Question

сократите дроби:
17a^3b + 17a^4c/ 51a^2b^2 - 51a^4c^2

c^2 - 18c + 81/ c - 9

Answer 1

$\frac{17a^3b+17a^4c}{51a^2b^2-51a^4c^2}= \frac{17a^3(b+ac)}{51a^2(b^2-a^2c^2)}= \frac{17a^3(b+ac)}{51a^2(b-ac)(b+ac)}= \frac{17a}{51(b-ac)}$

$\frac{c^2-18c+81}{c-9} = \frac{(c-9)^2}{c-9} =c-9$