Question

3(b-1)² +(b+2)(b²-2b+4)-(b+1)³ при b=⅓​

Answer 1

Ответ:

7

Объяснение:

3(b-1)² +(b+2)(b²-2b+4)-(b+1)³ = 3b² - 6b + 3 + b³ + 2b² - 2b² - 4b + 4b + 8 - b³ - 3b² - 3b - 1 =

= -9b + 10 = -9*1/3 + 10 = -3 + 10 = 7

Answer 2

To evaluate the expression 3(b-1)² +(b+2)(b²-2b+4)-(b+1)³ for b=⅓, we can substitute the value of b into the expression and perform the calculations.

Let's plug in b = ⅓:

3(⅓ - 1)² + (⅓ + 2)(⅓² - 2(⅓) + 4) - (⅓ + 1)³

First, let's simplify the terms inside the parentheses:

⅓ - 1 = -2/3

⅓² = 1/9

2(⅓) = 2/3

⅓ + 1 = 4/3

Now, let's substitute these values back into the expression:

3(-2/3)² + (⅓ + 2)(1/9 - 2/3 + 4) - (4/3)³

Next, let's simplify the squared term and the cube term:

(-2/3)² = 4/9

(4/3)³ = 64/27

Now, substitute these simplified values into the expression:

3(4/9) + (⅓ + 2)(1/9 - 2/3 + 4) - 64/27

Now, let's simplify the expression within the parentheses:

(⅓ + 2)(1/9 - 2/3 + 4) = (7/3)(1/9 - 2/3 + 4) = (7/3)(1/9 - 6/9 + 4) = (7/3)(-1/3 + 4) = (7/3)(-1/3 + 12/3) = (7/3)(11/3) = 77/9

Now, substitute this value back into the expression:

3(4/9) + 77/9 - 64/27

Let's find a common denominator and simplify further:

3(4/9) = 12/9

12/9 + 77/9 - 64/27 = (12 + 77 - 64)/9 = 25/9

Therefore, when b = ⅓, the value of the expression 3(b-1)² +(b+2)(b²-2b+4)-(b+1)³ is 25/9.