Question

formule de calcul prescurtate

Answer 1

$\boxed{(a+b)^2=a^2+2ab+b^2} \\ \\ \boxed{(a-b)^2=a^2-2ab+b^2} \\ \\ \boxed{(a-b)(a+b)=a^2-b^2} \\ \\ \boxed{(a+b+c)^2=a^2+b^2+c^2+2ab+2ac+bc} \\ \\ \\ \\ \boxed{(a+b)^3=a^3+3a^2b+3ab^2+b^3=a^3+b^3+3ab(a+b)} \\ \\ \boxed{a^3+b^3=(a+b)(a^2-ab+b^2)}$


$\star\star\star\boxed{1+a+a^2+...+a^n= \frac{a^{n+1}-1}{a-1}} \ \vee a\in\mathbb{R}\ /[1] , n\in\mathbb{N^*} \\ \\ Cu \ aceasta \ formula \ te \ vei \ intalnii \ rar$