Bài 1:

a, \(77^{n+1}=77^n.77+77^n\)

\(=77^n\left(77+1\right)=77^n.78⋮78\)

\(\Rightarrowđpcm\)

b, \(n^2\left(n-1\right)+\left(n^2-n\right)\)

\(=n^2\left(n-1\right)+n\left(n-1\right)\)

\(=\left(n^2+n\right)\left(n-1\right)=n\left(n+1\right)\left(n-1\right)\)

Vì 3 số liên tiếp chia hết cho 2, 3

Mà ( 2; 3 ) = 1

\(\Rightarrow n\left(n+1\right)\left(n-1\right)⋮6\)

\(\Rightarrowđpcm\)

c, tương tự

Bài 2:

a, \(x+y=xy\)

\(\Leftrightarrow x-xy+y=0\)

\(\Leftrightarrow x\left(1-y\right)-1+y=-1\)

\(\Leftrightarrow\left(x-1\right)\left(1-y\right)=1\)

\(\Leftrightarrow\left\{{}\begin{matrix}x-1=1\\1-y=-1\end{matrix}\right.\) hoặc \(\left\{{}\begin{matrix}x+1=-1\\1-y=1\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x=2\\y=2\end{matrix}\right.\) hoặc \(\left\{{}\begin{matrix}x=0\\y=0\end{matrix}\right.\)

Vậy x = y = 2 hoặc x = y = 0

b, tương tự

bài 2

a, x(y-1) +y-1=1 suy ra (x+1)(y-1)=1

suy ra (x,y)=(1;1),(-1;-1)