Вправи для повторення розділу 3 » 54 (5)

5) 5/(x^2- x-x) + 1/(x^2- x-5) = 2. Заміна x2 – x – 1 = t. Tоді 5/t + 1/(t-4) = 2; (5t-20+t)/(t(t-4)) = 2; 2t(t – 4) = 5t – 20; t ≠ 0, t ≠ 4. 2t2 – 8t – 6t + 20 = 0; 2t2 – 14t + 20 = 0; t2 – 7t + 10 = 0; D = (–7)2 – 4 • 10 = 9; t1 = (7+3)/2 = 5; t2 = (7-3)/2 = 2. 1) t1 = 5; x2 – x – 1 = 5; x2 – x – 6 = 0; D = (–1)2 – 4 • (–6) = 25; x1 = (1+5)/2 = 3; x2 = (1-5)/2 = –2. 2) t2 = 2; x2 – x – 1 = 2; x2 – x – 3 = 0; D = (–1)2 – 4 • (–3) = 13; x3,4 = (1 ± √13)/2. Відповідь: 3; –2; (1 ± √13)/2.