Wykaż że, jeżeli

[poczekaj]

Jeżeli \(\displaystyle{ \frac{a_{1}}{b_{1}}=\frac{a_{2}}{b_{2}}=...=\frac{a_{n}}{b_{n}} b_{1}+b_{2}+...+b_{n}\neq0}\)

to


\(\displaystyle{ \frac{a_{1}+a_{2}+a_{3}+...+a_{n}}{b_{1}+b_{2}+b_{3}+...+b_{n}}=\frac{a_{1}}{b_{1}}}\)


Poprostu

\(\displaystyle{ \frac{a_{1}+a_{1}+a_{1}+a_{1}+...}{b_{1}+b_{1}+b_{1}+b_{1}+...}}\) skoro \(\displaystyle{ \frac{a_{1}}{b_{1}}=\frac{a_{2}}{b_{2}}}\)

[mol_ksiazkowy]

\(\displaystyle{ a_i =cb_i}\) , i=1, 2, .....

\(\displaystyle{ \frac{a_{1}+a_{2}+a_{3}+...+a_{n}}{b_{1}+b_{2}+b_{3}+...+b_{n}}=\frac{cb_{1}+cb_{2}+cb_{3}+...+cb_{n}}{b_{1}+b_{2}+b_{3}+...+b_{n}}=c}\)

[poczekaj]

Dziękuję. Proste