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. $\frac{9 x^{2} -6x+4}{1-2x} = \frac{(3x-2) ^{2} }{1-2x}$