Решите систему уравнений {х-5у=2,х^2-у=10

{х-5у=2   ;   х²-у=10

{х=2+ 5у ; (2+ 5у)²- у=10

{х=2+ 5у ; 4+20у+ 25у²- у=10

{х=2+ 5у ;   25у²+ 19у-6 =0

                  d= 361 +600=961

                  y₁=(-19+31)/50= 12/50 =6/25     x₁=2+5*6/25= 16/5

                  y₂=(-19-31)/50 =-1                         x₂= 2+5*(-1)=-3

ответ : (16/5; 6/25), (-3 ; -1)

4х2 - 4х -1 = 0

d=16-4*4*(-1)=16+16=32

x1=(4+4sqr(2))^8=(1+sqr(2))/2=0.5+0.5sqr(2)

x2=(4-4sqr(2))^8=(1-sqr(2))/2=0.5-0.5sqr(2)

4х2 - 4х -1=4(x-0.5-0.5sqr(+0.5+0.5sqr(2))

х2 - 6х +1 = 0

d=36-4*1=32

x1=(6+4sqr(2))/2=3+2sqr(2)

  x2=(6-4sqr(2))/2=3-2sqr(2)

x2-6x+1=(x-3-2sqr(+3-2sqr(2))

4х2 -12х +7 = 0

d=144-4*4*7=32

x1=(12+4sqr(2))/8=1.5+0.5sqr(2)

x2=(12+4sqr(2))/8=1.5+0.5sqr(2)

4х2 -12х +7=4(x-1.5-0.5sqr(-1.5+0.5sqr(2))

2*sin^2(x)-cos(2x)=0

2*sin^2(-2sin^2(x))=0

4sin^2(x)=1

sin^2(x)=1/4

sin(x)=±1/2

x=(-1)^narcsin(±1/2)+pi*n

x1=pi/6+pi*n

x2=7*pi/6+pi*n