Lời giải:

Ta có:

\(x^2+2y+1=y^2+2z+1=z^2+2x+1=0\)

\(\Rightarrow x^2+2y+1+y^2+2z+1+z^2+2x+1=0+0+0=0\)

\(\Leftrightarrow (x^2+2x+1)+(y^2+2y+1)+(z^2+2z+1)=0\)

\(\Leftrightarrow (x+1)^2+(y+1)^2+(z+1)^2=0(*)\)

Ta thấy rằng \(\left\{\begin{matrix} (x+1)^2\geq 0\\ (y+1)^2\geq 0\\ (z+1)^2\geq 0\end{matrix}\right., \forall x,y,z\in\mathbb{R}\)

Do đó để $(*)$ xảy ra thì \((x+1)^2=(y+1)^2=(z+1)^2=0\)

\(\Leftrightarrow x=y=z=-1\)

Thử lại thấy thỏa mãn

Vậy \(x^{2017}+y^{2017}+z^{2017}=(-1)^{2017}.3=-3\)

cộng 3 vế lại cùng 1 lúc ta sẽ có (x+1)² +(y+1)²+(z+1)² = 0.

dấu bằng xảy ra khi cả 3 biểu thức bằng 0, suy ra x=y=z= -1

thế vào A thì A= -3

bài này dễ mà anh 

anh ko nhắn vs em à 

\(gt\Leftrightarrow x^2+y^2+z^2+2x+2y+2z+3=0\)

\(\Leftrightarrow\left(x^2+2x+1\right)+\left(y^2+2y+1\right)+\left(z^2+2z+1\right)=0\)

\(\Leftrightarrow\left(x+1\right)^2+\left(y+1\right)^2+\left(z+1\right)^2=0\)

\(\Leftrightarrow x=y=z=-1\)

Khi đó \(S=\left(-1\right)^{2017}+\left(-1\right)^{2018}+\left(-1\right)^{2019}=-1\)

Ta có : \(x^2+2y+1=0;y^2+2z+1=0;z^2+2x+1=0\)

\(\Rightarrow x^2+2y+1=y^2+2z+1=z^2+2x+1\)

\(\Rightarrow x^2+2y+1-y^2-2z-1-z^2-2x-1=0\)

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

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

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

Thay \(x=1;y=1;z=-1\)vào A ta có :

\(A=1^{2015}+1^{2016}+\left(-1\right)^{2017}=1+1-1=1\)

Vậy A = 1

Từ \(\hept{\begin{cases}x^2+2y+1=0\\y^2+2z+1=0\\z^2+2x+1=0\end{cases}}\)

\(\Rightarrow x^2+2y+1+y^2+2z+1+z^2+2x+1=0\)

\(\Rightarrow\left(x^2+2x+1\right)+\left(y^2+2y+1\right)+\left(z^2+2z+1\right)=0\)

\(\Rightarrow\left(x+1\right)^2+\left(y+1\right)^2+\left(z+1\right)^2=0\left(1\right)\)

Vì \(\hept{\begin{cases}\left(x+1\right)^2\ge0\forall x\\\left(y+1\right)^2\ge0\forall y\\\left(z+1\right)^2\ge0\forall z\end{cases}\left(2\right)}\)

Từ \(\left(1\right)\)và \(\left(2\right)\):

\(\Rightarrow\hept{\begin{cases}\left(x+1\right)^2=0\\\left(y+1\right)^2=0\\\left(z+1\right)^2=0\end{cases}}\)

\(\Rightarrow\hept{\begin{cases}x+1=0\\y+1=0\\z+1=0\end{cases}}\)

\(\Rightarrow x=y=z=-1\)

\(\Rightarrow A=\left(-1\right)^{2015}+\left(-1\right)^{2016}+\left(-1\right)^{2017}=-1+1-1=-1\)

Vậy \(A=-1\)

Ta có\(\hept{\begin{cases}x^2+2y+1=0\\y^2+2z+1=0\\z^2+2x+1=0\end{cases}}\)

\(\Rightarrow x^2+2y+1+y^2+2z+1+z^2+2x+1=0+0+0\)

\(\Leftrightarrow x^2+2y+1+y^2+2z+1+z^2+2x+1=0\)

