Question

помогите пожалуйста!​

Answer 1

$\displaystyle\bigg(\frac{a}{a^{2}-2a+1 } -\frac{a+4}{a^{2} -1}\bigg)\div\frac{a-2}{a^{2}-1 } =\frac{2}{1-a} \\\bigg(\frac{a}{(a-1)(a-1)}-\frac{a+4}{(a-1)(a+1)}\bigg)\div\frac{a-2}{(a-1)(a+1)}=\frac{2}{1-a}\\\frac{a}{(a-1)(a-1)}\div\frac{a-2}{(a-1)(a+1)}-\frac{a+4}{(a-1)(a+1)}\div\frac{a-2}{(a-1)(a+1)}=\frac{2}{1-a}\\\frac{a}{(a-1)(a-1)}\times\frac{(a-1)(a+1)}{a-2} -\frac{a+4}{(a-1)(a+1)} \times\frac{(a-1)(a+1)}{a-2} =\frac{2}{1-a} \\ \frac{a(a+1)}{(a-2)(a-1)} -\frac{a+4}{a-2}=\frac{2}{1-a}$

$\displaystyle\frac{a(a+1)-(a-1)(a+4)}{(a-2)(a-1)} =\frac{2}{1-a} \\\frac{a^{2}+a+a+4-a^{2} -4a }{(a-2)(a-1)} =\frac{2}{1-a}\\ - \frac{2(a-2)}{(a-2)(a-1)} =\frac{2}{1-a} \\\frac{2}{1-a} =\frac{2}{1-a}$

Ч.Т.Д.