Question

desenvolveu seguintes binomios (x-1)⁴​

Answer 1

Resposta:

$\sf \displaystyle ( x - 1)^4$

Resolução:

$\sf \displaystyle ( x - 1)^4 = [ x + (-\:1)]^4$

Lembrando que:

$\sf \displaystyle C_{n,p} = {n \choose p} = \frac{n!}{(n-p)! \cdot p!}$

$\framebox{ \boldsymbol{ \sf \displaystyle ( x - 1)^4 = x^4 -\;4x^3 + 6x^{2} -\: 4x + 1 }} \quad \gets \mathbf{ Resposta }$

Explicação passo-a-passo:

$\sf \displaystyle C_{4,0} = {4 \choose 0} = \frac{4!}{(4-0)! \cdot 0!} \cdot x^4 \cdot (-1)^0 = 1 \cdot x^4 \cdot 1 = x^4$

$\sf \displaystyle C_{4,1} = {4 \choose 1} = \frac{4!}{(4-1)! \cdot 1!} \cdot x^3 \cdot (-1)^1 = 4 \cdot x^3 \cdot (-1) = -\:4x^3$

$\sf \displaystyle C_{4,2} = {4 \choose2} = \frac{4!}{(4-2)! \cdot 2!} \cdot x^2 \cdot (-1)^2 = 6 \cdot x^2 \cdot 1 = 6 x^2$

$\sf \displaystyle C_{4,3} = {4 \choose 3} = \frac{4!}{(4-3)! \cdot 2!} \cdot x^1 \cdot (-1)^3 = 4 \cdot x \cdot (-1) = -4 x$

$\sf \displaystyle C_{4,4} = {4 \choose 4} = \frac{4!}{(4-4)! \cdot 4!} \cdot x^0 \cdot (-1)^4 = 1 \cdot 1 \cdot 1 = 1$