Question

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

Answer 1

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 :(

Answer 2

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