zapisz w postaci ilorazu sume odwrotnosci trzech kolejnych liczb naturalnych.
1 liczb --- x
2 liczba ---x +1
3 liczba ---x + 2
1/x + 1/(x +1) + 1/(x +2) =
[(x +1)(x +2) + x(x + 2) + x(x +1) ] /[ x(x +1)(x +2) ] =
[x² + 2x + x + 2 + x² + 2x +x² + x ] /[ x(x +1)(x +2) ] =
[3x² + 6x + 2 ] /[ x(x +1)(x +2) ]
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1 liczb --- x
2 liczba ---x +1
3 liczba ---x + 2
1/x + 1/(x +1) + 1/(x +2) =
[(x +1)(x +2) + x(x + 2) + x(x +1) ] /[ x(x +1)(x +2) ] =
[x² + 2x + x + 2 + x² + 2x +x² + x ] /[ x(x +1)(x +2) ] =
[3x² + 6x + 2 ] /[ x(x +1)(x +2) ]