b, \(\left(x^2+2015\right).\left(x-2016\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x^2+2015=0\\x-2016=0\end{cases}}\)\(\Leftrightarrow\orbr{\begin{cases}x^2==-2015\\x=2016\end{cases}}\)( \(x^2=-2015\)loại do \(x^2\ge0\))
Vậy x= 2016
a, \(xy+3x-7y=21\)
\(\Leftrightarrow x.\left(y+3\right)-7y-21=0\)
\(\Leftrightarrow x.\left(y+3\right)-7.\left(y+3\right)=0\)
\(\Leftrightarrow\left(y+3\right).\left(x-7\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}\hept{\begin{cases}y+3=0\\x-7\in Z\end{cases}}\\\hept{\begin{cases}x-7=0\\y+3\in Z\end{cases}}\end{cases}}\Leftrightarrow\orbr{\begin{cases}\hept{\begin{cases}y=-3\\x-7\in Z\end{cases}}\\\hept{\begin{cases}x=7\\y+3\in Z\end{cases}}\end{cases}}\)\(\orbr{\begin{cases}\hept{\begin{cases}y+3=0\\x-7\in Z\end{cases}}\\\hept{\begin{cases}x-7=0\\y+3\in Z\end{cases}}\end{cases}}\Leftrightarrow\orbr{\begin{cases}\hept{\begin{cases}y=-7\\x-7\in Z\end{cases}}\\\hept{\begin{cases}x=7\\y+3\in Z\end{cases}}\end{cases}}\)\(\Leftrightarrow\orbr{\begin{cases}\hept{\begin{cases}y+3=0\\x-7\in Z\end{cases}}\\\hept{\begin{cases}x-7=0\\y+3\in Z\end{cases}}\end{cases}}\Leftrightarrow\orbr{\begin{cases}\hept{\begin{cases}y=-3\\x-7\in Z\end{cases}}\\\hept{\begin{cases}x=7\\y+3\in Z\end{cases}}\end{cases}}\)
a, xy + 3x - 7y = 21
=> x(y + 3) - 7y - 21 = 21 - 21
=> x(y + 3) - (7y + 21) = 0
=> x(y + 3) - 7(y + 3) = 0
=> (x - 7)(y + 3) = 0
=> \(\orbr{\begin{cases}x-7=0\\x+3=0\end{cases}\Rightarrow\orbr{\begin{cases}x=7\\x=-3\end{cases}}}\)
Vậy x = {7;-3}
b, (x2 + 2015)(x - 2016) = 0
\(\Rightarrow\orbr{\begin{cases}x^2+2015=0\\x-2016=0\end{cases}\Rightarrow\orbr{\begin{cases}x^2=2015\left(loại\right)\\x=2016\end{cases}}}\)
Vậy x = 2016