Question

(ab-4)^5÷179+147-465

Answer 1

все просто и логично ровно 2565

Answer 2

Let's solve the given expression step by step:

1. Start with the expression: (ab-4)^5÷179+147-465

2. The first step is to simplify the term (ab-4)^5:

The fifth power of (ab-4) can be expanded using the binomial theorem:

(ab-4)^5 = C(5,0)(ab)^5(-4)^0 + C(5,1)(ab)^4(-4)^1 + C(5,2)(ab)^3(-4)^2 + C(5,3)(ab)^2(-4)^3 + C(5,4)(ab)^1(-4)^4 + C(5,5)(ab)^0(-4)^5

Where C(n, k) represents the binomial coefficient "n choose k."

The binomial coefficients for this expansion are: C(5,0) = 1, C(5,1) = 5, C(5,2) = 10, C(5,3) = 10, C(5,4) = 5, C(5,5) = 1

Now, we'll expand each term:

(ab-4)^5 = (1)(a^5b^5) + (5)(a^4b^4)(-4) + (10)(a^3b^3)(-4)^2 + (10)(a^2b^2)(-4)^3 + (5)(ab)(-4)^4 + (1)(-4)^5 = a^5b^5 - 20a^4b^4 + 160a^3b^3 - 640a^2b^2 + 1280ab - 1024

3. Now we have: (a^5b^5 - 20a^4b^4 + 160a^3b^3 - 640a^2b^2 + 1280ab - 1024) ÷ 179 + 147 - 465

4. Next, perform the division: Divide each term of the polynomial by 179:

(a^5b^5)/179 - (20a^4b^4)/179 + (160a^3b^3)/179 - (640a^2b^2)/179 + (1280ab)/179 - 1024/179 + 147 - 465

5. Finally, add and subtract the constants:

(a^5b^5)/179 - (20a^4b^4)/179 + (160a^3b^3)/179 - (640a^2b^2)/179 + (1280ab)/179 - 1024/179 + 147 - 465 = (a^5b^5)/179 - (20a^4b^4)/179 + (160a^3b^3)/179 - (640a^2b^2)/179 + (1280ab)/179 - (1024 + 179*465)/179 + 147

6. Further simplification:

(a^5b^5)/179 - (20a^4b^4)/179 + (160a^3b^3)/179 - (640a^2b^2)/179 + (1280ab)/179 - (1024 + 83235)/179 + 147 = (a^5b^5)/179 - (20a^4b^4)/179 + (160a^3b^3)/179 - (640a^2b^2)/179 + (1280ab)/179 - 1024/179 - 465 + 147

7. Final answer:

(a^5b^5)/179 - (20a^4b^4)/179 + (160a^3b^3)/179 - (640a^2b^2)/179 + (1280ab)/179 - 1024/179 - 465 + 147

The result is the simplified expression above.