Zad.1 4x² + (2x + 2)² + (2x + 4)² = 116 12x² + 24x + 20 = 116 /:4 3x² + 6x + 5 = 29 3x² + 6x - 24 = 0 ---------------------- a = 3, b = 6, c = -24 Δ = b² - 4ac Δ = 36 - 4 * 3 * (-24) Δ = 36 - (-288) Δ = 324 √Δ = 18 x₁ = -b - √Δ/2a x₁ = -6 - 18/ 2 * 3 x₁ = -6 - 18/6 x₁ = -24/6 x₁ = -4 x₂ = -b + √Δ/2a x₂ = -6 + 18/6 x₂ = 12/6 x₂ = 2 Odp: x ∈N zatem te liczby te to 4,6,8. ========================================================== Zad.2 Pierwsza liczba = 2n - 1 Druga liczba = 2n + 1 (2n + 1)² - (2n - 1)² = 368 (4n² + 4n + 1) - (4n² - 4n + 1) = 368 4n² + 4n + 1 - 4n² + 4n - 1 = 368 8n = 368/:8 n = 46 zatem: 2n - 1 = 2 * 46 - 1 = 92 - 1 = 91 2n + 1 = 2 * 46 + 1 = 92 + 1 = 93 Odp: Liczby te to 91 i 93. +++++++++++++++++++++++++++++++ Masz pytania to pisz na pw..
z.1 Kolejne liczby parzyste to: 2n, 2n +2, 2n + 4 Mamy I2n)² + (2n +2)² + (2n + 4)² = 116 4n² + 4n² + 8n + 4 + 4n² + 16n + 16 = 116 12 n² + 24 n +20 = 116 12 n² + 24 n -96 = 0 / : 12 n² + 2n - 8 = 0 Δ = 4 -4*1*(-8) = 4 + 32 = 36 √Δ = 6 n = [ -2 + 6]/2 = 2 zatem 2n = 4, 2n +2 = 6, 2n + 4 = 8 spr.4 ² + 6² + 8² = 16 + 36 + 64 = 116 Odp. Te liczby to: 4,6,8. ======================= z.2 Kolejne liczby nieparzyste to 2n -1 oraz 2n + 1 Mamy więc (2n +1)² - (2n -1)² = 368 4n² + 4n +1 - 4n² + 4n -1 = 368 8n = 368 / : 8 n = 46 zatem 2*46 -1 = 91 oraz 2*46 =1 = 93 Odp. Te liczby to: 91 oraz 93. =====================================
