a) 2(x−2)(2−x)−4x2−3x=−2+3x−x2
2−(x−2)2−4x2−3x=−2+3x−x2
4−2(x−2)2−4x2−3x=−2+3x−x2
4−2(x−2)2−x2+3x=−2+3x−x2
−2(x−2)2−x2+3x=−8+12x−4x2
−2(x−2)2=−8+9x−3x2
−2(x2−4x+4)=−8+9x−3x2
−2x2+8x−8=−8+9x−3x2
x2−x=0
x(x−1)=0
zatem:
x=0 ∨ x−1=0
x=0 ∨ x=1
b) 3(3−x)(x+3)−6(x+2)2−4=2531−x
39−x2−6(x+2)2−4=2531−x ∣⋅6
2(9−x2)−((x+2)2−4)=3⋅(531−x)
18−2x2−(x+2)2+4=16−3x
−2x2−(x2+4x+4)+22=16−3x
−2x2−x2−4x−4+22=16−3x
−3x2−x+2=0
Δ=(−1)2−4⋅(−3)⋅2=1+24=25, Δ=5
x1=2⋅(−3)1−5=−6−4=32
x1=2⋅(−3)1+5=−66=−1
c) 5(2x−3)(x+1)−2(x−1)(x+1)=x+3
102(2x−3)(x+1)−105(x−1)(x+1)=x+3
102(2x−3)(x+1)−5(x−1)(x+1)=x+3
10(x+1)(2(2x−3)−5(x−1))=x+3 ∣⋅10
(x+1)(4x−6−5x+5)=10x+30
(x+1)(−x−1)=10x+30
−x2−x−x−1=10x+30
−x2−2x−1=10x+30
−x2−12x−31=0
Δ=(−12)2−4⋅(−1)⋅(−31)=144−124=20, Δ=25
x1=−212−25=5−6
x1=−212+25=−5−6
d) 3(3x−1)2−4−310x2−25+(2−x)(x+2)=−x−7
3(3x−1)2−4−10x2−25+(4−x2)=−x−7
39x2−6x+1−4−10x2−29−x2=−x−7
3−x2−6x−3−29−x2=−x−7 ∣⋅6
2(−x2−6x−3)−3(9−x2)=−6x−42
−2x2−12x−6−27+3x2=−6x−42
x2−6x+9=0
(x−3)2=0
x−3=0
x=3
