Vì \(\left(x-y\right)^2\ge0\), \(\left(y-z\right)^2\ge0\), \(\left(z+1\right)^2\ge0\)mà \(\left(x-y\right)^2+\left(y-z\right)^2+\left(z+1\right)^2=0\).

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

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

\(\Rightarrow z=-1\)

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

Vậy \(x=y=z=-1\)

Ta có:

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

=>\(\left(x-1\right)^2=0\)và \(\left(y-z\right)^2=0\)và \(\left(z-1\right)^2=0\)

Với \(\left(x-1\right)^2=0\)=>\(x-1=0\)=>\(x=1\)

Với\(\left(z+1\right)^2=0\)=>\(z+1=0\)=>\(z=-1\)

Với \(\left(y-z\right)^2=0\)=>\(\left[y-\left(-1\right)\right]^2=0\)=>\(\left(y+1\right)^2=0\)=>\(y+1=0\)=>\(y=-1\)

Vậy x=1;y=z=-1

Trên AC lấy F sao cho AE=AF

Xét ΔAEI và ΔAFI co

AE=AF

góc EAI=góc FAI

AI chung

Do đó: ΔAEI=ΔAFI

=>EI=FI

góc IAC=180 độ-góc IAC-góc ICA

=180 độ-1/2*120

=120 độ

=>góc AIE=góc DIC=60 độ

góc AIF=góc AIE=60 độ

Xet ΔDIC và ΔFIC có

góc DCI=góc FCI 

CI chung

góc DIC=góc FIC

Do đó: ΔDIC=ΔFIC

=>ID=IF

=>ID=IE

=>ΔIDE cân tại I

(x-y)² + (y-1)² =0 

Mà (x-y) ²  \(\ge\)^{0 ,} (y-1)² \(\ge\)⁰

^nên 

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

Ta có:

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

\(\left(y-1\right)^2\ge0\forall y\)

=>\(\left(x-y\right)^2+\left(y-1\right)^2\ge0\forall x,y\)

Mà thao đề bài ta có:

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

=>\(\left(x-y\right)^2=0\)và \(\left(y-1\right)^2=0\)

Với \(\left(y-1\right)^2=0\)=>\(y-1=0\)=>\(y=1\)

Với \(\left(x-y\right)^2=0\)=>\(\left(x-1\right)^2=0\)=>\(x-1=0\)=>\(x=1\)

Vậy x=y=1

a)\(\frac{2016}{2017}< 1;\frac{2015}{2016}< 1\)

b)\(\frac{2017}{2016}>1;\frac{2016}{2015}>1\)

=> \(\frac{2016}{2017}\)và    

\(\frac{2016}{2017}< 1;\frac{2016}{2015}< 1\)

\(\frac{2017}{2016}>1;\frac{2016}{2015}>1\)

=> \(\frac{2016}{2017}\)và    \(\frac{2015}{2016}\)<    \(\frac{2017}{2016}\)và    \(\frac{2016}{2015}\)