Toán Chứng minh (4n-3)^2 – (3n – 4 )^2 chia hết cho 7 14/07/2021 By Serenity Chứng minh (4n-3)^2 – (3n – 4 )^2 chia hết cho 7
= (4n − 3)² − (3n − 4)²= [(4n − 3) + (3n − 4)] [(4n − 3) − (3n − 4)] =(4n-3+3n-4) (4n-3-3n+4)= (7n − 7) (n + 1) = 7 (n − 1) (n + 1) ⋮ 7 Vậy (4n − 3)² − (3n − 4)² ⋮ 7 Trả lời
$(4n-3)^2-(3n-4)^2$
$=(4n-3-3n+4)(4n-3+3n-4)$
$=(n+1)(7n-7)$
`=7(n+1)(n-1)\vdots 7` với `∀n`
= (4n − 3)² − (3n − 4)²
= [(4n − 3) + (3n − 4)] [(4n − 3) − (3n − 4)]
=(4n-3+3n-4) (4n-3-3n+4)
= (7n − 7) (n + 1)
= 7 (n − 1) (n + 1) ⋮ 7
Vậy (4n − 3)² − (3n − 4)² ⋮ 7