Вправи для повторення розділу 3 » 54 (6)

6) 2/(x^2- 11x+4) + 3/(x^2- 11x+1) = 8/(x^2- 11x-2). Заміна x2 – 11x = t. Тоді 2/(t+4) + 3/(t+1) – 8/(t-2) = 0; ОДЗ: t ≠ –4; t ≠ –1; t ≠ 2. 2(t + 1)(t – 2) + 3(t + 4)(t – 2) – 8(t + 4)(t + 1) = 0; 2t2 + 2t – 4t – 4 + 3t2 + 12t – 6t – 24 – 8t2 – 32t – 8t – 32 = 0; –3t2 – 36t – 60 = 0; t2 + 12t + 20 = 0; D = 122 – 4 • 20 = 64; t1 = (-12+8)/2 = –2; t2 = (-12-8)/2 = –10. 1) t1 = –2; x2 – 11x = –2; x2 – 11x + 2 = 0; D = (–11)2 – 4 = 113; x1,2 = (11 ± √113)/2. 2) t2 = –10; x2 – 11x = –10; x2 – 11x + 10 = 0; D = (–11)2 – 4 • 10 = 81; x3 = (11+9)/2 = 10; x4 = (11-9)/2 = 1. Відповідь: (11 ± √113)/2; 10; 1.