Question

Answer 1

a) \(\left(2k-1\right)^2-9\)

\(=\left[\left(2k-1\right)+3\right]\left[\left(2k-1\right)-3\right]\)

\(=\left(2k+2\right)\left(2k-4\right)\)

\(=4k^2-8k+4k-8\)

\(=4k^2-4k-8\)

\(=4\cdot\left(k^2-k-2\right)\)

Mà: \(4\cdot\left(k^2-k-2\right)\) ⋮ 4

Nên: \(\left(2k-1\right)^2-9\) ⋮ 4

b) \(4-\left(1+3k\right)^2\)

\(=\left[2-\left(1+3k\right)\right]\left[2+\left(1+3k\right)\right]\)

\(=\left(1-3k\right)\left(3+3k\right)\)

\(=3+3k-9k-9k^2\)

\(=3-6k-9k^2\)

\(=3\cdot\left(1-2k-3k^2\right)\)

Mà: \(3\cdot\left(1-2k-3k^2\right)\) ⋮ 3

Nên: \(4-\left(1+3k\right)^2\) ⋮ 3

Answer 2

a: =(2k-1-3)(2k-1+3)

=(2k-4)(2k+2)

=4(k-2)(k+1) chia hết cho 4

b: =(2-1-3k)(2+1+3k)

=(3k+3)(-3k+1)

=3(k+1)(-3k+1) chia hết cho 3