Question

Докажите ,что если сумма ab+cd делится на a-c, то сумма ad+bc также делится на a-c.

Answer 1

ab + cd = m*(a-c),
ab + cd = am - cm,
am - cm - ab - cd = 0, добавим к обеим частям (ad + bc),
ad + bc = (am - cm) - ab -cd + ad + bc = m*(a-c) + (bc - ab) + (ad - cd) = m*(a-c) - b*(a-c) + d*(a-c) = [ (a-c) выносим за скобку ] =
= (a-c)*(m - b + d), ч.т.д.

Answer 2

Proof of the Given Statement

To prove that if the sum of ab + cd is divisible by a-c, then the sum of ad + bc is also divisible by a-c, we can use the following approach:

Let's start by expressing ab + cd and ad + bc in terms of a and c and then try to establish a relationship between them.

We know that: 1. ab + cd 2. ad + bc

We also know that ab + cd is divisible by a-c. We need to prove that ad + bc is also divisible by a-c.

Proof:

Given: ab + cd is divisible by a-c.

We can express ab + cd as a(b) - c(d).

Now, let's express ad + bc in terms of a and c: ad + bc = a(d) - c(b).

We can see that ad + bc can also be expressed as a difference of two products involving a and c.

Therefore, if ab + cd is divisible by a-c, then ad + bc is also divisible by a-c.

This proves the given statement.

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