Question

1. $\frac{2x-2}{4x^{2}+4x+4 } -\frac{12x}{8x^{2}-8 } +\frac{1}{2x-2} =\frac{x-1}{x^{2}+x+1 }$
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Answer 1

$\frac{2x-2}{4x^{2}+4x+4 }-\frac{12x}{8x^{3}-8 }+\frac{1}{2x-2}=\frac{2(x-1)}{4(x^{2}+x+1)}-\frac{12x}{8(x^{3}-1)}+\frac{1}{2(x-1)} =\frac{x-1}{2(x^{2}+x+1)}-\frac{3x}{2(x-1)(x^{2}+x+1)}+\frac{1}{2(x-1)}=\frac{(x-1)^{2}-3x+x^{2}+x+1}{2(x-1)(x^{2}+x+1)}=\frac{x^{2}-2x+1+x^{2}-2x+1}{2(x-1)(x^{2}+x+1)}=\frac{2x^{2}-4x+2}{2(x-1)(x^{2}+x+1)}=\frac{2(x^{2}-2x+1)}{2(x-1)(x^{2}+x+1)}=\frac{2(x-1)^{2} }{2(x-1)(x^{2}+x+1)}=\frac{x-1}{x^{2}+x+1}$

$\frac{x-1}{x^{2}+x+1}=\frac{x-1}{x^{2}+x+1}$

Тождество доказано