Question

Ayuda porfavor xc reporto las respuestas que no tengan que ver:)

Resolver con teorema del binomio


(2x-6)^{2}

(3x+ 1)^{3}

(2a−3b)^{4}

(1 +y)^{3}

(r−6)^{8}

(x+ 3y)^{2}

(a−2b)^{6}

(s−7)^{7}

(4p−q)^{5}

(x−1)^{4}

Answer 1

Explicación paso a paso:

1. (2x - 6)^ 2 = (2x)^2 - 2(2x)(6)+(2x)^0 (6)^2

=4x^2 - 24x + 36

2. (3x + 1)^3 = (3x)^3+ 3(3x)^2 (1) + 3(3x)^1 (1)^2 + (3x)^0 (1)^3

=27x^3 + 27x^2 + 9x +1

3. (2a - 3b)^4 = (2a)^4 - 4(2a)^3 (3b) + 6(2a)^2 (3b)^2 - 4 ( 2a) ^1 (3b) ^3 + ( 2a)^0 (3b)^4

=16a^4 - 96 a^3 b + 216 a^2 b^2 - 216a b^3 + 81 b^4

4. (1 + y) ^ 3 = (1)^3 + 3(1)^2(y) + 3 (1)^1 (y) ^2 + y^3

=1+3y + 3y^2 + y^3

5. (r- 6) ^ 8 = r^8 - 8r^7 (6) +28r^6 (6)^2 - 56r^5 (6)^3 + 70r^4 (6)^4 - 56 r^3 (6)^5 + 28 r^2 (6)^6 - 8r (6)^7 + 6^8

= r^8 - 48r^7 + 1008r^6 - 12096r^5 +90720r^4 - 435456 r^3 +1306368 r^2−2239488 r +1679616

6. ( x + 3y)^2 = x^2 + 2(x)^1 ( 3y)+ x^0 ( 3y)^2

= x^2 + 6xy + 9y^2

7. ( a - 2b) ^ 6 = a^6 - 6 a^5 ( 2b) + 15 a^4 ( 2b)^2 - 20 a^3 ( 2b)^3 + 15 a^2 ( 2b)^4 - 6a (2b)^5 + (2b)^6

= a^6 - 12a^5 b + 60 a^4 - 160a^3 b^3 + 240 a^2 b^4 - 192 a b^5 +64 b^6