Задание Б. Найдите значение выражения при данных значениях переменных а и в:

Решение:

При a = 1, b = 1:

• 1) \( 2a + b = 2(1) + 1 = 2 + 1 = 3 \)
• 2) \( 2(a + b) = 2(1 + 1) = 2(2) = 4 \)
• 3) \( 2a + b^2 = 2(1) + 1^2 = 2 + 1 = 3 \)
• 4) \( 2(a + b^2) = 2(1 + 1^2) = 2(1 + 1) = 2(2) = 4 \)
• 5) \( 2(a + b)^2 = 2(1 + 1)^2 = 2(2)^2 = 2(4) = 8 \)
• 6) \( 2a^2 + b = 2(1)^2 + 1 = 2(1) + 1 = 2 + 1 = 3 \)
• 7) \( 2(a^2 + b) = 2(1^2 + 1) = 2(1 + 1) = 2(2) = 4 \)
• 8) \( 2a^2 + b^2 = 2(1)^2 + 1^2 = 2(1) + 1 = 2 + 1 = 3 \)

При a = 2, b = -1:

• 1) \( 2a + b = 2(2) + (-1) = 4 - 1 = 3 \)
• 2) \( 2(a + b) = 2(2 + (-1)) = 2(1) = 2 \)
• 3) \( 2a + b^2 = 2(2) + (-1)^2 = 4 + 1 = 5 \)
• 4) \( 2(a + b^2) = 2(2 + (-1)^2) = 2(2 + 1) = 2(3) = 6 \)
• 5) \( 2(a + b)^2 = 2(2 + (-1))^2 = 2(1)^2 = 2(1) = 2 \)
• 6) \( 2a^2 + b = 2(2)^2 + (-1) = 2(4) - 1 = 8 - 1 = 7 \)
• 7) \( 2(a^2 + b) = 2(2^2 + (-1)) = 2(4 - 1) = 2(3) = 6 \)
• 8) \( 2a^2 + b^2 = 2(2)^2 + (-1)^2 = 2(4) + 1 = 8 + 1 = 9 \)

При a = -1, b = -3:

• 1) \( 2a + b = 2(-1) + (-3) = -2 - 3 = -5 \)
• 2) \( 2(a + b) = 2(-1 + (-3)) = 2(-4) = -8 \)
• 3) \( 2a + b^2 = 2(-1) + (-3)^2 = -2 + 9 = 7 \)
• 4) \( 2(a + b^2) = 2(-1 + (-3)^2) = 2(-1 + 9) = 2(8) = 16 \)
• 5) \( 2(a + b)^2 = 2(-1 + (-3))^2 = 2(-4)^2 = 2(16) = 32 \)
• 6) \( 2a^2 + b = 2(-1)^2 + (-3) = 2(1) - 3 = 2 - 3 = -1 \)
• 7) \( 2(a^2 + b) = 2((-1)^2 + (-3)) = 2(1 - 3) = 2(-2) = -4 \)
• 8) \( 2a^2 + b^2 = 2(-1)^2 + (-3)^2 = 2(1) + 9 = 2 + 9 = 11 \)

При a = 1, b = -1:

• 1) \( 2a + b = 2(1) + (-1) = 2 - 1 = 1 \)
• 2) \( 2(a + b) = 2(1 + (-1)) = 2(0) = 0 \)
• 3) \( 2a + b^2 = 2(1) + (-1)^2 = 2 + 1 = 3 \)
• 4) \( 2(a + b^2) = 2(1 + (-1)^2) = 2(1 + 1) = 2(2) = 4 \)
• 5) \( 2(a + b)^2 = 2(1 + (-1))^2 = 2(0)^2 = 0 \)
• 6) \( 2a^2 + b = 2(1)^2 + (-1) = 2(1) - 1 = 2 - 1 = 1 \)
• 7) \( 2(a^2 + b) = 2(1^2 + (-1)) = 2(1 - 1) = 2(0) = 0 \)
• 8) \( 2a^2 + b^2 = 2(1)^2 + (-1)^2 = 2(1) + 1 = 2 + 1 = 3 \)

Ответ: см. расчеты выше.