Question

 sobre Binomios 

4) simplifica  a)(n-5)!    b)( 2n+3)
                      (n-3)!       (2n+1)
5) desenvolver os binomios 
a) 2x+3)5      b) (x²-3x)³

Answer 1

$=x^{3} . ( x^{3} -9 x^{2}+27x - 27 )$$4)a) \frac{(n-5)!}{(n-3)!} =\frac{(n-5)!}{(n-3).(n-4).(n-5)!} = \frac{1}{(n-3).(n-4)}$

b) Como não foi colocado sinal de fatorial, temos:

$\frac{(2n+3)}{(2n+1)} =\frac{(2n+1+2)}{(2n+1)}=\frac{(2n+1)}{(2n+1)}+\frac{2}{(2n+1)}=1+\frac{2}{2n+1}$

5) a) $(2x+3).5=10x+15$

b) $( x^{2} -3x)^{3} = x^{3} . (x-3)^{3} = x^{3} . ( x^{3} -3 x^{2} .3+3x. 3^{2} - 3^{3} )$

Answer 2

Binômio de Newton:

$\boxed{(a+b)^{n}=\sum\limits_{k=0}^{n}C_{n,k}*a^{n-k}*y^{k}}$

$\boxed{C_{n,k}=\frac{n!}{k!(n-k)!}}$

Fatorial:

$n!=n*(n-1)*(n-2)*(n-3)*...*3*2*1$
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4)
a)

$\frac{(n-5)!}{(n-3)!}=\frac{(n-5)!}{(n-3)*(n-4)*(n-5)!}\\\\\frac{(n-5)!}{(n-3)!}=\frac{1}{(n-3)(n-4)}$

b)

$\frac{(2n+3)!}{(2n+1)!}=\frac{(2n+3)*(2n+2)*(2n+1)!}{(2n+1)!}\\\\ \frac{(2n+3)!}{(2n+1)!}=\frac{(2n+3)*(2n+2)*1}{1}\\\\\frac{(2n+3)!}{(2n+1)!}=(2n+3)(2n+2)$

5)

a)

$(2x+3)^{5}=\sum\limits_{k=0}^{5}C_{5,k}*(2x)^{5-k}*3^{k}$

Resolvendo o somatório:

Vou resolver de uma maneira que dê pra escrever pelo brainly:

$(2x+3)^{5}=\sum\limits_{k=0}^{5}C_{5,k}*(2x)^{5-k}*3^{k}=u_{0}+u_{1}+u_{2}+u_{3}+u_{4}+u_{5}$

$u_{0}=C_{5,0}*(2x)^{5-0}*3^{0}\\u_{0}=1*(2x)^{5}*1\\u_{0}=1*2^{5}x^{5}\\u_{0}=1*32x^{5}\\u_{0}=32x^{5}$

$u_{1}=C_{5,1}*(2x)^{5-1}*3^{1}\\u_{1}=5*(2x)^{4}*3\\u_{1}=15*2^{4}x^{4}\\u_{1}=15*16x^{4}\\u_{1}=240x^{4}$

$u_{2}=C_{5,2}*(2x)^{5-2}*3^{2}\\u_{2}=10*(2x)^{3}*9\\u_{2}=90*2^{3}x^{3}\\u_{2}=90*8x^{3}\\u_{2}=720x^{3}$

$u_{3}=C_{5,3}*(2x)^{5-3}*3^{3}\\u_{3}=10*(2x)^{2}*27\\u_{3}=270*2^{2}x^{2}\\u_{3}=270*4x^{2}\\u_{3}=1080x^{2}$

$u_{4}=C_{5,4}*(2x)^{5-4}*3^{4}\\u_{4}=5*(2x)^{1}*81\\u_{4}=405*2x\\u_{4}=810x$

$u_{5}=C_{5,5}*(2x)^{5-5}*3^{5}\\u_{5}=1*(2x)^{0}*3^{5}\\u_{5}=1*1*243\\u_{5}=243$

$(2x+3)^{5}=u_{0}+u_{1}+u_{2}+u_{3}+u_{4}+u_{5}\\\\\boxed{\boxed{(2x+3)^{5}=32x^{5}+240x^{4}+720x^{3}+1080x^{2}+810x+243}}$

b)

Cubo da soma: $(a+b)^{3}=a^{3}+3a^{2}b+3ab^{2}+b^{3}$

$(x^{2}-3x)^{3}=(x^{2})^{3}-[3(x^{2})^{2}.3x]+[3.x^{2}.(3x)^{2}]-(3x)^{3}\\(x^{2}-3x)^{3}=x^{6}-3x^{4}.3x+3x^{2}.9x^{2}-27x^{3}\\(x^{2}-3x)^{3}=x^{6}-9x^{5}+27x^{4}-27x^{3}$