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

1)a)[latex]q= \frac{ b_{2}}{b_{2}}= \frac{27}{9} =3[/latex] б) [latex]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} } [/latex] 2)  [[latex]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} =[/latex]27,125 3)[latex]b_{1} =x[/latex] [latex]b_{2} =1[/latex] [latex]q= \frac{b_{2}}{b_{1}} = \frac{1}{x} [/latex] [latex]S_{6} = \frac{ b_{1}(q^6-1) }{q-1}= [/latex][latex]\frac{x(( \frac{1}{x})^6-1) }{ \frac{1}{x}-1 } =[/latex][latex]\frac{ \frac{1-x^6}{x^5} }{ \frac{1-x}{x} } =\frac{1-x^6}{x^5}:\frac{1-x}[latex]\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} [/latex]{x}=\frac{1-x^6}{x^5}*\frac{x}{1-x}=[/latex]