Question

(2x-y^2) в 6 степени

Answer 1

(2x)⁶+6(2x)⁵·(-y²)+15· (2x)⁴·(-y²)²+20·(2x)³·(-y²)³+15·(2x)²·(-y²)⁴+6· (2x)·(-y²)⁵+(-y²)⁶=

=64x⁶-192x⁵·y²+240x⁴·y⁴-160x³·y⁶+60·x²·y⁸-12x·y¹⁰+y¹²

Answer 2

(2x - y^2)^6 can be expanded using the binomial theorem. The general form of the binomial theorem is:

(a + b)^n = C(n, 0)a^n b^0 + C(n, 1)a^(n-1) b^1 + C(n, 2)a^(n-2) b^2 + ... + C(n, n-1)a^1 b^(n-1) + C(n, n)a^0 b^n

where C(n, k) represents the binomial coefficient, given by C(n, k) = n! / (k!(n-k)!).

In this case, we have (2x - y^2)^6. Let's expand it using the binomial theorem:

(2x - y^2)^6 = C(6, 0)(2x)^6 (-y^2)^0 + C(6, 1)(2x)^5 (-y^2)^1 + C(6, 2)(2x)^4 (-y^2)^2 + C(6, 3)(2x)^3 (-y^2)^3 + C(6, 4)(2x)^2 (-y^2)^4 + C(6, 5)(2x)^1 (-y^2)^5 + C(6, 6)(2x)^0 (-y^2)^6

Expanding each term:

C(6, 0)(2x)^6 (-y^2)^0 = 1(2x)^6 = 64x^6 C(6, 1)(2x)^5 (-y^2)^1 = 6(2x)^5 (-y^2) = -192x^5 y^2 C(6, 2)(2x)^4 (-y^2)^2 = 15(2x)^4 (-y^2)^2 = 720x^4 y^4 C(6, 3)(2x)^3 (-y^2)^3 = 20(2x)^3 (-y^2)^3 = -960x^3 y^6 C(6, 4)(2x)^2 (-y^2)^4 = 15(2x)^2 (-y^2)^4 = 720x^2 y^8 C(6, 5)(2x)^1 (-y^2)^5 = 6(2x)^1 (-y^2)^5 = -192x y^10 C(6, 6)(2x)^0 (-y^2)^6 = 1(2x)^0 (-y^2)^6 = 64y^12

Putting it all together:

(2x - y^2)^6 = 64x^6 - 192x^5 y^2 + 720x^4 y^4 - 960x^3 y^6 + 720x^2 y^8 - 192x y^10 + 64y^12

So, the expansion of (2x - y^2)^6 is 64x^6 - 192x^5 y^2 + 720x^4 y^4 - 960x^3 y^6 + 720x^2 y^8 - 192x y^10 + 64y^12.