Question

Alguém me ajuda a resolver essa questão por favor?

(x-1).(x+1)+3.(x-1).(x-1)+3.(x-1)+1

Answer 1

Dada as propriedades:

◾ $\mathsf{\left(a+b\right)\cdot \left(a-b\right)=a^2-b^2}$
◾ $\mathsf{\left(a+b\right)^2=a^2+2ab+b^2}$
◾ $\mathsf{\left(a-b\right)^2=a^2-2ab+b^2}$

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Logo:

$\mathsf{\left(x-1\right)\cdot \left(x+1\right)+3\cdot \left(x-1\right).\left(x-1\right)+3\cdot \left(x-1\right)+1=}\\\\\mathsf{=\left[\left(x-1\right)\cdot \left(x+1\right)\right]+\left[3\cdot \left(x-1\right).\left(x-1\right)\right]+\left[3\cdot \left(x-1\right)\right]+1}\\\\\mathsf{=\left(x^2-1\right)+3\cdot \left(x-1\right)^2+\left[3\cdot \left(x-1\right)\right]+1}\\\\\mathsf{=x^2-\diagup \!\!\!\! 1+3\cdot \left(x^2-2x+1\right)+\left(3x-3\right)+\diagup \!\!\!\!1}\\\\\mathsf{=x^2+\left(3x^2-6x+3\right)+\left(3x-3\right)}\\\\\mathsf{=x^2+3x^2-6x+\diagup \!\!\!\! 3+3x-\diagup \!\!\!\! 3}\\\\\mathsf{=4x^2-6x+3x}\\\\\mathsf{=\:}\boxed{\mathsf{4x^2-3x}}\: \: \checkmark$