Question

(3a^7+b ^-1)^5 Заранее спасибо

Answer 1

59 049а^35+в^-5

я думаю, что так

Answer 2

To expand the expression (3a^7+b^-1)^5, you can use the binomial theorem, which states that:

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(a + b)^n = C(n, 0)*a^n*b^0 + C(n, 1)*a^(n-1)*b^1 + ... + C(n, n-1)*a^1*b^(n-1) + C(n, n)*a^0*b^n

where C(n, k) represents the binomial coefficient "n choose k", given by:

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C(n, k) = n! / (k!*(n-k)!)

In your case, a = 3a^7 and b = b^-1. So, applying the binomial theorem with n = 5, you get:

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(3a^7+b^-1)^5 = C(5, 0)*(3a^7)^5*(b^-1)^0 + C(5, 1)*(3a^7)^4*(b^-1)^1 + ... + C(5, 4)*(3a^7)^1*(b^-1)^4 + C(5, 5)*(3a^7)^0*(b^-1)^5

Now you just need to simplify each term by expanding the powers of 3a^7 and b^-1, and then simplify the fractions. Note that b^-1 raised to an odd power will become negative, while b^-1 raised to an even power will be positive. Here's the expanded expression:

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(3a^7+b^-1)^5 = 243a^35 + 405a^28b^-1 - 270a^21b^-2 + 90a^14b^-3 - 15a^7b^-4 + b^-5

So, (3a^7+b^-1)^5 = 243a^35 + 405a^28b^-1 - 270a^21b^-2 + 90a^14b^-3 - 15a^7b^-4 + b^-5.