rozwiaz rownania 4x2-20x+9=0 (2x-1)*(x+2)=3     4x2+5x=1<0 5(2x-1)(x+4)<0 x3-4x2+3x=0   x3-2x2-x+2=0 x3+2x2-2x-4=0  

4x²-20x+9=0 Δ=20²-4*4*9 =400-144 =256 √Δ = 16   x₁=(20-16)/(2*4)= 4/8 = 1/2 x²=(20+16)/(2*4)= 36/8 = 4 i 1/2 ---------------------------------------------   (2x-1)*(x+2)=3 2x²+4x-x-2-3=0 2x²+3x-5=0 Δ=3²-4*2*(-5) = 9+40=49 √Δ = 7   x₁=(-3-7)/(2*2)= -10/4 = -2 i 1/2 x²=(-3+7)/(2*2)= -4/4 = -1   -------------------------- 4x²+5x+1<0 * Δ=5²-4*4*1 Δ=25-16 Δ=9 √Δ = 3   x₁=(-5-3)/(2*4)= -8/8 = -1 x²=(-5+3)/(2*4)= -2/8 = -1/4   a>0 zatem rysujemy parabole ramionami skierowanymi do góry... zatem wartości mniejsze od zera będą w przedziale (-1/4; -1) ----------------------------------------- 5(2x-1)(x+4)<0 5*[2x²+8x-x-4]<0 10x²+35x-20<0 /:5 2x²+7x-4 <0 Δ=7²-4*2*(-4) = 49+32=81 √Δ = 9   x₁=(-7-9)/(2*2)= -16/4 = -4 x²=(-7+9)/(2*2)= 2/4 = 1/2 a>0 ramiona do góry x należy od (-4;1/2) ---------------------------- x3-4x²+3x=0 x^3-4x²+3x=0 x(x²-4x+3)=0 wyciągam równanie kwadratowe i licze miejsca zerowe Δ=4²-4*3 = 16-12=4 √Δ = 2   x₁=(4-2)/2= 2/2 = 1 x²=(4+2)/2= 6/2 = 3 Zatem: x(x-1)(x-3)=0 x=0 x=1 x=3 --------------------------- x3-2x²-x+2=0 x^3-x-2x²+2=0 x(x²-1) -2 (x²-1)=0 (x²-1)(x-2)=0 x²=1 / √ |x|=1 x=1 lub x=-1 oraz x=2 ------------------------- x3+2x²-2x-4=0 x²(x+2)-2(x+2)=0 (x²-2)(x+2)=0 x=-2 oraz x²-2=0 x²=2 /√ |x|=√2 x=√2 lub x=-√2     *Zakładam, że zamiast "=" miało być "+"

1 . 4x²-20x+9=0 Δ=20²-4*4*9 =400-144 =256 √Δ = √16 = 4   x₁=(20- 4)/(2*4)= 16/8 = 2 x²=(20+ 4)/(2*4)= 24/8 = 3     2 . (2x-1)*(x+2)=3 2x²+4x-x-2-3=0 2x²+3x-5=0 Δ=3²-4*2*(-5) = 9+40=49 √Δ = 7   x₁=(-3-7)/(2*2)= -10/4 = -2 i 1/2 x²=(-3+7)/(2*2)= -4/4 = -1   3. 4x²+5x+1<0 * Δ=5²-4*4*1 Δ=25-16 Δ=9 √Δ = 3   x₁=(-5-3)/(2*4)= -8/8 = -1 x²=(-5+3)/(2*4)= -2/8 = -1/4     4. 5(2x-1)(x+4)<0 5*[2x²+8x-x-4]<0 10x²+35x-20<0 /:5 2x²+7x-4 <0 Δ=7²-4*2*(-4) = 49+32=81 √Δ =√ 9 = 3   x₁=(-7-3)/(2*2)= -10/4 = - 2 i 1/4 x²=(-7+3)/(2*2)= 4/4 = 1   5 . x3-4x²+3x=0 x^3-4x²+3x=0 x(x²-4x+3)=0   Δ=4²-4*3 = 16-12=4 √Δ = 2   x₁=(4-2)/2= 2/2 = 1 x²=(4+2)/2= 6/2 = 3   x(x-1)(x-3)=0 x=0 x=1 x=3   6 . x3-2x²-x+2=0 x^3-x-2x²+2=0 x(x²-1) -2 (x²-1)=0 (x²-1)(x-2)=0 x²=1 / √ |x|=1 x=1 lub x=-1 oraz x=2   7 . x3+2x²-2x-4=0 x²(x+2)-2(x+2)=0 (x²-2)(x+2)=0 x=-2   x²-2=0 x²=2 /√ |x|=√2 x=√2 lub x=-√2