[tex]\displaystyle\bf\\b_{1} =x+1\\\\b_{2} =x-1\\\\b_{3}=5x+1\\\\\\b_{2} ^{2} =b_{1} \cdot b_{3} \\\\\\(x-1)^{2} =(x+1)\cdot(5x+1)\\\\\\x^{2}-2x+1=5x^{2} +x+5x+1\\\\\\5x^{2} +6x+1-x^{2} +2x-1=0\\\\\\4x^{2} +8x=0 \ |:4\\\\\\x^{2} +2x=0\\\\\\x\cdot(x+2)=0\\\\\\\left[\begin{array}{ccc}x=0\\x+2=0\end{array}\right\\\\\\\left[\begin{array}{ccc}x_{1} =0\\x_{2} =-2\end{array}\right\\\\\\Otvet \ : \ 0 \ , \ -2[/tex]

[tex]\displaystyle\bf\\1) \ x= 0 \ \ \Rightarrow \ \ 1 \ ; \ -1 \ ; \ 1\\\\2) \ x=-2 \ \ \Rightarrow \ \ -1 \ ; \ -3 \ \ ; \ -9[/tex]