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

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

1) x² + 7y² - 4xy - 2x - 2y + 4 = 0

\(\Leftrightarrow\)[ x² - 2x.( 2y + 1 ) + 4y² + 4y +1 ] - 4y² - 4y - 1 + 7y² - 2y +4 = 0

\(\Leftrightarrow\) [ x² - 2x.( 2y +1 ) + ( 2y +1 )² ] + 3y² - 6y +3 = 0

\(\Leftrightarrow\) ( x - 2y - 1 )² + 3.( y² - 2y + 1 ) = 0

\(\Leftrightarrow\)( x - 2y - 1 )² + 3.( y - 1 )² = 0

\(\Leftrightarrow\)\(\hept{\begin{cases}\left(x-2y-1\right)^2=0\\\left(y-1\right)^2=0\end{cases}}\)\(\Leftrightarrow\)\(\hept{\begin{cases}x-2y-1=0\\y-1=0\end{cases}}\)\(\Leftrightarrow\)\(\hept{\begin{cases}x=2y+1\\y=1\end{cases}}\)\(\Leftrightarrow\)\(\hept{\begin{cases}x=3\\y=1\end{cases}}\)

Vậy x = 3 , y = 1 thì x² + 7y² - 4xy - 2x - 2y + 4 = 0

2) 11x² + y² - 6xy - 14x + 2y +9 = 0

\(\Leftrightarrow\)[ y² - 2y.( 3x - 1 ) + 9x² - 6x +1 ] + 2x² - 8x + 8 = 0

\(\Leftrightarrow\)[ y² - 2y.( 3x - 1 ) + ( 3x - 1 )² ] + 2.( x² - 4x + 4 ) = 0

\(\Leftrightarrow\)( y - 3x + 1 )² + 2.( x - 2 )² = 0

\(\Leftrightarrow\)\(\hept{\begin{cases}\left(y-3x+1\right)^2=0\\\left(x-2\right)^2=0\end{cases}}\)\(\Leftrightarrow\)\(\hept{\begin{cases}y-3x+1=0\\x-2=0\end{cases}}\)\(\Leftrightarrow\)\(\hept{\begin{cases}y=3x-1\\x=2\end{cases}}\)\(\Leftrightarrow\)\(\hept{\begin{cases}y=5\\x=2\end{cases}}\)

Vậy x = 2 , y = 5 thì 11x² + y² - 6xy - 14x + 2y + 9 = 0

Cảm ơn bạn

Đáp án D

Toán lớp 0 ?????  \(\text{ 🤔 }\text{ 🤔 }\text{ 🤔 }\text{ 😅 }\text{ 😅 }\text{ 😅 }\)

sao giống câu hỏi của mình thế chỉ khác số bạn biết làm ko chỉ mình đi

\(8x^2+14xy+8y^2+2x-2y+2=0\)

\(\Leftrightarrow7\left(x^2+2xy+y^2\right)+\left(x^2+2x+1\right)+\left(y^2-2y+1\right)=0\)

\(\Leftrightarrow7\left(x+y\right)^2+\left(x+1\right)^2+\left(y-1\right)^2=0\)

Do \(\left\{{}\begin{matrix}7\left(x+y\right)^2\ge0\\\left(x+1\right)^2\ge0\\\left(y-1\right)^2\ge0\end{matrix}\right.\) ; \(\forall x;y\)

Nên \(7\left(x+y\right)^2+\left(x+1\right)^2+\left(y-1\right)^2\ge0;\forall x;y\)

Dấu "=" xảy ra khi và chỉ khi:

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