Question

1) Desenvolva os seguintes binômios:
a) (y+1)^6                                b) (2x-1)^4                               c) (x²+3y)^5

2) Resolva os binominais
A) (raiz quadrada de 7 -3)^3
B) (1+ raiz quadrada de 2)^4

3) Desenvolva a expressão
[(x-2)² + 4x -3)³]

Answer 1

1) 

a) $(y+1)^6=\dbinom{6}{0}y^6+\dbinom{6}{1}y^5+\dbinom{6}{2}y^4+\dbinom{6}{3}y^3+\dbinom{6}{4}y^2+\dbinom{6}{5}y+\dbinom{6}{6}$

$(y+1)^6=y^6+6y^5+15y^4+20y^3+15y^2+6y+1$

b) $(2x-1)^{4}=\dbinom{4}{0}(2x)^4-\dbinom{4}{1}(2x)^3+\dbinom{4}{2}(2x)^2-\dbinom{4}{3}(2x)^1+\dbinom{4}{4}(2x)^0$

$(2x-1)^{4}=16x^4-32x^3+24x^2-8x+1$

c) $(x^2+3y)^5=\dbinom{5}{0}(x^2)^5.(3y)^0+\dbinom{5}{1}(x^2)^4.(3y)^1+\dbinom{5}{2}(x^2)^3.(3y)^2+\dbinom{5}{3}(x^2)^2.(3y)^3+\dbinom{5}{4}(x^2)^1.(3y)^4+\dbinom{5}{5}(x^2)^0.(3y)^5$

$(x^2+3y)^5=x^{10}+15x^{8}y+90x^{6}y^2+27x^2y^3+405x^2y^4+243y^5$

2)

a) $(\sqrt{7}-3)^{3}=\dbinom{3}{0}(\sqrt{7})^3.3^0-\dbinom{3}{1}(\sqrt{7})^2.3^1+\dbinom{3}{2}(\sqrt{7})^1.3^2-\dbinom{3}{3}(\sqrt{7})^0.3^3$

$(\sqrt{7}-3)^3=7\sqrt{7}-63+27\sqrt{7}-27=34\sqrt{7}-90$


b) $(1+\sqrt{2})^4=\dbinom{4}{0}1^4.(\sqrt{2})^0+\dbinom{4}{1}1^3.(\sqrt{2})^1+\dbinom{4}{2}1^2.(\sqrt{2})^2+\dbinom{4}{3}1^1.(\sqrt{2})^3+\dbinom{4}{4}1^0.(\sqrt{2})^4$

$(1+\sqrt{2})^4=1+4\sqrt{2}+12+8\sqrt{2}+4=17+12\sqrt{2}$

3) $[(x-2)^2+4x-3]^3=[(x^2-4x+4)-4x-3]^3=[x^2+1]^3=x^6+3x^4+3x^2+1$

Answer 2

Resposta:

Eu nunca vi isso, mas pode ser qualquer coisa