| a1 | a2 | an | ||||
Wykaż, że jeśli | = | =.... | i b1+b2+....bn≠0 | |||
| b1 | b2 | bn |
| a1+a2+.....an | a1 | |||
to | = | |||
| b1+b2+....bn | b1 |
| a1 | a2 | ||
= | |||
| b1 | b2 |
| b1a2 | ||
a1= | mam takie coś | |
| b2 |
| a1 | a2 | ||
= | = c..jakaś stała, z tego wynika, że: | ||
| b1 | b2 |
| a1 | |
= c ⇒ a1 = c*b1 | |
| b1 |
| a1 | a1 + a2 | ||
= | , bo: | ||
| b1 | b1 + b2 |
| cb1 + cb2 | |
= c...czyli rownosc sie zgadza | |
| b1 + b2 |
mojego archiwum np. tak :
| a1 | a k | |||
niech | = ... = | = c= const. ⇒ a k= cb k i k=1,2,3, ... , n, | ||
| b1 | b k |
| a1+a2+a3+ ... +an−1+an | ||
wtedy | = | |
| b1+b2+b3 + ... + bn−1+bn |
| cb1+cb2+cb3 + ... + cbn−1+cbn | ||
= | = | |
| b1+b2+b3+ ... +bn−1+bn |
| c(b1+b2+b3 + ... + bn−1+bn) | ||
= | = c= | |
| b1+b2+b3+ ... +bn−1+bn |
| a1 | ||
= | , c.n.w. ... ![]() | |
| b1 |
słyszeć ...
To juz na rozwiazanie nie mozna wpasc samemu
?
Myślalem, ze posądzasz mnie o plagiat − zgoda