Question

Докажите что при любом целом п значение выражения (2п-3)^2-(4п-1)(п+6) кратно 5.

Answer 1

$(2n-3)^2-(4n-1)(n+6)=4n^2-12n+9-(4n^2+23n-6)=$
$=-35n+15=5(-7n+3)$

Answer 2

Checking the Expression for Divisibility by 5

To prove that the expression \((2p-3)^2 - (4p-1)(p+6)\) is divisible by 5 for any integer \(p\), we can expand the expression and then check for divisibility by 5.

Expanding the Expression

The given expression is \((2p-3)^2 - (4p-1)(p+6)\).

Expanding the expression: \((2p-3)^2 = 4p^2 - 12p + 9\) \((4p-1)(p+6) = 4p^2 + 24p - p - 6\)

Substituting these expansions back into the original expression: \(4p^2 - 12p + 9 - (4p^2 + 24p - p - 6)\)

Simplifying the expression: \(4p^2 - 12p + 9 - 4p^2 - 24p + p + 6\) \(-35p + 15\)

Checking for Divisibility by 5

To prove that the expression is divisible by 5 for any integer \(p\), we need to show that \(-35p + 15\) is divisible by 5.

We can rewrite \(-35p + 15\) as \(5(-7p + 3)\). This clearly shows that the expression is a multiple of 5 for any integer \(p\).

Therefore, the expression \((2p-3)^2 - (4p-1)(p+6)\) is indeed divisible by 5 for any integer \(p\).

This conclusion is based on the mathematical operations performed and the properties of divisibility by 5.