Lim 3x / tan3x x 0 Kako ga riješiti? Mislim da će odgovor biti 1 ili -1 tko ga može riješiti?
Ograničenje je 1. Lim_ (x -> 0) (3x) / (tan3x) = Lim_ (x -> 0) (3x) / ((sin3x) / (cos3x)) = Lim_ (x -> 0) (3xcos3x) ) / (sin3x) = Lim_ (x -> 0) (3x) / (sin3x) .cos3x = Lim_ (x -> 0) boja (crvena) ((3x) / (sin3x)). cos3x = Lim_ (x - > 0) cos3x = Lim_ (x -> 0) cos (3 * 0) = Cos (0) = 1 Zapamtite: Lim_ (x -> 0) boja (crvena) ((3x) / (sin3x)) = 1 i Lim_ (x -> 0) boja (crvena) ((sin3x) / (3x)) = 1
Što je (sqrt (5+) sqrt (3)) / (sqrt (3+) sqrt (3+) sqrt (5)) - (sqrt (5-) sqrt (3)) / (sqrt (3+) sqrt) (3) sqrt (5))?
2/7 Primamo, A = (sqrt5 + sqrt3) / (sqrt3 + sqrt3 + sqrt5) - (sqrt5-sqrt3) / (sqrt3 + sqrt3-sqrt5) = (sqrt5 + sqrt3) / (2sqrt3 + sqrt5) - (sqrt5) -sqrt3) / (2sqrt3-sqrt5) = (sqrt5 + sqrt3) / (2sqrt3-sqrt5) = ((sqrt5 + sqrt3) (2sqrt3-sqrt5) - (sqrt5-sqrt3) (2sqrt3 + sqrt5) (2sqrt3 + sqrt5) / ((2sqrt15-5 + 2 * 3-sqrt15) - (2sqrt15 + 5-2 * 3-sqrt15)) / ((2sqrt3) ^ 2- (sqrt5) ^ 2) = (poništi (2sqrt15) -5 + 2 * 3kkazati (-sqrt15) - otkazati (2sqrt15) -5 + 2 * 3 + otkazati (sqrt15)) / (12-5) = ( Imajte na umu da, ako su u nazivnicima (sqrt3 + sqrt (3 + sqrt5)) i (sqrt3 + sqrt (3-sqrt5)), odgovor će biti promijenjen.
Kako pojednostavljujete (1 / sqrt (a-1) + sqrt (a + 1)) / (1 / sqrt (a + 1) -1 / sqrt (a-1)) div sqrt (a + 1) / ( (a-1) sqrt (a + 1) - (a + 1) sqrt (a-1)), a> 1?
Ogromno formatiranje matematike ...> boja (plava) (((1 / sqrt (a-1) + sqrt (a + 1)) / (1 / sqrt (a + 1) -1 / sqrt (a-1)) ) / (sqrt (a + 1) / ((a-1) sqrt (a + 1) - (a + 1) sqrt (a-1))) = boja (crvena) (((1 / sqrt (a- 1) + sqrt (a + 1)) / ((sqrt (a-1) -sqrt (a + 1)) / (sqrt (a + 1) cdot sqrt (a-1)))) / (sqrt (a +1) / (sqrt (a-1) cdot sqrt (a-1) cdot sqrt (a + 1) -sqrt (a + 1) cdot sqrt (a + 1) sqrt (a-1))) = boja ( plava) (((1 / sqrt (a-1) + sqrt (a + 1)) / ((sqrt (a-1) -sqrt (a + 1)) / (sqrt (a + 1) cdot sqrt (a -1)))) / (sqrt (a + 1) / (sqrt (a 1) cdot sqrt (a-1) (sqrt (a-1) -sqrt (a + 1))) = boja (crvena) ((1 / sqrt (a