cho A= 999993^1999 – 555557^1997. chứng minh A chia hất cho 5 10/11/2021 Bởi Emery cho A= 999993^1999 – 555557^1997. chứng minh A chia hất cho 5
Ta có : `A=999993^{1999}-555557^{1997}` `⇒ A=999993^{1998}.999993-555557^{1996}.555557` $⇒ A=\left(999993^2\right)^{999}.999993-\left(555557^2\right)^{998}.555557$ $⇒ A=\left(…9\right)^{999}.999993-\left(…1\right).555557$ `⇒ A = (…7) – (…7)` `⇒ A = (…0)` `⇒ A \vdots 5` Bình luận
`A=999993^{1999}-555557^{1997}` `⇒ A=999993^{1998}.999993-555557^{1996}.555557` $⇒ A=\left(999993^2\right)^{999}.999993-\left(555557^2\right)^{998}.555557$ $⇒ A=\left(….9\right)^{999}.999993-\left(….1\right).555557$ `⇒ A = (….7) – (….7)` `⇒ A = (….0)` Vì `A` có tận cùng là `0` `⇒ A \vdots 5` `(Đpcm)` Bình luận
Ta có :
`A=999993^{1999}-555557^{1997}`
`⇒ A=999993^{1998}.999993-555557^{1996}.555557`
$⇒ A=\left(999993^2\right)^{999}.999993-\left(555557^2\right)^{998}.555557$
$⇒ A=\left(…9\right)^{999}.999993-\left(…1\right).555557$
`⇒ A = (…7) – (…7)`
`⇒ A = (…0)`
`⇒ A \vdots 5`
`A=999993^{1999}-555557^{1997}`
`⇒ A=999993^{1998}.999993-555557^{1996}.555557`
$⇒ A=\left(999993^2\right)^{999}.999993-\left(555557^2\right)^{998}.555557$
$⇒ A=\left(….9\right)^{999}.999993-\left(….1\right).555557$
`⇒ A = (….7) – (….7)`
`⇒ A = (….0)`
Vì `A` có tận cùng là `0`
`⇒ A \vdots 5` `(Đpcm)`