\(\left(x+y-1\right)^2-2\left(x+y-1\right)\left(x+y\right)+\left(x+y\right)^2\)
\(=\left(x+y-1-x-y\right)^2\)
\(=\left(-1\right)^2\)
\(=1\)
\(2x^3-18x=0\)
\(2x\left(x^2-9\right)=0\)
\(\Rightarrow\orbr{\begin{cases}2x=0\\x^2-9=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=0\\x=\pm3\end{cases}}\)
\(\left(x+y-1\right)^2-2\left(x+y-1\right)\left(x+y\right)+\left(x+y\right)^2\)
\(=\left(x+y-1-x-y\right)^2=\left(-1\right)^2=1\)
Áp dụng hằng đẳng thức: \(a^2+2ab+b^2=\left(a+b\right)^2\)
\(2x^3-18x=0\Leftrightarrow2x\left(x^2-9\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}2x=0\\x^2-9=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=0\\x^2=9\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=0\\x=\left\{-3;3\right\}\end{cases}}}\)
Vậy x = {-3;0;3}
