Rozwiąż nierownosci:
a. 2x^2-4x mniejsze rowne3(1-x)-5
b. 3x^2 -4x<-1,33
c. (x+3)(x-2) wieksze rowne 6(x-1)
d. (x-5)^2 wieksze rowne 2x-11
e. (x+5)(x-3)>(x-4)(x+4)
f. (2x-3)^2>-(x+3)(x-3)
Prosze o szybka odpowiedz. Dam NAJ!
" Life is not a problem to be solved but a reality to be experienced! "
© Copyright 2013 - 2024 KUDO.TIPS - All rights reserved.
Rozwiąż nierownosci:
a. 2x^2-4x<=3(1-x)-5
2x^2-4x<=3-3x-5 / + 3x+2
2x^2-x +2 <=0
Δ = b^2 - 4ac = 1-4*2*2 < 0 brak pierwiastków rzeczywistych
b. 3x^2 -4x<-1,33
3x^2 -4x+1,33 < 0
Δ = b^2 - 4ac = 16-4*3*1.33=0,04
√Δ = 0,2
x₁ = (-b-√Δ)/2a = (4-0,2)/2*3 = 38/60=19/30
x₂ = (-b-√Δ)/2a = (4+0,2)/2*3 = 42/60=7/10
x∈(19/30 ; 7/10)
c. (x+3)(x-2) >= 6(x-1)
x^2 -2x +3x -6 >=6x - 6 / -6x+6
x^2 - 5x >=0
x(x-5) >=0
x₁ = 0
x₂ = -5
x∈ <-5 ; 0>
d. (x-5)^2 >= 2x-11 /-2x+11
x^2 - 10x +25 -2x +11 >=0
x^2 -12x +36 >=0
Δ = b^2 - 4ac = 144-4*36 = 0
x= -b/2a = 12/2 = 6
rozwiązaniem jest x = 6
e. (x+5)(x-3)>(x-4)(x+4)
x^2 -3x +5x -15 > x^2 - 16
2x - 15 +16 > 0
2x +1 >0
2x > -1
x > -1/2
f. (2x-3)^2>-(x+3)(x-3)
4x^2 - 12x +9 > -x^2 + 9 / +x^2 - 9
5x^2 - 12x > 0
5x(x - 12/5) >0
x₁ = 0
x₂ = 12/5
x∈(-∞;0 ) u (12/5 ; ∞)