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: (2x-1)²⁰¹⁸≥0 ; (y-2/5)²⁰¹⁸≥0 ; |x+y-z|≥0

=>\(\hept{\begin{cases}\left(2x-1\right)^{2018}=0\\\left(y-\frac{2}{5}\right)^{2018}=0\\\left|x+y-z\right|=0\end{cases}\Leftrightarrow\hept{\begin{cases}x=\frac{1}{2}\\y=\frac{2}{5}\\z=\frac{9}{10}\end{cases}}}\)

Chúc bạn học tốt!

Ta có : 

\(\left(2x-1\right)^{2018}\ge0\)

\(\left(y-\frac{2}{5}\right)^{2018}\ge0\)

\(\left|x+y-z\right|\ge0\)

Mà \(\left(2x-1\right)^{2018}+\left(y-\frac{2}{5}\right)^{2018}+\left|x+y-z\right|=0\) ( Giả thiết ) 

\(\Rightarrow\)\(\hept{\begin{cases}\left(2x-1\right)^{2018}=0\\\left(y-\frac{2}{5}\right)^{2018}=0\\\left|x+y-z\right|=0\end{cases}\Leftrightarrow\hept{\begin{cases}x=\frac{1}{2}\\y=\frac{2}{5}\\z=\frac{9}{10}\end{cases}}}\)

Vậy \(x=\frac{1}{2}\)\(;\)\(y=\frac{2}{5}\) và \(z=\frac{9}{10}\)

Chúc bạn học tốt ~ 

1. Vì \(\left(x+6\right)^2\ge0\forall x\); \(\left|y-\frac{1}{2}\right|\ge0\forall y\); \(\left|x+y+z\right|\ge0\forall x,y,z\)

\(\Rightarrow\left(x+6\right)^2+\left|y-\frac{1}{2}\right|+\left|x+y+z\right|\ge0\)

mà \(\left(x+6\right)^2+\left|y-\frac{1}{2}\right|+\left|x+y+z\right|\le0\)( đề bài )

\(\Rightarrow\left(x+6\right)^2+\left|y-\frac{1}{2}\right|+\left|x+y+z\right|=0\)\(\Leftrightarrow\hept{\begin{cases}x+6=0\\y-\frac{1}{2}=0\\x+y+z=0\end{cases}}\Leftrightarrow\hept{\begin{cases}x=-6\\y=\frac{1}{2}\\-6+\frac{1}{2}+z=0\end{cases}}\Leftrightarrow\hept{\begin{cases}x=-6\\y=\frac{1}{2}\\z=\frac{11}{2}\end{cases}}\)

Vậy \(x=-6\); \(y=\frac{1}{2}\); \(z=\frac{11}{2}\)

2. \(B=\left|x-2016\right|+\left|x-2018\right|=\left|x-2016\right|+\left|2018-x\right|\ge\left|x-2016+2018-x\right|=\left|2\right|=2\)

Dấu " = " xảy ra \(\Leftrightarrow\left(x-2016\right)\left(2018-x\right)\ge0\)

TH1: \(\hept{\begin{cases}x-2016< 0\\2018-x< 0\end{cases}}\Leftrightarrow\hept{\begin{cases}x< 2016\\2018< x\end{cases}}\Leftrightarrow\hept{\begin{cases}x< 2016\\x>2018\end{cases}}\)( vô lý )

TH2: \(\hept{\begin{cases}x-2016\ge0\\2018-x\ge0\end{cases}}\Leftrightarrow\hept{\begin{cases}x\ge2016\\2018\ge x\end{cases}}\Leftrightarrow\hept{\begin{cases}x\ge2016\\x\le2018\end{cases}}\Leftrightarrow2016\le x\le2018\)( thoả mãn )

Vậy \(minB=2\Leftrightarrow2016\le x\le2018\)

min mình có ra rồi. nhưng chỉ không biết là khi x=y hay x,y bằng bao nhiêu thôi. 

\(\sqrt{x\left(x+3y\right)}\ge\frac{x+x+3y}{2}=\frac{2x+3y}{2}\)

\(\sqrt{y\left(y+3x\right)}\le\frac{y+y+3x}{2}=\frac{2y+3x}{2}\)

\(\Rightarrow\sqrt{x\left(x+3y\right)}+\sqrt{y\left(y+3x\right)}\le\frac{5}{2}\left(x+y\right)\)

=> \(A\ge2016\left(x+y\right):\frac{5}{2}\left(x+y\right)=\frac{2016\cdot2\left(x+y\right)}{5\left(x+y\right)}=\frac{4032}{5}\)

nhưng không biết x,y bằng bao nhiêu. ai làm đc ghi hẳn cách giải ra nha