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ta có \(2x^2+2xy+2y^2+2x-2y+2=0\)
<=>\(x^2+2xy+y^2+x^2+2x+1+y^2-2y+1=0\)
<=>\(\left(x+y\right)^2+\left(x+1\right)^2+\left(y-1\right)^2=0\)
<=>\(\hept{\begin{cases}x=-1\\y=1\end{cases}}\)
thay vào, ta có M=\(0^{30}+\left(-1+2\right)^{12}+\left(1-1\right)^{2017}=1\)
Vậy M=1
^_^
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
Ta có : x2 + xy + y2 + x - y + 1 = 0
=> 2( x2 + xy + y2 + x - y + 1) = 0
=> 2x2 + 2xy + 2y2 + 2x - 2y + 2 = 0
=> x2 + 2xy + y2 + x2 + 2x + 1 + y2 - 2y + 1 = 0
=> ( x + y)2 + ( x + 1)2 + ( y - 1)2 = 0
Suy ra :
* x + y = 0 => x = -y
* x + 1 = 0 => x = -1
* y - 1 = 0 => y = 1
Từ đó , ta có :
M = ( x + y)30 + ( x + 2)12 + ( y - 1)2017
M = ( -y + y )30 + ( 2 - 1)12 + ( 1 - 1)2017
M = 1
\(5x^2+5y^2+8xy-2x+2y+2=0\)
\(\Leftrightarrow\left(4x^2+8xy+4y^2\right)+\left(x^2-2x+1\right)+\left(y^2+2y+1\right)=0\)
\(\Leftrightarrow4\left(x+y\right)^2+\left(x-1\right)^2+\left(y+1\right)^2=0\)
Ta thấy \(VT\ge VP\forall x;y\) để đấu "=" xảy ra \(\Leftrightarrow x=1;y=-1\) thay vào M :
\(M=\left(-1+1\right)^{2015}+\left(1-2\right)^{2016}+\left(-1+1\right)^{2017}=1\)
\(5x^2+5y^2+8xy-2x+2y+2=0\)
=>\(4x^2+8xy+4y^2+x^2-2x+1+y^2+2y+1=0\)
=>\(4\left(x+y\right)^2+\left(x-1\right)^2+\left(y+1\right)^2=0\)
=>x=1 và y=-1
\(M=\left(1-1\right)^{2023}+\left(1-2\right)^{2024}+\left(-1+1\right)^{2025}=1\)
Đẳng thức: \(5x^2+5y^2+8xy-2x+2y+2=0\)
\(\Leftrightarrow\left(4x^2+8xy+4y^2\right)+\left(x^2-2x+1\right)+\left(y^2+2y+1\right)=0\)
\(\Leftrightarrow\left(2x+2y\right)^2+\left(x-1\right)^2+\left(y+1\right)^2=0\)
\(\Leftrightarrow4\left(x+y\right)^2+\left(x-1\right)^2+\left(y+1\right)^2=0\)
\(\Rightarrow\left\{{}\begin{matrix}x+y=0\\x-1=0\\y+1=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=1\\y=-1\end{matrix}\right.\)
Thay vào \(M=\left(x+y\right)^{2007}+\left(x-2\right)^{2008}+\left(y+1\right)^{2009}\) ta được:
\(M=\left(1-1\right)^{2007}+\left(1-2\right)^{2008}+\left(-1+1\right)^{2009}=\left(-1\right)^{2008}=1\)
Ta có:
\(5x^2+5y^2+8xy-2x+2y+2=0\)
\(\Leftrightarrow x^2+4x^2+y^2+4y^2+8xy-2x+2y+1+1=0\)
\(\Leftrightarrow\left(x^2-2x+1\right)+\left(y^2+2y+1\right)+\left(4x^2+8xy+4y^2\right)=0\)
\(\Leftrightarrow\left(x-1\right)^2+\left(y+1\right)^2+\left(2x+2y\right)^2=0\)
