Unordered triplet
Author: mathforces
Problem has been solved: 52 times
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For an unordered triplet of positive integer numbers $ a $, $ b $ and $ c $, the following equalities hold: $$ a + b + c = 25 $$ $$\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=\frac{40}{99}-\frac{1}{abc}$$ Find $ ab + bc + ca $.
Для неупорядочнной тройки натуральных чисел $a$, $b$ и $c$ выполнены равенства: $$a+b+c=25$$ $$\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=\frac{40}{99}-\frac{1}{abc}$$ Найдите $ab+bc+ca$.
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