Вправи для повторення розділу 3 » 54 (4)

4) (x + 2)(x – 7) = 19/((x-1)(x-4)); x2 – 5x – 14 = 19/(x^2- 5x+4). Заміна x2 – 5x = t, тоді t – 14 = 19/(t+4). t2 – 14t + 4t – 56 = 19, t2 – 10t – 75 = 0, t ≠ –4; t ≠ – 4. D = (–10)2 – 4 • (–75) = 400; t1 = (10+20)/2 = 15; t2 = (10-20)/2 = –5. 1) t1 = 15; x2 – 5x = 15; x2 – 5x – 15 = 0; D = (–5)2 – 4 • (–15) = 85; x1,2 = (5 ± √85)/2. 2) t2 = –5; x2 – 5x = –5; x2 – 5x + 5 = 0; D = (–5)2 – 4 • 5 = 5; x3,4 = (5 ± √5)/2 Відповідь: (5 ± √85)/2; (5 ± √5)/2.