Решить уравнение (2/х)+(10/(х²—2х))=(1+2x)/(x—2)

2/х + 10/(х^2 - 2 * х) = (1 + 2 * x)/(x - 2).2/х + 10/(х * (х – 2)) - (1 + 2 * x)/(x - 2) = 0.(2 * (x – 2) + 10 – x * (1 + 2 * x))/(x * (x – 2))= 0.2 * x – 4 + 10 – x – 2 * x^2 = 0; x ≠ 0, x ≠ 2.2 * x^2 – x – 6 = 0.D = 1 - 4 * 2 * (- 6) = 49 = 7^2.X1,2 = (1 ± √49)/(2 * 2) = (1 ± 7)/4.X1 = (1 + 7)/4 = 8/4 = 2 — не является корнем, так как x ≠ 2.X2 = (1 – 7)/4 = - 6/4 = - 3/2.Проверка:2/(-3/2) + 10/((- 3/2)^2 - 2 * (- 3/2)) = (1 + 2 * (- 3/2))/(- 3/2 - 2).- 4/3 + 10/(9/4 + 3) = (1 - 3))/(- 7/2).- 4/3 + 10/(21/4) = 4/7.- 4/3 + 40/21 = 4/7.12/21 = 4/7.4/7 = 4/7.Ответ: х = -3/2.