Question

Bardzo prosze o pomoc:) Zadania w zalaczniku:)

Answer 1

1) [(3x-3)/(5x-10)] :[(3x²-3)/(10x²-40)] =

  = [3(x-1)/5(x-2)] · [10(x²-4) / 3(x²-1)] =

  = [(x-1)·2·(x-2)(x+2)]/[(x-2)(x-1)(x+1)] = [2(x+2)]/(x+1)

        D = R \{-2,-1,1,2}

Dla x=2 nie można wyliczyć wartości, gdyż  2∉D.

2) a)  (x-1)/(x-2) - (x+1)/(x+2) =                         D = R\{2, -2}

         = [(x-1)(x+2) -(x+1)(x-2)]/[(x-2)(x+2)] =

         = (x²+2x-x-2-x²+2x-x+2)/(x²-4) =2x/(x²-4)

   b) (4x+1)/(x²-3x) + (x²-4)/(2x-6) =

        = (4x+1)/[x(x-3)] + (x²-4)/[2(x-3)] =

        = [2(4x+1)+x(x²-4)]/[2x(x-3)] =

        = (8x+2+x³-4x)/[2x(x-3)] =  (x³+4x+2) / [2x(x-3)]           D = R\{0,3}

Answer 2

1.Zał. 5(x-2)≠0   i   10(x²-4)≠0    i   3(x²-1)≠0

            x≠2         x≠2       x≠-2         x≠1    x≠-1

D=R\{-2,-1,1,2}

 

2. a) Zał. x-2≠0   i   x+2≠0

               x≠2          x≠-2

D=R\{-2,2}

 

 

b) Zał. x(x-3)≠0   i   2x-6≠0

           x≠0   x≠3     x≠3

D=R\{0,3}