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»1) ALGEBRA ELEMENTARA

Orice ramură o matematicii presupune cunoaşterea aprofundată a calculului elementar cu numere reale şi complexe (reguli, formule, identităţi remarcabile, teoreme).

Lista de mai jos conţine informaţiile esenţiale din acest punct de vedere:

TEORIE

Data publicarii: 16.10.2008

Identitati remarcabile:

$\bf{{({a}\pm{b})^2}={a^2}\pm{2ab}+{b^2}};$ \bf{{({a}\pm{b})^2}={a^2}\pm{2ab}+{b^2}};
$\bf{{({a}\pm{b})^3}={a^3}\pm{3{a^2}{b}}+3a{b^2}\pm{b^3}};$ \bf{{({a}\pm{b})^3}={a^3}\pm{3{a^2}{b}}+3a{b^2}\pm{b^3}};
$\bf{{a^2-b^2}=(a-b)(a+b)};$ \bf{{a^2-b^2}=(a-b)(a+b)};
$\bf{{a^3-b^3}=(a-b)(a^2+ab+b^2)};$ \bf{{a^3-b^3}=(a-b)(a^2+ab+b^2)};
$\bf{{a^3+b^3}=(a+b)(a^2-ab+b^2)};$ \bf{{a^3+b^3}=(a+b)(a^2-ab+b^2)};
$\bf{{a}^{n}-{b}^{n}=(a-b)({a}^{n-1}+{a}^{n-2}{b}+{a}^{n-3}{b^2}+\cdots+a{{b}^{n-2}}+{b}^{n-1})},$ \bf{{a}^{n}-{b}^{n}=(a-b)({a}^{n-1}+{a}^{n-2}{b}+{a}^{n-3}{b^2}+\cdots+a{{b}^{n-2}}+{b}^{n-1})}, $\bf{\forall{n}}\in{\mathbb{\bf{N}}},\bf{{n}\geq{2}};$ \bf{\forall{n}}\in{\mathbb{\bf{N}}},\bf{{n}\geq{2}};
$\bf{{a}^{n}+{b}^{n}=(a+b)({a}^{n-1}-{a}^{n-2}{b}+{a}^{n-3}{b}^{2}-...-{a}{b}^{n-2}+{b}^{n-1})},$ \bf{{a}^{n}+{b}^{n}=(a+b)({a}^{n-1}-{a}^{n-2}{b}+{a}^{n-3}{b}^{2}-...-{a}{b}^{n-2}+{b}^{n-1})}, $\bf{\forall{n}}\in{\mathbb{\bf{N}}},\bf{n\:{impar},{n}\geq{3}};$ \bf{\forall{n}}\in{\mathbb{\bf{N}}},\bf{n\:{impar},{n}\geq{3}};
$\bf{{(a+b+c)}^2=a^2+b^2+c^2+2ab+2bc+2ca};$ \bf{{(a+b+c)}^2=a^2+b^2+c^2+2ab+2bc+2ca};
$\bf{(x_1+x_2+\cdots+x_n)^2=}$ \bf{(x_1+x_2+\cdots+x_n)^2=} $\bf{{x_1}^2+{x_2}^2+\cdots+{x_n}^2+2({x_1}{x_2}+{x_1}{x_3}+}$ \bf{{x_1}^2+{x_2}^2+\cdots+{x_n}^2+2({x_1}{x_2}+{x_1}{x_3}+} $\bf{\cdots+{x_1}{x_n}+{x_2}{x_3}+\cdots+{x_2}{x_n}+\cdots+{x_{n-1}}{x_n})}.$ \bf{\cdots+{x_1}{x_n}+{x_2}{x_3}+\cdots+{x_2}{x_n}+\cdots+{x_{n-1}}{x_n})}.
$\bf{{a^3+b^3+c^3-3abc}=(a+b+c)(a^2+b^2+c^2-ab-bc-ca)=}$ \bf{{a^3+b^3+c^3-3abc}=(a+b+c)(a^2+b^2+c^2-ab-bc-ca)=} $\bf{\frac{1}{2}(a+b+c)[(a-b)^2+(b-c)^2+(c-a)^2]};$ \bf{\frac{1}{2}(a+b+c)[(a-b)^2+(b-c)^2+(c-a)^2]};
$\bf{(a+b+c)^3=a^3+b^3+c^3+3(a+b)(b+c)(c+a)}.$ \bf{(a+b+c)^3=a^3+b^3+c^3+3(a+b)(b+c)(c+a)}.