Вправи 201 - 342 » 277

277. Подайте у вигляді добутку вираз: 1) a–2 – 4 = (a–1)2 – 22 = (a–1 – 2)(a–1 + 2); 2) a–4b–6 – 1 = (a–2b–3)2 – 12 = (a2b–3 – 1)(a–2b–3 + 1); 3) 25х–8y–12 – z–2 = (5х–4у–6)2 – (z–1)2 = (5x–4y–6 – z–1)(5х–4y–6 +z–1); 4) а–3 + b–3 = (а–1)3 + (b–1)3 = (а–1 + b–1)(а–2 + а–1b–1 + b– 2); 5) m–4 – 6m–2р–1 + 9р–2 = (m–2)2 – 2 • m–2 • 3 • р–1 + (Зр–1)2 = (m–2 – Зр–1)2; 6) а–8 – 49а–2 = (а–4)2 – (7а–1)2 = (а–4 – 7a–1)(а–4 + 7а–1).