Prosilbym o rozwiązanie tych zadan(najlepiej na jutro daje naj tylko za rozwiązanie całosci )

1.x/x+3 + 2/x+1 = 2 / × x x × x/x+3 + x × 2/x+1 = 2x x/3 +2/1 = 2x /×3 x+6 =2x 6=x x-1/x - x+1/x-1 < 2 / × x x × x-1/x -x x+1/x-1 < 2x x-1 - x-1/-1 < 2x / ×(-1) -x +1-x+1< -2x 2<0 /-kreska ułamkowa sory ale tylko tyle umiałam :(

1) a) x/x+3 + 2/x+1 = 2 x=/= -3 i x=/= -1 mnozymy stronami (x+3)(x+1) x(x+1) + 2(x+3) = 2(x+3)(x+1) x² + x + 2x + 6 = 2 ( x² + 4x + 3) x² + x + 2x + 6 = 2x² + 8x + 6 x² + 5x = 0 x ( x+5 ) = 0 x =0 lub x = -5 b) x-1/x - x+1/x-1 < 2 Z nr 1 )jesli x e (-oo; 0) u (1; +oo) to x(x-1) >0 mnozym stronami przez x(x-1) (x-1)(x-1) - x(x+1)< 2 x² - 2x + 1 - x²- x < 2 -3x + 1 - 2 < 0 3x > -1 x > -1/3 z Z => x e (-1/3; 0) u (1; + oo) Z nr 2) jesli x e (0;1) to x (x-1) < 0 mnozymy stronami przez x(x-1) (x-1)(x-1) - x(x+1) >2 x² - 2x + 1 - x²- x > 2 -3x + 1 - 2 > 0 3x < -1 x < -1/3 sprzeczne z Z nr 2 wiec x e ( (-1/3; 0 ) u ( 1; + oo) 2) a(n) = 0 <=> n² - 9n + 14 = 0 Δ = 9² - 4 * 1 * 14 = 81 - 56 = 25 √Δ = 5 n1 = (9 - 5)/2 = 2 n2 = (9+5)/2 = 7 Odp Drugi i siodmy 3) a(n) = n+2 / 3n-1 n1 > n2 a(n1) /a(n2) = [(n1 + 2) / (3n1 - 1)] / [ (n2 + 2) / (3n2 - 1)] = = (n1+2) (3n2-1) / (3n1-1)(n2+2) jesli to >0 (n1+2) (3n2-1) / (3n1-1)(n2+2) > 0 wiec (n1+2) (3n2-1) / (3n1-1)(n2+2) > 0 dokoncz ;p 4) a2 + a5 - a3 = 10 a1 + r + a1 + 4r - a1 - 2r =10 a1 + 3r = 10 a2 + a9 = 17 a1 + r + a1 + 8r = 17 2a1 + 9r = 17 wiec a1 + 3r = 10 2a1 + 9r = 17 a1 = 10 - 3r 2 ( 10 - 3r) + 9r =17 20 - 6r + 9r= 17 3r = -3 r = -1 a1 = 10 - 3r = 10 - 3 * (-1) = 13