Question

Решите пожалуйста, очень надо

Answer 1

a)$frac{3x-1}{x^2} + frac{x-9}{3x} = frac{3(3x-1)}{3x^2} + frac{x(x-9)}{3x^2} =frac{9x-3+x^2-9x}{3x^2} = \ frac{x^2-3}{3x^2}$

b)$frac{1}{2a-b}- frac{1}{2a+b}= frac{2a+b}{(2a-b)(2a+b)}- frac{2a-b}{(2a+b)(2a-b)}=frac{2a+b-(2a-b)}{4a^2-b^2}= \ frac{2a+b-2a+b}{4a^2-b^2}= frac{2b}{4a^2-b^2}$

в) $frac{5}{c+3} - frac{5c-2}{c^2+3c} = frac{5}{c+3} - frac{5c-2}{c(c+3)} =frac{5c}{c(c+3)} - frac{5c-2}{c(c+3)} = \ frac{5c-(5c-2)}{c(c+3)} =frac{5c-5c+2}{c(c+3)} =frac{2}{c^2+3c}$

Answer 2

1)(3х-1)/х² + (х-9)/3х=(9х-3+х²-9х)/3х²=(х²-3)/3х²
2)1/(2a-b) -1/(2a+b)=(2a+b-2a+b)/(4a²-b²)=2b/(4a²-b²)
3)5/(c+3) - (5c-2)/(c²+3c)=5/(c+3) - (5c-2)/c(c+3)=(5c-5c+2)/c(c+3)=2/c(c+3)