Điều kiện x \(\ge0\)từ đó ta có

x + 2015 + x + 2016 + x + 2017 = 6x 

<=> x = 2016

Ta có \(\left|x+2015\right|+\left|x+2016\right|+\left|x+2017\right|\ge0\Rightarrow6x\ge0\Rightarrow x\ge0\)

=> \(\left|x+2015\right|+\left|x+2016\right|+\left|x+2017\right|=3x+6048=6x\Rightarrow3x=6048\Rightarrow x=2016\)

Vậy x=2016

Ta có:

\(\left|x+2015\right|\ge0\)

\(\left|x+2016\right|\ge0\)

\(\left|x+2017\right|\ge0\)

\(\Rightarrow\left|x+2015\right|+\left|x+2016\right|+\left|x+2017\right|\ge0\)

\(\Rightarrow6x\ge0\)

\(\Rightarrow x\ge0\)

\(\Rightarrow\left|x\right|=x\)

\(\Rightarrow\left|x+2015\right|+\left|x+2016\right|+\left|x+2017\right|=\left(x+2015\right)+\left(x+2016\right)+\left(x+2017\right)=6x\)

\(\Rightarrow3x+6048=6x\)

\(\Rightarrow3x=6048\)

\(\Rightarrow x=2016\)

Vậy \(x=2016\)

|x+2015|+|x+2016|+|x+2017| = 6x

<=> \(\left[\begin{array}{nghiempt}x+2015+x+2016+x+2017=6x\\x+2015+x+2016+x+2017=-6x\end{array}\right.\)

<=> \(\left[\begin{array}{nghiempt}3x+6048=6x\\3x+6048=-6x\end{array}\right.\)

<=> \(\left[\begin{array}{nghiempt}3x=6048\\-9x=6048\end{array}\right.\)

<=> \(\left[\begin{array}{nghiempt}x=2016\\x=-672\end{array}\right.\)

h mk di minh tra loi noi that

đặt t=x+y

x^2 +2xy+6x+6y+2y^2+8=0

x^2+2xy+y^2+6(x+y)+8= -y^2

(x+y)^2 + 6(x+y)+8 = -y^2

t^2 +6t +8= -y^2

(t+2)(t+4) = -y^2

do y^2 >=0 với mọi y

-y^2 <=0 với mọi y

t^2+6t+8<=0

(t+2)(t+4)<=0

* Trường hợp 1:   t+2<=0 và t+4>=0        (1)

t<=-2 và t>=4

* trường hợp 2:  t+2>=0 và t+4<=0           (2)

t>= -2 và t<= -4   ( vô nghiệm)

 Từ (1), (2) ta có:

-4<= t <=-2 

-4 <= x+y <= -2

-4 + 2016 <= x+y+ 2016 <= -2 +2016

2012 <= x+y +2016 <= 2014

Bmin= 2012

Bmax= 2014

 *Bmin= 2012 khi x+y+2016 = 2012 và -y^2= 0

thì x=-4 và y=0

* Bmax= 2014 khi x+y+2016 = 2014 và -y^2= 0

thì x=-2 và y=0

vậy Bmin= 2012 khi (x,y) = (-4, 0)

Bmax= 2014 khi (x,y)= (-2,0)

\(x^2+2xy+6x+6y+2y^2+8=0\\ \Leftrightarrow\left(x+y\right)^2+6\left(x+y\right)+y^2=-8\)

Ta có \(y^2\ge0\Leftrightarrow\left(x+y\right)^2+6\left(x+y\right)\le-8\)

\(\Leftrightarrow\left(x+y\right)^2+6\left(x+y\right)+9\le1\\ \Leftrightarrow\left(x+y+3\right)^2\le1\\ \Leftrightarrow\left|x+y+3\right|\le1\\ \Leftrightarrow-1\le x+y+3\le1\\ \Leftrightarrow2012\le B\le2014\)

\(B_{min}=2012\Leftrightarrow\left\{{}\begin{matrix}x+y+2016=2012\\y=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-4\\y=0\end{matrix}\right.\)

\(B_{max}=2014\Leftrightarrow\left\{{}\begin{matrix}x+y+2016=2014\\y=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-2\\y=0\end{matrix}\right.\)

\(x^2+2xy+6x+6y+2y^2+8=0\)

\(\Leftrightarrow\left(x+y\right)^2+6\left(x+y\right)+9=1-y^2\)

\(\Leftrightarrow\left(x+y+3\right)^2=1-y^2\)

Ta thấy : \(1-y^2\le1\forall y\) \(\Rightarrow\left(x+y+3\right)^2\le1\)

\(\Rightarrow-1\le x+y+3\le1\)

\(\Rightarrow-1+2013\le x+y+3+2013\le1+2013\)

\(\Rightarrow2012\le x+y+2016\le2014\)

Vậy ta có : 

+) Min \(B=2012\) . Dấu "=" xảy ra \(\Leftrightarrow\hept{\begin{cases}y=0\\x+y+3=-1\end{cases}\Leftrightarrow}\hept{\begin{cases}y=0\\x=-4\end{cases}}\)

+) Max \(M=2014\). Dấu "=" xảy ra \(\Leftrightarrow\hept{\begin{cases}y=0\\x+y+3=1\end{cases}\Leftrightarrow}\hept{\begin{cases}y=0\\x=-2\end{cases}}\)