\(2xy+y-4x=6\)

\(\Rightarrow2xy+y-4x-6=0\)

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

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

\(\Rightarrow\left(y-2\right)\left(2x+1\right)=4\)

Lập bảng , tìm các cặp (x;y) rồi thế vào M để tính

Ta có : \(x^4-7x^2+y^2+16=2xy\)

=> \(\left(x^2-8x^2+16\right)+\left(x^2-2xy+y^2\right)=0\)

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

Vì \(\left(x-4\right)^2\ge0 \forall x ,\left(x-y\right)^2 \ge0 \forall x,y \)

=> \(\left(x-4\right)^2+\left(x-y\right)^2\ge0 \forall x,y\)

=> \(\hept{\begin{cases}x-4=0\\x-y=0\end{cases}\Rightarrow\hept{\begin{cases}x=4\\x=y=4\end{cases}}}\)

Thay vào \(A=4^{2016}.4^{2017}-4^{2017}.4^{2016}+4+4=8\)

Vậy A=8

https://olm.vn/thanhvien/nguyentrangth8 bạn giỏi thế

Dễ như 1+1=3

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