Спараметрами реши уравнение (относительно x): m2⋅(x−2)−3m=x+1

1) 3х²-124х-84=0

      d = 15376 + 1008 = 16384

  √d = 128

      x₁ = (124+128)\6 = 42

      x₂ = (124 - 128)\6 = -⅔

2) 7х²+6х+1=0

    d = 8

  √d = 2√2

x₁ = (-3 + √2) \ 7

x₂ = (-3 - √2) \ 7

 

3)(х²+9х+14)/(х²-49) = (x+2)(x+7) \ (x-7)(x+7) = (x+2) \ (x-7)

х²+9х+14 -

по т. виета

x₁ + x₂ = -9

x₁ · x₂ = 14

x₁ = -2

x₂ = -7

следовательно выражение х²+9х+14 раскладывается на множители (x+2)(x+7)

 

4) (х^2+4х-21)/(2х^2+11х-21) = (x-3)(x+7) \ (x+7)(2x-3) = (x-3)\(2x-3)

a) х^2+4х-21 = 0

d = 100

√d = 10

x₁=3

x₂= -7

х^2+4х-21 = (x - 3)(x+7)

б) 2х^2+11х-21 = (x+7)(2x-3)

 

  у astragorta во втором уравнении ошибка.

 

1) 1/cos^2(a) – tg^2(a)-sin^2(a)= 1/cos^2(a) – sin^2(a)/cos^2((a)= =[1 – sin^2(a) ]/cos^2((a)=cos^2(a)/cos^2((a)= =1-sin^2(a)=cos^2(a) 2) cos^2(a)+ctg^2(a)-1/sin^2(a)=cos^2(a)+[cos^2(a)-1]/sin^2(a)= =cos^2(a)-sin^2(a)]/sin^2(a)=cos^2(a)-1 = -sin^2(a) 3) 1/cos^2(a) – tg^2(a)(cos^2(a)+1)=1/cos^2(a) – sin^2(a)-sin^2(a)/(cos^2(a)= =(1 – sin^2(a))/cos^2(a)-sin^2(a)=1-sin^2(a)=cos^2(a) 4) (1+sin^2(a))ctg^2(a) – 1/sin^2(a)=cos^2/sin^2 +cos^2 – 1/sin^2(a)= =(cos^2 - 1)/sin^2 +cos^2= -sin^2/sin^2 +cos^2= -1+cos^2=  -sin^2(a)   5)sin(a)/(1+cos(a)) + sin(a)/(1-cos(a))=sin(a) *  ((1-cosa)+(1+cosa)) / (1-cos^2)=2/sin(a)   6)cos(a)/(1+sin(a))+ cos(a)/(1-sin(a))=cos(a)* ((1-sina+1+sina)) / (1-sin^2a)= 2/cos(a)