Question

Dau 20 puncte. Am nevoie de rezolvare completa! Multumesc anticipat!​

Answer 1

Răspuns:

E = 1

Explicație pas cu pas:

$\frac{a}{b} + \frac{c}{d} = 1 < = > \frac{c}{d} = 1 - \frac{a}{b}$

$E = \frac{ {a}^{2010} }{ {b}^{2010} } + \frac{ {a}^{2009} \cdot c}{ {b}^{2009}\cdot d } + \frac{ {a}^{2008} \cdot c}{ {b}^{2008}\cdot d } + ... + \frac{ {a}^{3} \cdot c}{ {b}^{3}\cdot d } + \frac{ {a}^{2} \cdot c}{ {b}^{2}\cdot d } + \frac{a \cdot c}{ b\cdot d } + \frac{c}{d} =$

$= \frac{ {a}^{2010} }{ {b}^{2010} } + \frac{ {a}^{2009}}{ {b}^{2009}}\left (1 - \frac{a}{b} \right) + \frac{ {a}^{2008}}{ {b}^{2008}}\left(1 - \frac{a}{b} \right) + ... + \frac{ {a}^{3}}{ {b}^{3}}\left(1 - \frac{a}{b} \right) + \frac{ {a}^{2}}{ {b}^{2}}\left(1 - \frac{a}{b}\right) + \frac{a}{b}\left(1 - \frac{a}{b} \right) + \left(1 - \frac{a}{b} \right)$

$= \frac{ {a}^{2010} }{ {b}^{2010} } + \left(\frac{ {a}^{2009}}{ {b}^{2009}} - \frac{ {a}^{2010} }{ {b}^{2010} } \right) + \left(\frac{ {a}^{2008}}{ {b}^{2008}} - \frac{ {a}^{2009}}{ {b}^{2009}} \right) + ... + \left(\frac{ {a}^{3}}{ {b}^{3}} - \frac{ {a}^{4}}{ {b}^{4}} \right) + \left(\frac{ {a}^{2}}{ {b}^{2}} - \frac{ {a}^{3}}{ {b}^{3}} \right) + \left(\frac{a}{b} - \frac{ {a}^{2}}{ {b}^{2}} \right) + \left(1 - \frac{a}{b} \right)$

$= \frac{ {a}^{2010} }{ {b}^{2010} } + \frac{ {a}^{2009}}{ {b}^{2009}} - \frac{ {a}^{2010} }{ {b}^{2010} } + \frac{ {a}^{2008}}{ {b}^{2008}} - \frac{ {a}^{2009}}{ {b}^{2009}} + ... + \frac{ {a}^{3}}{ {b}^{3}} - \frac{ {a}^{4}}{ {b}^{4}} + \frac{ {a}^{2}}{ {b}^{2}} - \frac{ {a}^{3}}{ {b}^{3}} + \frac{a}{b} - \frac{ {a}^{2}}{ {b}^{2}} + 1 - \frac{a}{b}$

$= 1$