Помогите пожалуйста решить первый вариант из 3-х заданий!!!

1)a) $q= \frac{ b_{2}}{b_{2}}= \frac{27}{9} =3$
б) $q= \frac{ b_{2}}{b_{2}}= \frac{1}{4} : \frac{1}{ \sqrt{2} } =\frac{ \sqrt{2}}{4}=\frac{ \sqrt{2}}{ \sqrt{16} }= \frac{1}{ \sqrt{8} } = \frac{1}{ 2\sqrt{2} }$
2) [ $S_{5} = \frac{ b_{1}(q^5-1) }{q-1} = \frac{14( (\frac{1}{2})^5-1)}{ \frac{1}{2}-1 } = \frac{14( \frac{1}{32}- \frac{32}{32} ) }{- \frac{1}{2} } = \frac{ \frac{14}{1}*(- \frac{31}{32}) }{- \frac{1}{2} } = \frac{7*31}{16} : \frac{1}{2} = \frac{7*31}{8} =$ 27,125
3) $b_{1} =x$
$b_{2} =1$
$q= \frac{b_{2}}{b_{1}} = \frac{1}{x}$
$S_{6} = \frac{ b_{1}(q^6-1) }{q-1}=$ $\frac{x(( \frac{1}{x})^6-1) }{ \frac{1}{x}-1 } =$ $\frac{ \frac{1-x^6}{x^5} }{ \frac{1-x}{x} } =\frac{1-x^6}{x^5}:\frac{1-x}[tex]\frac{1^3-(x^2)^3}{x^4(1-x)}= \frac{(1-x^2)(x^4+x^2+1)}{x^4(1-x)} =\frac{(1-x)(1+x)(x^4+x^2+1)}{x^4(1-x)} =\frac{(1+x)(x^4+x^2+1)}{x^4}$ {x}=\frac{1-x^6}{x^5}*\frac{x}{1-x}=[/tex]