1) 5x^2 - 15x - x - 3 =0
Объяснение:
1) D= b^2-4ac= 196+60= 256 = 16^2
x1= -14+16/10= 1/5
x2= - 14-16/10= - 3
2) D=25-16=9=3^2
x1= 5+3/2=8/2=4
x2= 5-3/2=2/2=1
3) 10x^2 + 5x-3/5=0
50x^2 +25x-3=0
D= 625+600=1225=35^2
x1= -25+35/100= 1/10
x2= - 25-35/100= - 3/5
4) x^2 +6x-5x=0
x^2+x=0
D= 1=1^2
x1= - 1+1/2=0
x2= - 1-1/2=-1
5) 2-3x-5x^2=0
-5x^2-3x+2=0
5x^2+3x-2=0
D=40+9=49=7^2
x1=-3+7/10= 2/5
x2=-3-7/10= - 1
6) x^2-4x+4=3x-8
x^2-4x+4-3x+8=0
x^2-7x+12=0
D=49-48=1=1^2
x1=7+1/2= 4
x2=7-1/2= 3
7) 5(x^2+4x+4)= -50
5x^2+20x+20= -50
5x^2+20x+20-50=0
5x^2+20x+70=0
x^2+4x+14=0
D= 16-56=-40 - корней нет
1) Cosx = t
6t² + t -1 = 0
D = b² -4ac = 1 - 4*6*(-1) = 25 > 0
t₁ = (-1+5)/12 = 4/12 = 1/3
t₂ = (-1 -5)/12 = -1/2
a) Cosx = 1/3 б) Сosx = -1/2
x = +-arcCos(1/3) + 2πk , k ∈Z x = +-arcCos(-1/2) + 2πn , n ∈Z
x = +- 2π/3 +2πn , n ∈ Z
2) учтём, что Cosx = 2Cos²x/2 -1
наше уравнение:
Cosx/2 = 1 + 2Cos²x/2 -1
Cosx/2 = t
2Cos²x/2 - Cosx/2 = 0
Cosx/2(2Cosx/2 -1) = 0
Cosx/2 = 0 или 2Cosx/2 -1 = 0
x/2 = π/2 + 2πk , k ∈Z Cosx/2 = 1/2
x = π + 4πk , k ∈ Z x/2 = +-arcCos(1/2) + 2πn , n ∈ Z
x/2= +- π/3+ 2πn , n ∈ Z
x = +-2π/3 + 4 πn , n ∈ Z
