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:

\(M=\left(x+y\right)^{2007}+\left(x-2\right)^{2008}+\left(y+1\right)^{2009}\)

\(M=\left(1-1\right)^{2007}+\left(1-2\right)^{2008}+\left(-1+1\right)^{2009}\)

\(M=0^{2007}+\left(-1\right)^{2008}+0^{2009}\)
\(M=\left(-1\right)^{2008}\)

\(M=1\)

Ta có:  5x² + 5y² + 8xy - 2x + 2y + 2 = 0

\(\Leftrightarrow\)(4x² + 8xy + 4y²) + (x² - 2x + 1) + (y² + 2y + 1) = 0

\(\Leftrightarrow\)(2x + 2y)² + (x - 1)² + (y + 1)² = 0

\(\Leftrightarrow\)\(\hept{\begin{cases}2x+2y=0\\x-1=0\\y+1=0\end{cases}}\)\(\Leftrightarrow\)\(\hept{\begin{cases}x+y=0\\x=1\\y=-1\end{cases}}\)

Thay x = 1; y = -1; x + y = 0 vào M ta được:

 M = 0 + (1 + 2)²⁰⁰⁸ + ( - 1 + 1)²⁰⁰⁹

     = 0 + 3²⁰⁰⁸ + 0 = 3²⁰⁰⁸

Ta có\(5x^2+5y^2+8xy-2x+2y+2=0\Leftrightarrow4x^2+8xy+4y^2+x^2-2x+1+y^2+2y+1=0\)

<=>\(4\left(x+y\right)^2+\left(x-1\right)^2+\left(y+1\right)^2=0\)

mà \(\hept{\begin{cases}4\left(x+y\right)^2\ge0\\\left(y+1\right)^2\ge0\\\left(x-1\right)^2\ge0\end{cases}\Rightarrow}4\left(x+y\right)^2+\left(y+1\right)^2+\left(x-1\right)^2\ge0\)

dâu = xảy ra <=>\(\hept{\begin{cases}x=1\\y=1\end{cases}}\)

rồi bạn thay vào và tự tính M nhé !

^_^

ban lam dung roi day

\(5x^2+5y^2+8xy-2x+2y+2=0\)

\(\Leftrightarrow\left(4x^2+8xy+4y^2\right)+\left(x^2-2x+1\right)+\left(y^2+2y+1\right)=0\)

\(\Leftrightarrow4\left(x+y\right)^2+\left(x-1\right)^2+\left(y+1\right)^2=0\)

Mà \(\left\{{}\begin{matrix}4\left(x+y\right)^2\ge0\\\left(x-1\right)^2\ge0\\\left(y+1\right)^2\ge0\end{matrix}\right.\Leftrightarrow4\left(x+y\right)^2+\left(x-1\right)^2+\left(y+1\right)^2\ge0\)

\(\Rightarrow\left\{{}\begin{matrix}4\left(x+y\right)^2=0\\\left(x-1\right)^2=0\\\left(y+1\right)^2=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=-y\\x=1\\y=-1\end{matrix}\right.\)

Ta có: \(M=\left(x+y\right)^{2017}+\left(x-2\right)^{2008}+\left(y+1\right)^{2009}\)

\(=\left(-1\right)^{2008}=1\)

Vậy M = 1

5x^2+5y^2+8xy-2x+2y+2=0

=>(4x^2+8xy+4y^2)+(x^2-2x+1)+(y^2+2y+1)=0

=>(2x+2y)^2+(x-1)^2+(y+1)^2=0

tổng 3 biểu thức không âm = 0 <=> chúng đều = 0

<=>2(x+y)=x-1=y+1=0

=>x=1;y=-1

Thay vào M ........

thanks hoàng phúc nha!

\(5x^2+5y^2+8xy-2x+2y+2=0\)

=>\(4x^2+8xy+4y^2+x^2-2x+1+y^2+2y+1=0\)

=>\(4\left(x+y\right)^2+\left(x-1\right)^2+\left(y+1\right)^2=0\)

=>x=1 và y=-1

\(M=\left(1-1\right)^{2023}+\left(1-2\right)^{2024}+\left(-1+1\right)^{2025}=1\)

3x^2+3y^2+4xy-2x+2y+2=0

=>2x^2+4xy+2y^2+x^2-2x+1+y^2+2y+1=0

=>x=1 và y=-1

M=(1-1)^2017+(1-2)^2018+(-1+1)^2015=1