Na dobranoc :3
Dobranocka: Wykaz ze jesli a b i c sa roznymi od zera liczbami rzeczywistymi spelniajacymi warunek a+b+c=0,
to prawdziwa jest rownosc ((a2)/(b+c))+((b2)/(a+c))+((c2)/(a+b))=0
23 lut 23:59
Dobranocka: Hmm. Prosze!
24 lut 00:59
Bogdan:
| a2 | | b2 | | c2 | |
| + |
| + |
| = 0 /*(a + b)(a + c)(b + c) |
| b + c | | a + c | | a + b | |
a
2(a + b)(a + c) + b
2(a + b)(b + c) + c
2(a + c)(b + c) = 0
a
2(a
2 + ab + ac + bc) + b
2(ab + b
2 + bc + ac) + c
2(ac + bc + c
2 + ab) = 0
a
2[ a(a + b + c) + bc] + b
2[ b(a + b + c) + ac] + c
2[ c(a + b + c) + ab] = 0
a
2( a*0 + bc) + b
2( b*0 + ac) + c
2[ c*0 + ab) = 0
a
2bc + ab
2c + abc
2 = 0 ⇒ abc(a + b + c) = 0 ⇒ 0 = 0
24 lut 01:09
Eta:
a+b+c=0 ⇒ b+c= −a , a+c= −b , a+b= −c
| | a2 | | b2 | | c2 | |
L= |
| + |
| + |
| = −a−b−c=−(a+b+c)= 0 |
| | −a | | −b | | −c | |
24 lut 23:31