tìm x,y thỏa mãn /2x-2/^2017+3y+10)^2018=0
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Ta có: \(3x^2+3y^2+4xy+2x-2y+2=0\)
\(\Leftrightarrow x^2+2x+1+y^2-2y+1+2x^2+4xy+2y^2=0\)
\(\Leftrightarrow\left(x+1\right)^2+\left(y-1\right)^2+2\left(x^2+2xy+y^2\right)=0\)
\(\Leftrightarrow\left(x+1\right)^2+\left(y-1\right)^2+2\left(x+y\right)^2=0\)
Ta có: \(\left(x+1\right)^2\ge0\forall x\)
\(\left(y-1\right)^2\ge0\forall y\)
\(2\left(x+y\right)^2\ge0\forall x,y\)
Do đó: \(\left(x+1\right)^2+\left(y-1\right)^2+2\left(x+y\right)^2\ge0\forall x,y\)
Dấu '=' xảy ra khi
\(\left\{{}\begin{matrix}x+1=0\\y-1=0\\x+y=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-1\\y=1\\-1+1=0\left(đúng\right)\end{matrix}\right.\)
Thay x=-1 và y=1 vào biểu thức \(M=\left(x+y\right)^{2016}+\left(x+2\right)^{2017}+\left(y-1\right)^{2018}\), ta được:
\(M=\left(-1+1\right)^{2016}+\left(-1+2\right)^{2017}+\left(1-1\right)^{2018}\)
\(=0^{2016}+1^{2017}+0^{2018}=1\)
Vậy: M=1
![](https://rs.olm.vn/images/avt/0.png?1311)
\(x^2+y^2+xy+3x-3y+9=0\)
\(\Leftrightarrow2x^2+2y^2+2xy+6x-6y+18=0\)
\(\Leftrightarrow\left(x^2+2xy+y^2\right)+\left(x^2+6x+9\right)+\left(y^2-6y+9\right)=0\)
\(\Leftrightarrow\left(x+y\right)^2+\left(x+3\right)^2+\left(y-3\right)^2=0\)
\(\Leftrightarrow x=-3;y=3\)
Thay vào:\(Q=\left(3-3+1\right)^{2017}+\left(2-3\right)^{2018}=2\)
![](https://rs.olm.vn/images/avt/0.png?1311)
1) (x + 2016)2016 + |y - 2017|2017 = 0
\(\Leftrightarrow\hept{\begin{cases}\left(x+2016\right)^{2016}=0\\\left|y-2017\right|^{2017}=0\end{cases}}\Leftrightarrow\hept{\begin{cases}x+2016=0\\y-2017=0\end{cases}}\Leftrightarrow\hept{\begin{cases}x=-2016\\y=2017\end{cases}}\)
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\(\left(2x-5\right)^{2018}+\left(3y+4\right)^{2020}\le0\\ \Leftrightarrow\left(2x-5\right)^{2018}+\left(3y+4\right)^{2020}=0\\ \Leftrightarrow\left\{{}\begin{matrix}\left(2x-5\right)^{2018}=0\\\left(3y+4\right)^{2020}=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{5}{2}\\y=-\dfrac{4}{3}\end{matrix}\right.\\ \Leftrightarrow M=6x^2+9xy-y^2-5x^2+2xy=x^2+11xy-y^2\\ \Leftrightarrow M=\dfrac{25}{4}-11\cdot\dfrac{4}{3}\cdot\dfrac{5}{2}-\dfrac{16}{9}=\dfrac{25}{4}-\dfrac{110}{3}-\dfrac{16}{9}=-\dfrac{1159}{36}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\left|2x-2\right|^{2017}+\left(3y+10\right)^{2018}=0\left(1\right)\)
Ta thấy: \(\begin{cases}\left|2x-2\right|^{2017}\ge0\\\left(3y+10\right)^{2018}\ge0\end{cases}\)
\(\Rightarrow\left|2x-2\right|^{2017}+\left(3y+10\right)^{2018}\ge0\left(2\right)\)
Từ (1) và (2) suy ra \(\begin{cases}\left|2x-2\right|^{2017}=0\\\left(3y+10\right)^{2018}=0\end{cases}\)\(\Rightarrow\begin{cases}2x-2=0\\3y+10=0\end{cases}\)
\(\Rightarrow\begin{cases}2x=2\\3y=-10\end{cases}\)\(\Rightarrow\begin{cases}x=1\\y=-\frac{10}{3}\end{cases}\)
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