Объяснение:

Пусть b₁; b₂; b₃ - геометрическая прогрессия,

знаменатель которой равен q.       ⇒          b₁,   b₂=b₁q,   b₃=b₁q².

[tex]\displaystyle\\\left \{ {{b_2+b_3=4} \atop {b_3-b_2=b_2-\frac{5}{9} b_1}} \right. \ \ \ \ \left \{ {{b_1q+b_1q^2=4} \atop {b_3=2b_2-\frac{5}{9}b_1 }} \right.\ \ \ \ \ \left \{ {{b_1*(q+q^2)=4} \atop {b_1q^2=2b_1q-\frac{5}{9}b_1\ |:b_1\neq 0 }} \right. \\\\\\\left \{ {{b_1*(q+q^2)=4} \atop {q^2=2q-\frac{5}{9} } \right. \ \ \ \ \left \{ {{b_1*(q+q^2)=4} \atop {q^2-2q+\frac{5}{9} =0\ |*9}} \right. \ \ \ \ \left \{ {{b_1*(q+q^2)=4} \atop {9q^2-18q+5=0}} \right. \\\\\\\[/tex]

[tex]\displaystyle\\9q^2-18q+5=0\\\\D=18^2-4*9*5=324-180=144\\\\\sqrt{D} =\sqrt{144}=б 12\\\\q_{1,2}=\frac{18б12}{2*9} \\\\q_1=\frac{18-12}{18} =\frac{6}{18} =\frac{1}{3} .\\\\q_2=\frac{18+12}{18} =\frac{30}{18}=\frac{5}{3} .\ \ \ \ \Rightarrow[/tex]

[tex]\displaystyle\\1)\\\\b_1*(\frac{1}{3} +(\frac{1}{3})^2)=4\\\\b_1*(\frac{1}{3}+\frac{1}{9} )=4\\\\b_1*\frac{4}{9} =4\ |*\frac{9}{4}\\\\b_1=9.\ \ \ \ \Rightarrow\\\\b_2=9*\frac{1}{3}=3.\\\\b_3=3*\frac{1}{3}=1.\ \ \ \ \Rightarrow\\\\9;\ 3;\ 1.[/tex]

[tex]\displaystyle\\2)\\\\b_1*(\frac{5}{3}+(\frac{5}{3})^2)=4\\\\ b_1*(\frac{5}{3} +\frac{25}{9})=4 \\\\b_1*\frac{40}{9} =4\ |*\frac{9}{40} \\\\b_1=\frac{9}{10}=0,9 .\\\\b_2=\frac{9}{10}*\frac{5}{3} =\frac{3}{2}=1,5.\\\\b_3=\frac{3}{2}*\frac{5}{3} =\frac{5}{2}=2,5 .\ \ \ \ \Rightarrow\\\\0,9;\ 1,5;\ 2,5.[/tex]