\(\Leftrightarrow\left(x^2+2x+1\right)+\left(y^2+2y+1\right)+\left(z^2+2z+1\right)=0\)

\(\Leftrightarrow\left(x+1\right)^2+\left(y+1\right)^2+\left(z+1\right)^2=0\)

Ta thấy \(\hept{\begin{cases}\left(x+1\right)^2\\\left(y+1\right)^2\\\left(z+1\right)^2\end{cases}\ge0\forall x,y,z}\)

Dấu"=" xảy ra khi \(\left(x+1\right)^2=\left(y+1\right)^2=\left(z+1\right)^2=0\)

\(\Rightarrow x=y=z=-1\)

Vậy giá trị của biểu thức A là:

\(A=\left(-1\right)^{2015}+\left(-1\right)^{2016}+\left(-1\right)^{2017}=-1\)

Vậy A=-1

1) \(E^2=\frac{x^2-2xy+y^2}{x^2+2xy+y^2}=\frac{2\left(x^2+y^2\right)-4xy}{2\left(x^2+y^2\right)+4xy}=\frac{5xy-4xy}{5xy+4xy}=\frac{xy}{9xy}=\frac{1}{9}\)

\(\Rightarrow E=\frac{1}{3}\)(vì x>y>0)

2) Ta có \(x+y+z=0\Rightarrow x+y=1-z\)

Lại có : \(1=\left(x+y+z\right)^2=1+2\left(xy+yz+xz\right)\Rightarrow2xy+2yz+2xz=0\Rightarrow2xy=-2z\left(x+y\right)=-2z\left(1-z\right)\)Thay vào \(x^2+y^2+z^2=1\) được : 

\(\left(x+y\right)^2-2xy+z^2=1\)\(\Leftrightarrow\left(1-z\right)^2-2z\left(1-z\right)+z^2=1\Leftrightarrow4z^2-4z=0\Leftrightarrow z\left(z-1\right)=0\Leftrightarrow\orbr{\begin{cases}z=0\\z=1\end{cases}}\)

Với z = 0 => x + y = 1 và x²+y² = 1 => x = 0 , y = 1 hoặc x = 1 , y =0

=> A = 1

Tương tự với z = 1 , ta cũng có x = 0 , y = 0 => A = 1

ta có x²+2y+1+y²+2z+1+z²+2x+1=0

=>(x²+2x+1)+(y²+2y+1)+(z²+2z+1)=0

=>(x+1)²+(y+1)²+(z+1)²=0

Vì (x+1)²> hoặc = 0

.......

=> x=-1,y=-1,z=-1

sau đó thay vào nha

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

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

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

=>x=1 và y=-1

M=(1-1)^2015+(1-2)^2016+(-1+1)^2017=1

Ta có : \(4x^2+2y^2+2z^2-4xy-4zx+2yz-6y-10z+34=0\)

\(\Rightarrow\left(4x^2+y^2+z^2-4xy-4zx+2yz\right)+\left(y^2-6y+9\right)+\left(z^2-10z+25\right)=0\)

\(\Rightarrow\left(2x-y-z\right)^2+\left(y-3\right)^2+\left(z-5\right)^2=0\)

Vì \(\hept{\begin{cases}\left(2x-y-z\right)^2\ge0\forall x,y,z\\\left(y-3\right)^2\ge0\forall y\\\left(z-5\right)^2\ge0\forall z\end{cases}}\)

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

\(\Rightarrow\hept{\begin{cases}2x-y-z=0\\y-3=0\\z-5=0\end{cases}}\)

\(\Rightarrow\hept{\begin{cases}2x-3-5=0\\y=3\\z=5\end{cases}}\)

\(\Rightarrow\hept{\begin{cases}2x=8\\y=3\\z=5\end{cases}}\)\(\Rightarrow\hept{\begin{cases}x=4\\y=3\\z=5\end{cases}}\left(1\right)\)

Lại có : \(S=\left(x-4\right)^{2017}+\left(y-4\right)^{2017}+\left(z-4\right)^{2017}\)

Thay \(\left(1\right)\)vào \(S\),ta được :

\(S=0^{2017}+\left(-1\right)^{2017}+1^{2017}\)

    \(=0-1+1=0\)

Vậy \(S=0\)