Problema
(Indicado a partir do 7º ano do E. F.)
Simplificar a expressão
[tex]\qquad \qquad \dfrac{a-b}{a+b}+\dfrac{b-c}{b+c}+\dfrac{c-a}{c+a}+\dfrac{(a-b)(b-c)(c-a)}{(a+b)(b+c)(c+a)}, [/tex]
admitindo que nenhum denominador seja zero.
Solução 1
Observe o seguinte desenvolvimento:
[tex]\quad S=\dfrac{a-b}{a+b}+\dfrac{b-c}{b+c}+\dfrac{c-a}{c+a}+\dfrac{(a-b)(b-c)(c-a)}{(a+b)(b+c)(c+a)}\\
\quad S=\dfrac{(a-b)(b+c)+(b-c)(a+b)}{(a+b)(b+c)}+\dfrac{c-a}{c+a} \left(1+\dfrac{(a-b)(b-c)}{(a+b)(b+c)} \right)\\
\quad S=\dfrac{ab+\cancel{ac}-\cancel{b^2}-bc+ab+\cancel{b^2}-\cancel{ac}-bc}{(a+b)(b+c)}+\\
\quad \quad \quad + \dfrac{c-a}{c+a} \left(\dfrac{ab+\cancel{ac}+\cancel{b^2}+bc+ab-\cancel{ac}-\cancel{b^2}+bc}{(a+b)(b+c)} \right)\\
\quad S= \dfrac{2b(a-c)}{(a+b)(b+c)}+\dfrac{c-a}{\cancel{c+a}} \left(\dfrac{2b\cancel{(c+a)}}{(a+b)(b+c)}\right)\\
\quad S=\dfrac{2b(a-c)}{(a+b)(b+c)}+\dfrac{2b(c-a)}{(a+b)(b+c)}\\
\quad S=\dfrac{2b(a-c)}{(a+b)(b+c)}-\dfrac{2b(a-c)}{(a+b)(b+c)}=0\,.[/tex]
Portanto,
[tex]\qquad \boxed{\dfrac{a-b}{a+b}+\dfrac{b-c}{b+c}+\dfrac{c-a}{c+a}+\dfrac{(a-b)(b-c)(c-a)}{(a+b)(b+c)(c+a)}=0 }\,.[/tex]
Solução elaborada pelos Moderadores do Blog.
Solução 2
Note que:
[tex]\qquad \dfrac{a-b}{a+b}+\dfrac{b-c}{b+c}+\dfrac{c-a}{c+a}+\dfrac{(a-b)(b-c)(c-a)}{(a+b)(b+c)(c+a)}=\\
\qquad \qquad \dfrac{(a-b)(b+c)(c+a)+(b-c)(a+b)(c+a)}{(a+b)(b+c)(c+a)}+\\
\qquad \qquad +\dfrac{(c-a)(a+b)(b+c)+(a-b)(b-c)(c-a)}{(a+b)(b+c)(c+a)}=\\
\qquad \qquad \dfrac{(c+a)[(a-b)(b+c)+(b-c)(a+b)]}{(a+b)(b+c)(c+a)}+\\
\qquad \qquad +\dfrac{(c-a)[(a+b)(b+c)+(a-b)(b-c)]}{(a+b)(b+c)(c+a)}=\\
\qquad \qquad \dfrac{(c+a)(ab\cancel{-b^2}\cancel{+ac}-bc+ab\cancel{-ac}\cancel{+b^2}-bc)}{(a+b)(b+c)(c+a)}+\\
\qquad \qquad +\dfrac{(c-a)(ab\bcancel{+b^2}\bcancel{+ac}+bc+ab\bcancel{-b^2}\bcancel{-ac}+bc)}{(a+b)(b+c)(c+a)}=[/tex]
[tex]\qquad \qquad \dfrac{2b(c+a)(a-c)+2b(c-a)(a+c)}{(a+b)(b+c)(c+a)}=[/tex]
[tex]\qquad \qquad \dfrac{2b(a^2-c^2)+2b(c^2-a^2)}{(a+b)(b+c)(c+a)}=[/tex]
[tex]\qquad \qquad \dfrac{2b(a^2-a^2+c^2-c^2)}{(a+b)(b+c)(c+a)}=0[/tex]
Assim,
[tex]\qquad \boxed{\dfrac{a-b}{a+b}+\dfrac{b-c}{b+c}+\dfrac{c-a}{c+a}+\dfrac{(a-b)(b-c)(c-a)}{(a+b)(b+c)(c+a)}=0}\,. [/tex]
Solução elaborada pelos Moderadores do Blog.