Решение:
Подставим значения a = 1 и b в выражения.
Случай 1: a = 1, b = 1
| b=1 |
| 1) 2a + b | \(2(1) + 1 = 3\) |
| 2) 2(a + b) | \(2(1 + 1) = 2(2) = 4\) |
| 3) 2a + b² | \(2(1) + 1² = 2 + 1 = 3\) |
| 4) 2(a + b²) | \(2(1 + 1²) = 2(1 + 1) = 2(2) = 4\) |
| 5) 2(a + b)² | \(2(1 + 1)² = 2(2)² = 2(4) = 8\) |
| 6) 2a² + b | \(2(1)² + 1 = 2(1) + 1 = 3\) |
| 7) 2(a² + b) | \(2(1² + 1) = 2(1 + 1) = 2(2) = 4\) |
| 8) 2a² + b² | \(2(1)² + 1² = 2(1) + 1 = 3\) |
Случай 2: a = 1, b = 2
| b=2 |
| 1) 2a + b | \(2(1) + 2 = 4\) |
| 2) 2(a + b) | \(2(1 + 2) = 2(3) = 6\) |
| 3) 2a + b² | \(2(1) + 2² = 2 + 4 = 6\) |
| 4) 2(a + b²) | \(2(1 + 2²) = 2(1 + 4) = 2(5) = 10\) |
| 5) 2(a + b)² | \(2(1 + 2)² = 2(3)² = 2(9) = 18\) |
| 6) 2a² + b | \(2(1)² + 2 = 2(1) + 2 = 4\) |
| 7) 2(a² + b) | \(2(1² + 2) = 2(1 + 2) = 2(3) = 6\) |
| 8) 2a² + b² | \(2(1)² + 2² = 2(1) + 4 = 6\) |
Случай 3: a = 1, b = -1
| b=-1 |
| 1) 2a + b | \(2(1) + (-1) = 1\) |
| 2) 2(a + b) | \(2(1 + (-1)) = 2(0) = 0\) |
| 3) 2a + b² | \(2(1) + (-1)² = 2 + 1 = 3\) |
| 4) 2(a + b²) | \(2(1 + (-1)²) = 2(1 + 1) = 2(2) = 4\) |
| 5) 2(a + b)² | \(2(1 + (-1))² = 2(0)² = 0\) |
| 6) 2a² + b | \(2(1)² + (-1) = 2(1) - 1 = 1\) |
| 7) 2(a² + b) | \(2(1² + (-1)) = 2(1 - 1) = 2(0) = 0\) |
| 8) 2a² + b² | \(2(1)² + (-1)² = 2(1) + 1 = 3\) |
Случай 4: a = 1, b = -3
| b=-3 |
| 1) 2a + b | \(2(1) + (-3) = -1\) |
| 2) 2(a + b) | \(2(1 + (-3)) = 2(-2) = -4\) |
| 3) 2a + b² | \(2(1) + (-3)² = 2 + 9 = 11\) |
| 4) 2(a + b²) | \(2(1 + (-3)²) = 2(1 + 9) = 2(10) = 20\) |
| 5) 2(a + b)² | \(2(1 + (-3))² = 2(-2)² = 2(4) = 8\) |
| 6) 2a² + b | \(2(1)² + (-3) = 2(1) - 3 = -1\) |
| 7) 2(a² + b) | \(2(1² + (-3)) = 2(1 - 3) = 2(-2) = -4\) |
| 8) 2a² + b² | \(2(1)² + (-3)² = 2(1) + 9 = 11\) |
Случай 5: a = 1, b = -1 (повторно)
| b=-1 |
| 1) 2a + b | \(2(1) + (-1) = 1\) |
| 2) 2(a + b) | \(2(1 + (-1)) = 2(0) = 0\) |
| 3) 2a + b² | \(2(1) + (-1)² = 2 + 1 = 3\) |
| 4) 2(a + b²) | \(2(1 + (-1)²) = 2(1 + 1) = 2(2) = 4\) |
| 5) 2(a + b)² | \(2(1 + (-1))² = 2(0)² = 0\) |
| 6) 2a² + b | \(2(1)² + (-1) = 2(1) - 1 = 1\) |
| 7) 2(a² + b) | \(2(1² + (-1)) = 2(1 - 1) = 2(0) = 0\) |
| 8) 2a² + b² | \(2(1)² + (-1)² = 2(1) + 1 = 3\) |
Ответ: значения выражений внесены в таблицы для каждого случая.