Question

[2- (x^2 - x + 1)/(4x^3 + 4) * (4x^2 + 8x + 4)/ (3 - x)]: (10 - x - 3x^2)/ (x^2 - 9) =

(te znaczki "/" prosze traktować jako kreski ułamkowe dotyczące funkcji w "()"; zaś "^" potegi)

Answer 1

[2 - (x²-x+1)/(4x³+4) · (4x²+8x+4)/(3-x)] : (10-x-3x²)/(x²-9) =

= {2 - (x²-x+1)/[4(x+1)(x²-x+1)] · [4(x²+2x+1)]/(3-x) } · [(x-3)(x+3)]/(-3x²-x+10) =

= { 2 -(x²+2x+1)/[(x+1)(3-x)]} · [(x-3)(x+3)]/[-(x+2)(x-1⅔)] =                 Δ=1+120=121

= [2(x+1)(3-x) -x²-2x-1]/[(x+1)(3-x)] · [(x-3)(x+3)]/[-(x+2)(x-1⅔)]=           √Δ=11

= [2(3x-x²+3-x)-x²-2x-1]/[(x+1)(3-x)] · [(x-3)(x+3)]/[-(x+2)(x-1⅔)] =      x₁=-2, x₂=1⅔

= (6x-2x²+6-2x-x²-2x-1)/[(x+1)(x-3) · [(x-3)(+3)]/[(x+2)(x-1⅔)] =

= [(-3x²+2x+5)(x+3)] / [(x+1)(x+2)(x-1⅔)] =                      -3x²+2x+5=0

= [-(x+1)(x-1⅔)(x+3)]/[(x+1)(x+2)(x-1⅔)] =                           Δ=4+60 = 64,    √Δ=8

= [-(x+3)]/(x+2) =                                                                   x₁= -1,   x₂ = 1⅔

= (-x-3)/(x+2)                          D=R\{3, -3, -2, -1, 1⅔}