
Răspuns:
Explicaţie:
Noi avem
Ce este (sqrt (5 +) sqrt (3)) / (sqrt (3+) sqrt (3+) sqrt (5)) / sqrt (3+) sqrt (3) sqrt (5))?

2/7 Luăm, A = (sqrt5 + sqrt3) / (sqrt3 + sqrt3 + sqrt5) - (sqrt5-sqrt3) / (sqrt3 + sqrt3-sqrt5) = (sqrt5 + sqrt3) -sqrt3-sqrt5) = (sqrt5 + sqrt3) / (2sqrt3 + sqrt5) - (sqrt5-sqrt3) / (2sqrt3-sqrt5) = (sqrt5 + sqrt3) ) (2sqrt3 + sqrt5)) / (2sqrt3 + sqrt5) (2sqrt3-sqrt5) = (2sqrt15-5 + 2 * 3-sqrt15) - (2sqrt15 + 5-2 * 3-sqrt15) ^ 2 (sqrt5) ^ 2) = (anulați (2sqrt15) -5 + 2 * 3cancel (-sqrt15) - anulați (2sqrt15) -5 + -10 + 12) / 7 = 2/7 Rețineți că dacă în numitori există (sqrt3 + sqrt (3 + sqrt5)) și (sqrt3 + sqrt (3 sqrt5)), atunci răspunsul se va schimba.
Cum verificați (tan ^ 2x) / (secx-1) -1 = secx?

"Left Hand Side" = tan ^ 2x / (secx-1) -1 Utilizați identitatea: cos ^ 2x + sin ^ 2x = 1 => 1 + tan ^ 2x = sec ^ 2x = (Secx-1)) / (secx-1) / -1 (> secx-1) -1 => secx + 1-1 = culoare (albastru) secx = "Partea dreaptă"
Cum simplificați (1 / sqrt (a-1) + sqrt (a + 1)) / (1 / sqrt (a + 1) -1 / sqrt (a-1) (a-1) sqrt (a + 1) - (a + 1) sqrt (a-1)), a>

Format mare de matematică ...> culoare (albastru) (((1 / sqrt (a-1) + sqrt (a + 1)) / (1 / sqrt (a + 1) -1 / sqrt ) / (sqrt (a + 1) / ((a-1) sqrt (a + 1) - (a + 1) sqrt (a-1))) = (A + 1)) / (sqrt (a-1) -sqrt (a + 1)) / (sqrt (a +1) cdot sqrt (a-1) +1) / (sqrt (a-1) cdot sqrt (a-1) cdot sqrt (a +1) -sqrt (a + 1) cdot 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) (A + 1) / sqrt (a-1) -sqrt (a + 1)) / (sqrt (a + 1) )) xx (a + 1) cdot sqrt (a-1) (sqrt (a-1) -sqrt (a + 1))) / sqrt (a + 1)) xx (sqrt (a + 1) cdot sqrt (a-1)) / sqrt (a-1) (a + 1))) c