Решите уравнение x/х-1 + 3/х+1 = 2/х^2-1

X/(x-1)+3/(x+1)=2/(x²-1); |x≠1 x(x+1)+3(x-1)=2; |x≠-1 x²+x+3x-3-2=0; x²+4x-5=0; d=16+20=36; x1=(-4+√36)/2=(-4+6)/2=2/2=1; x2=(-4-√36)/2=(-4-6)/2=-10/2=-5; ответ: -5, т.к. одз: x≠1.

x^2+2x+y^2=16

x+y=16   => y=2-x

 

x^2+2x+(2-x)^2=16

x^2+2x+4-4x+x^2=16

2x^2-2x-12=0

x^2-x-6=0

 

d=b^2-4ac=1-4*1(-6)=25

 

x1,2=(-b±sqrt(d))/2a

x1=(1+5)/2=3

x2=(1-5)/2=-2

 

при x1=3   y1=2-3=-1

при x2=-2   y2=)=4

b1+b2=-10

b2+b3=-5

 

b1+b1q=-10

b1q+b1q^2=-5

 

b1(1+q)=-10

b1q(1+q)=-5

 

b1=-10/(1+q)

-10q(1+q)/(1+q)=-5

 

b1=-10/(1+q)

-10q=-5

 

b1=-10/(1+q)

q=1/2

 

b1=-10(1+1/2)

q=1/2

 

b1=-20/3

q=1/2

 

b2=b1q=-20/3*1/2=-10/3

b3=b2q=-10/3*1/2=-5/3

 

-20/3; -10/3; -5/3