a) Ta có: \(\left(2n-1\right)^3-\left(2n-1\right)\)
\(=\left(2n-1\right)\left[\left(2n-1\right)^2-1\right]\)
\(=\left(2n-1\right)\left(2n-1-1\right)\left(2n-1+1\right)\)
\(=\left(2n-1\right)\left(2n-2\right)\cdot2n\)
\(=2n\cdot2\cdot\left(n-1\right)\cdot\left(2n-1\right)\)
\(=4n\cdot\left(n-1\right)\cdot\left(2n-1\right)⋮4\)
mà n và n-1 là hai số nguyên liên tiếp
nên \(4\cdot n\cdot\left(n-1\right)\cdot\left(2n-1\right)⋮4\cdot2\)
hay \(\left(2n-1\right)^3-\left(2n-1\right)⋮8\)
b) Ta có: \(\left(n+7\right)^2-\left(n-5\right)^2\)
\(=\left(n+7-n+5\right)\left(n+7+n-5\right)\)
\(=12\cdot\left(2n+2\right)\)
\(=12\cdot2\cdot\left(n+1\right)=24\left(n+1\right)⋮24\)(đpcm)