\(\Leftrightarrow\left(x-1\right)^2+\left(y+1\right)^2+4\left(x+y\right)^2=0\)
Mà: \(\left\{{}\begin{matrix}\left(x-1\right)^2\ge0\\\left(y+1\right)^2\ge0\\4\left(x+y\right)^2\ge0\end{matrix}\right.\Leftrightarrow\left(x-1\right)^2+\left(y+1\right)^2+4\left(x+y\right)^2\ge0\)
\(\Leftrightarrow\left\{{}\begin{matrix}x-1=0\\y+1=0\\x+y=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=1\\y=-1\\x=-y\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=1\\y=-1\end{matrix}\right.\)
Thay giá trị x và y vào M ta có:
\(M=\left(x+y\right)^{2007}+\left(x-2\right)^{2008}+\left(y+1\right)^{2009}\)
\(M=\left(1-1\right)^{2007}+\left(1-2\right)^{2008}+\left(-1+1\right)^{2009}\)
\(M=0^{2007}+\left(-1\right)^{2008}+0^{2009}\)
\(M=\left(-1\right)^{2008}\)
\(M=1\)
\(\left(x+y+1\right)\left(xy+x+y\right)=5+2\left(x+y\right)\)
\(\Leftrightarrow\left(x+y+1\right)\left(xy+x+y\right)=3+2\left(x+y+1\right)\)
\(\Leftrightarrow\left(x+y+1\right)\left(xy+x+y-2\right)=3\)
Từ đây bạn xét các trường hợp và giải ra nghiệm.
mk ko vt lại đề
=> (4x^2+8xy+4y^2)+(x^2-2x+1)+(y^2+2y+1)=0
=>(2x+2y)^2+(x-1)^2+(y+1)^2=0
...... phần này bn tự làm đc
=>x=1,y=-1
thay vào là dc
Ta có : \(5x^2+5y^2+8xy-2x+2y+2=0\)
=> \(\left(4x^2+8xy+4y^2\right)+\left(x^2-2x+1\right)+\left(y^2-2y+1\right)=0\)
=> \(\left(2x+2y\right)^2+\left(x-1\right)^2+\left(y+1\right)^2=0\)
Ta có \(\left(2x+2y\right)^2\ge0\forall x,y\) , \(\left(x-1\right)^2\ge0\forall x\) , \(\left(y+1\right)^2\ge0\forall x\)
=> \(4\left(x+y\right)^2+\left(x-1\right)^2+\left(y+1\right)^2\ge0\forall x,y\)
=> \(\hept{\begin{cases}x+y=0\\x-1=0\\y+1=0\end{cases}\Rightarrow\hept{\begin{cases}x+y=0\\x=1\\y=-1\end{cases}}}\)
Thay vào M ta có:
\(M=0^{2016}+\left(1-2\right)^{2018}+\left(-1+1\right)^{2019}=1\)
\(x^2+xy+y^2+x-y+1=0\)
\(\Leftrightarrow2x^2+2xy+2y^2+2x-2y+2=0\)
\(\Leftrightarrow\left(x^2+2xy+y^2\right)+\left(x^2+2x+1\right)+\left(y^2-2y+1\right)=0\)
\(\Leftrightarrow\left(x+y\right)^2+\left(x+1\right)^2+\left(y-1\right)^2=0\) (*)
Vì \(\left(x+y\right)^2\ge0;\left(x+1\right)^2\ge0;\left(y-1\right)^2\ge0\)
(*) \(\Leftrightarrow\left(x+y\right)^2=0;\left(x+1\right)^2=0;\left(y-1\right)^2=0\)
\(\Leftrightarrow x+y=0;x+1=0;y-1=0\)
\(\Rightarrow x+2=1\)
\(\Rightarrow\left(x+y\right)^{30}+\left(x+2\right)^{12}+\left(y-1\right)^{2017}=0+1+0=1\)