Question

(4a^2-20a+25):(2a-5)=2a-5

Answer 1

4A^2 - 20A + 25 = ( 2A - 5)*( 2A - 5 )
4A^2 - 20A + 25 = 4A^2 - 20A + 25 
25 = 25 
Ответ любые значения переменной А 

Answer 2

To verify whether $\frac{4a^2 - 20a + 25}{2a - 5}$ is equal to $2a - 5$, we can use long division to simplify the expression:

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         2a - 5
____________________
2a - 5 | 4a^2 - 20a + 25

Let's go through the steps of long division:

1. Divide the first term of the numerator (4a^2) by the first term of the denominator (2a). This gives us 2a, which we write on top.

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         2a - 5
____________________
2a - 5 | 4a^2 - 20a + 25
         4a^2 - 10a

2. Multiply the entire denominator (2a - 5) by 2a:

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         2a - 5
____________________
2a - 5 | 4a^2 - 20a + 25
         4a^2 - 10a
         ___________

3. Subtract the result from step 2 (4a^2 - 10a) from the numerator (4a^2 - 20a + 25):

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         2a - 5
____________________
2a - 5 | 4a^2 - 20a + 25
         4a^2 - 10a
         ___________
                  -10a + 25

4. Bring down the next term, which is +25:

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         2a - 5
____________________
2a - 5 | 4a^2 - 20a + 25
         4a^2 - 10a
         ___________
                  -10a + 25
                    + 25

5. Now, divide (-10a + 25) by (2a - 5). This gives us -5:

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         2a - 5
____________________
2a - 5 | 4a^2 - 20a + 25
         4a^2 - 10a
         ___________
                  -10a + 25
                    + 25
         ___________
                      0

Since we have a remainder of 0, the division is exact, and the result of $\frac{4a^2 - 20a + 25}{2a - 5}$ is indeed $2a - 5$. So, your statement is correct:

$\frac{4a^2 - 20a + 25}{2a - 5} = 2a - 5$.