Question

Ejercicio 128 resuelto del álgebra de Baldor numero 8 3/5 - 2a+1/10a - 4a2+1/20a2

Answer 1

3/5 - (2a+1)/10a - (4a²+1)/20a²
(30a - 10a - 5)/50a - (4a²+1)/20a²
(20a - 5)/50a - (4a²+1)/20a²
5(4a - 1)/5(10a) - (4a²+1)/20a²
(4a - 1)/10a - (4a²+1)/20a²
(80a³ - 20a² - 40a³ - 10a)/200a³
(40a³ - 20a² - 10a)/200a³
10a(4a² - 2a - 1)/10(20a³)
(4a² - 2a - 1)/20a³