightarrow\hept{\begin{cases}\left(2x-5\right)^{2020}=0\\\left(3y+4\right)^{2018}=0\end{cases}}\Leftrightarrow\hept{\begin{cases}x=\frac{5}{2}\\y=\frac{-4}{3}\end{cases}}\)
Vậy...
27 tháng 9 2016
[2x-5]^2016+[3y+4]^2014<hoặc=0
=>2x-5=0 và 3y+4=0 (vì [2x-5]^2016+[3y+4]^2014>hoặc=0 với mọi x;y)
=>x=5/2 và y=-4/3
vậy x=5/2 và y=-4/3
VM
16 tháng 10 2019
\(\left(2x-5\right)^{2020}+\left(3y+4\right)^{2018}\le0\)
Ta có:
\(\left\{{}\begin{matrix}\left(2x-5\right)^{2020}\ge0\\\left(3y+4\right)^{2018}\ge0\end{matrix}\right.\forall xy.\)
\(\Rightarrow\left(2x-5\right)^{2020}+\left(3y+4\right)^{2018}\ge0\) \(\forall xy.\)
Mà \(\left(2x-5\right)^{2020}+\left(3y+4\right)^{2018}\le0.\)
\(\Rightarrow\left(2x-5\right)^{2020}+\left(3y+4\right)^{2018}=0\)
\(\Rightarrow\left(2x-5\right)+\left(3y+4\right)=0\)
\(\Rightarrow\left\{{}\begin{matrix}2x-5=0\\3y+4=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}2x=5\\3y=-4\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=5:2\\y=\left(-4\right):3\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}x=\frac{5}{2}\\y=-\frac{4}{3}\end{matrix}\right.\)
Vậy \(\left(x;y\right)\in\left\{\frac{5}{2};-\frac{4}{3}\right\}.\)
Chúc bạn học tốt!
DQ
1
VM
1 tháng 10 2019
\(\left(2x-5\right)^{2016}+\left(3y+4\right)^{2018}\le0\)
Ta có:
\(\left\{{}\begin{matrix}\left(2x-5\right)^{2016}\ge0\\\left(3y+4\right)^{2018}\ge0\end{matrix}\right.\forall x.\)
\(\Rightarrow\left(2x-5\right)^{2016}+\left(3y+4\right)^{2018}=0\)
\(\Leftrightarrow\left[{}\begin{matrix}\left(2x-5\right)^{2016}=0\\\left(3y+4\right)^{2016}=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x-5=0\\3y+4=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=0+5=5\\3y=0-4=-4\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=5:2\\y=\left(-4\right):3\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\frac{5}{2}\\y=-\frac{4}{3}\end{matrix}\right.\)
Vậy \(\left(x;y\right)\in\left\{\frac{5}{2};-\frac{4}{3}\right\}.\)
Chúc bạn học tốt!
VT
0
23 tháng 11 2021
\(\left(2x-5\right)^{2018}+\left(3y+4\right)^{2020}\le0\\ \Leftrightarrow\left(2x-5\right)^{2018}+\left(3y+4\right)^{2020}=0\\ \Leftrightarrow\left\{{}\begin{matrix}\left(2x-5\right)^{2018}=0\\\left(3y+4\right)^{2020}=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{5}{2}\\y=-\dfrac{4}{3}\end{matrix}\right.\\ \Leftrightarrow M=6x^2+9xy-y^2-5x^2+2xy=x^2+11xy-y^2\\ \Leftrightarrow M=\dfrac{25}{4}-11\cdot\dfrac{4}{3}\cdot\dfrac{5}{2}-\dfrac{16}{9}=\dfrac{25}{4}-\dfrac{110}{3}-\dfrac{16}{9}=-\dfrac{1159}{36}\)
1 tháng 8 2017
\(\left(5x-2\right)^{10}=\left(5x-2\right)^{100}\)
\(\Rightarrow\left(5x-2\right)^{100}-\left(5x-2\right)^{10}=0\)
\(\Rightarrow\left(5x-2\right)^{10}\left[\left(5x-2\right)^{90}-1\right]=0\)
\(\Rightarrow\left[{}\begin{matrix}\left(5x-2\right)^{10}=0\\\left(5x-2\right)^{90}-1=0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}\left(5x-2\right)^{10}=0\\\left(5x-2\right)^{90}=1\Rightarrow5x-2=\pm1\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}5x-2=0\Rightarrow5x=2\Rightarrow x=\dfrac{2}{5}\\5x=1;3\Rightarrow x=\dfrac{1}{5};\dfrac{3}{5}\end{matrix}\right.\)
\(\left(\dfrac{2x-3}{4}\right)^{2016}+\left(\dfrac{3y+4}{5}\right)^{2018}=0\)
\(\left\{{}\begin{matrix}\left(\dfrac{2x-3}{4}\right)^{2016}\ge0\forall x\\\left(\dfrac{3y+4}{5}\right)^{2018}\ge0\forall y\end{matrix}\right.\)
\(\Rightarrow\left(\dfrac{2x-3}{4}\right)^{2016}+\left(\dfrac{3y+4}{5}\right)^{2014}\ge0\)
Dấu "=" xảy ra khi:
\(\left\{{}\begin{matrix}\left(\dfrac{2x-3}{4}\right)^{2016}=0\Rightarrow\dfrac{2x-3}{4}=0\Rightarrow2x-3=0\Rightarrow2x=3\Rightarrow x=\dfrac{3}{2}\\\left(\dfrac{3y+4}{5}\right)^{2018}=0\Rightarrow\dfrac{3y+4}{5}=0\Rightarrow3y+4=0\Rightarrow3y=-4\Rightarrow y=\dfrac{-4}{3}\end{matrix}\right.\)