Question

Сократить дробь 81x ^4 + 9x ^2 + 1 ------------------------- 9x ^2 +3x + 1

Answer 1

$\dfrac{81x^4+9x^2+1}{9x^2+3x+1}=\dfrac{81x^4+18x^2+1-9x^2}{9x^2+3x+1}=\dfrac{(9x^2+1)^2-9x^2}{9x^2+3x+1}=\\ \\ \\ =\dfrac{(9x^2-3x+1)(9x^2+3x+1)}{9x^2+3x+1}=9x^2-3x+1$

Answer 2

To simplify the fraction, you can perform polynomial division. The expression can be simplified as follows:

Divide 81x^4 by 9x^2: 81x^4 / 9x^2 = 9x^2

So, the expression becomes:

(81x^4 + 9x^2 + 1) / (9x^2 + 3x + 1) = (9x^2 + 9x^2 + 1) / (9x^2 + 3x + 1)

Now, combine like terms in the numerator:

(9x^2 + 9x^2 + 1) = 18x^2 + 1

So, the simplified fraction is:

(81x^4 + 9x^2 + 1) / (9x^2 + 3x + 1) = (18x^2 + 1) / (9x^2 + 3x + 1)