a, Xét tứ giác ABDC có: AM=MD (gt) ; BM=MC (gt)

=> ABDC là hình bình hành

b,Để ABDC là hình thoi => AB = AC => \(\Delta ABC\)cân

c, I đâu ra vậy bạn?

\(a^2-2a+1-b^2\)

\(=\left(a-1\right)^2-b^2\)

\(=\left(a-1-b\right)\left(a-1+b\right)\)

Xét \(\frac{x}{y^3-1}+\frac{y}{x^3-1}=\frac{1-y}{y^3-1}+\frac{1-x}{x^3-1}=-\frac{1}{x^2+x+1}-\frac{1}{y^2+y+1}\)

\(=-\frac{x^2+y^2+x+y+2}{\left(x^2+x+1\right)\left(y^2+y+1\right)}=-\frac{x^2+y^2+3}{x^2y^2+xy\left(x+y\right)+x^2+y^2+xy+x+y+1}\)

\(=-\frac{\left(x+y\right)^2-2xy+3}{x^2y^2+x^2+y^2+2xy+2}=-\frac{4-2xy}{x^2y^2+3}=\frac{2\left(xy-2\right)}{x^2y^2+3}\)

từ đó ta có đpcm

\(x\left(x-7\right)-x+7=0\)

\(\Leftrightarrow x\left(x-7\right)-\left(x-7\right)=0\)

\(\Leftrightarrow\left(x-1\right)\left(x-7\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x-1=0\\x-7=0\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x=1\\x=7\end{cases}}\)

Ta có: x(x-7)-(x-7)=0

<=> (x-7)(x-1) = 0 <=> \(\orbr{\begin{cases}x-7=0\\x-1=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=7\\x=1\end{cases}}}\)

\(\frac{x^2+y^2+z^2}{a^2+b^2+c^2}=\frac{x^2}{a^2}+\frac{y^2}{b^2}+\frac{z^2}{c^2}\)

\(\Leftrightarrow\left(\frac{x^2}{a^2}-\frac{x^2}{a^2+b^2+c^2}\right)+\left(\frac{y^2}{b^2}-\frac{y^2}{a^2+b^2+c^2}\right)+\left(\frac{z^2}{c^2}-\frac{z^2}{a^2+b^2+c^2}\right)=0\)

\(\Leftrightarrow\left(x^2.\frac{b^2+c^2}{a^2+b^2+c^2}\right)+\left(y^2.\frac{a^2+c^2}{a^2+b^2+c^2}\right)+\left(z^2.\frac{a^2+b^2}{a^2+b^2+c^2}\right)=0\)

Vì a,b,c khác 

=>Dấu bằng xảy ra khi x=y=z=0

\(\Rightarrow x^{2014}+y^{2015}+z^{2016}=0^{2014}+0^{2015}+0^{2016}=0\)