Question

Докажите тождество (6x^2/3x+2 - 2x)*27x^3+8/8x^2-4x=9x^2-6x+4/1-2x

Answer 1

(6x²:(3x+2)-2x)*27x³+8:8x²-4x=(9x²-6x+4):(1-2х)

1. $frac{6 x^{2} }{3x+2} -2x=frac{6 x^{2} }{3x+2}-frac{6 x^{2}+4x }{3x+2}=frac{6 x^{2}-6 x^{2} -4x }{3x+2}=frac{-4x }{3x+2}$
2. $frac{-4x }{3x+2}*27 x^{3} =frac{-108 x^{4} }{3x+2}$
3. $frac{8}{8 x^{2} -4x}= frac{8}{4(2 x^{2} -x)}= frac{2}{2 x^{2} -x}$
4. $frac{-108 x^{4} }{3x+2}: frac{2}{2 x^{2} -x}=frac{-108 x^{4} }{3x+2}* frac{2 x^{2} -x}{2}= frac{-108 x^{4}*x(2x-1) }{(3x+2)*2}=frac{-54 x^{5}*(2x-1) }{3x+2}=frac{54 x^{5}*(1-2x) }{3x+2}$=$frac{-108 x^{6}+54 x^{5} }{3x+2} = frac{54 x^{5} -108 x^{6} }{3x+2}$
5. $frac{9 x^{2} -6x+4}{1-2x} = frac{(3x-2) ^{2} }{1-2x}$