Có: \(a+b+c=11\)

\(\Leftrightarrow a^2+b^2+c^2+2ab+2ac+2ca=121\)

\(\Leftrightarrow87+2\left(ab+bc+ca\right)=121\)

\(\Leftrightarrow2\left(ab+bc+ca\right)=34\)

\(\Leftrightarrow ab+bc+ca=17\)

Có: \(a+b+c=0\)

\(\Leftrightarrow a+b=-c\)

\(\Leftrightarrow a^3+b^3+3ab\left(a+b\right)=-c^3\)

\(\Leftrightarrow a^3+b^2-3abc=c^3\) (Vì a+b=-c)

\(\Leftrightarrow a^3+b^3+c^3=3abc=3\cdot\left(-2\right)=-6\)

\(\left|x-2016\right|+\left|1008-\frac{1}{2}y\right|=0\)

\(\Leftrightarrow\begin{cases}x-2016=0\\1008-\frac{1}{2}y=0\end{cases}\)\(\Leftrightarrow x=y=2016\)

\(6x\left(1-3x\right)+9x\left(2x-7\right)+171=0\)

\(\Leftrightarrow6x-18x^2+18x^2-63x+171=0\)

\(\Leftrightarrow-57x=-171\)

\(\Leftrightarrow x=3\)

\(\frac{x+1}{2015}+\frac{x+2}{2014}=\frac{x+3}{2013}+\frac{x+4}{2012}\)

\(\Leftrightarrow\left(\frac{x+1}{2015}+1\right)+\left(\frac{x+2}{2014}+1\right)-\left(\frac{x+3}{2013}+1\right)-\left(\frac{x+4}{2012}+1\right)=0\)

\(\Leftrightarrow\)\(\frac{x+2016}{2015}+\frac{x+2016}{2014}-\frac{x+2016}{2013}+\frac{x+2016}{2012}=0\)

\(\Leftrightarrow\left(x+2016\right)\left(\frac{1}{2015}+\frac{1}{2014}-\frac{1}{2013}-\frac{1}{2012}\right)=0\)

\(\Leftrightarrow x+2016=0\) ( vì \(\frac{1}{2015}+\frac{1}{2014}-\frac{1}{2013}-\frac{1}{2012}\ne0\) )

\(\Leftrightarrow x=-2016\)

ui , viết ko gõ viết lại nhá
 

\(\left(x-2016\right)^2+\left(y+2017\right)^2=0\)

\(\Leftrightarrow\begin{cases}x-2016=0\\y+2017=0\end{cases}\)\(\Leftrightarrow\begin{cases}x=2016\\y=-2017\end{cases}\)

\(D=x^2+2xy+y^2-6x-6y-15\)

\(=\left(x+y\right)^2-6\left(x+y\right)-15=-9^2-6\cdot\left(-9\right)-15=120\)