Question

Chứng minh rằng (4n-3)^2-(3n-4)^2 chia hết cho 7 với mọi n thuộc Z

 

Answer 1

\(=\left(4n-3\right)^2-\left(3n-4\right)^2\)

\(=\left[\left(4n-3\right)+\left(3n-4\right)\right]\left[\left(4n-3\right)\right]-\left(3n-4\right)\)

\(=\left(7n-7\right)\left(n+1\right)=7\left(n-1\right)\left(n+1\right)\)

Vậy \(\left(4n-3\right)^2-\left(3n-4\right)^2\)   Chia hết cho 7 với mọi n thuộc Z

Answer 2

t.i.c.k mik mik t.i.c.k lại

Answer 3

\(\left(4n-3\right)^2-\left(3n-4\right)^2=\left[\left(4n-3\right)+\left(3n-4\right)\right]\left[\left(4n-3\right)-\left(3n-4\right)\right]=\left(7n-7\right)\left(n+1\right)=7\left(n-1\right)\left(n+1\right)\)