Rozwiąż nierówność; ( zadania w załączniku)
Prosze o założenia.
log x ( 5) > 1
log x (5) > log x (x)
1) [ 0 < x < 1 i log x (5) > log x ( x) ] <=> 5 < x
Nie ma rozwiązań.
2)
[ x > 1 i log x ( 5) > log x ( x) ] <=> 5 > x
Odp.
x > 1 i x < 5
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log x ( x +2) > 2
log x ( x + 2) > log x ( x^2)
x + 2 > 0 <=> x > - 2
1)
[ 0 < x < 1 i x > - 2 i log x ( x +2) > log x ( x^2 ) ] <=> x + 2 < x^2
-x^2 + x + 2 < 0
delta = 1 - 4*(-1)*2 = 1 + 8 = 9
x1 = [ -1 + 3]/ (-2) = 2/ (-2) = - 1
x2 = [ -1 - 3]/(-2) = -4/(-2) = 2
a = -1 < 0 , zatem - x^2 + x + 2 < 0 <=> x < -1 lub x > 2
Brak rozwiazań.
[ x > 1 i x > - 2 i log x ( x +2) > log x ( x^2) ] <=> x+2 > x^2
- x^2 + x + 2 > 0
delta = 1 -4*(-1)*2 = 1 + 8 = 9
x1 = [-1 +3]/(-2) = 2/(-2) = -1
x2 = ( -1 -3]/(-2) = -4/(-2) = 2
a = -1 < 0 zatem -x^2 +x + 2 > 0 <=> -1 < x < 2
Odp. 1 < x < 2
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log x ( 5) > 1
log x (5) > log x (x)
1) [ 0 < x < 1 i log x (5) > log x ( x) ] <=> 5 < x
Nie ma rozwiązań.
2)
[ x > 1 i log x ( 5) > log x ( x) ] <=> 5 > x
Odp.
x > 1 i x < 5
================
log x ( x +2) > 2
log x ( x + 2) > log x ( x^2)
x + 2 > 0 <=> x > - 2
1)
[ 0 < x < 1 i x > - 2 i log x ( x +2) > log x ( x^2 ) ] <=> x + 2 < x^2
-x^2 + x + 2 < 0
delta = 1 - 4*(-1)*2 = 1 + 8 = 9
x1 = [ -1 + 3]/ (-2) = 2/ (-2) = - 1
x2 = [ -1 - 3]/(-2) = -4/(-2) = 2
a = -1 < 0 , zatem - x^2 + x + 2 < 0 <=> x < -1 lub x > 2
Brak rozwiazań.
2)
[ x > 1 i x > - 2 i log x ( x +2) > log x ( x^2) ] <=> x+2 > x^2
- x^2 + x + 2 > 0
delta = 1 -4*(-1)*2 = 1 + 8 = 9
x1 = [-1 +3]/(-2) = 2/(-2) = -1
x2 = ( -1 -3]/(-2) = -4/(-2) = 2
a = -1 < 0 zatem -x^2 +x + 2 > 0 <=> -1 < x < 2
Odp. 1 < x < 2
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