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ГДЗ Алгебра 7 клас Тарасенкова (2024)
Роздiл 3. Одночлени i многочлени
Реклама:
Спростіть вираз:
1) (3 + b)3 – b3= ((3 + b) – b)((3 + b)2 + b • (3 + b) + b2) = 3 • (9 + 6b + b2 + 3b + b2 + b2) = 3 • (3b2 + 9b + 9) = 9b2 + 27b + 27;
2) (5 – 4a)3 + 64a3 = (5 – 4a)3 + (4a)3 = (5 – 4a + 4a)((5 – 4a)2 – (5 – 4a) • 4a + (4a)2) = 5 • (25 – 40a + 16a2 – 20a + 16a2 + 16a2) = 5(25 – 60a + 48a2) = 125 – 300a + 240a2;
3) 343y6 – (5 + 7y2)3 = (7y2)3 – (5 + 7y2)3 = (7y2 – (5 + 7y2))((7y2)2 + 7y2 • (5 + 7y2) + (5 + 7y2)2) = (7у2 – 5 – 7y2)(49y4 + 35у2 + 49у2 + 25 + 70у2 + 49y4) = –5(147y4 + 105y2 + 25) = –735 + 25) = –735у4 – 525y2 – 125;
4) 125x3 – (5х + 1)3 = (5х)3 – (5х + 1)3 = (5х – 5х – 1)(25x2 + 5х(5х + 1) + (5х + 1)2) = –1(25х2 + 25х2 + 5х + 25x2 + 10x + 1) = –(75x2 + 15x + 1) = –75x2 – 15x – 1.
1) (3 + b)3 – b3= ((3 + b) – b)((3 + b)2 + b • (3 + b) + b2) = 3 • (9 + 6b + b2 + 3b + b2 + b2) = 3 • (3b2 + 9b + 9) = 9b2 + 27b + 27;
2) (5 – 4a)3 + 64a3 = (5 – 4a)3 + (4a)3 = (5 – 4a + 4a)((5 – 4a)2 – (5 – 4a) • 4a + (4a)2) = 5 • (25 – 40a + 16a2 – 20a + 16a2 + 16a2) = 5(25 – 60a + 48a2) = 125 – 300a + 240a2;
3) 343y6 – (5 + 7y2)3 = (7y2)3 – (5 + 7y2)3 = (7y2 – (5 + 7y2))((7y2)2 + 7y2 • (5 + 7y2) + (5 + 7y2)2) = (7у2 – 5 – 7y2)(49y4 + 35у2 + 49у2 + 25 + 70у2 + 49y4) = –5(147y4 + 105y2 + 25) = –735 + 25) = –735у4 – 525y2 – 125;
4) 125x3 – (5х + 1)3 = (5х)3 – (5х + 1)3 = (5х – 5х – 1)(25x2 + 5х(5х + 1) + (5х + 1)2) = –1(25х2 + 25х2 + 5х + 25x2 + 10x + 1) = –(75x2 + 15x + 1) = –75x2 – 15x – 1.