Giúp em vs

Để $\displaystyle x+1,\ x+3,3x+7$ thành 1 cấp số nhân thì:
+ TH1: $\displaystyle u_{1} =x+1,u_{2} =x+3,u_{3} =3x+7$
$\displaystyle  \begin{array}{{>{\displaystyle}l}}
\Rightarrow \frac{x+3}{x+1} =\frac{3x+7}{x+3}\\
\Leftrightarrow x^{2} +6x+9=3x^{2} +10x+7\\
\Leftrightarrow 2x^{2} +4x-2=0\\
\Leftrightarrow x=-1\pm \sqrt{2}
\end{array}$
+ TH2: $\displaystyle u_{1} =x+1,\ u_{2} =3x+7,\ u_{3} =x+3$
$\displaystyle  \begin{array}{{>{\displaystyle}l}}
\Rightarrow \frac{3x+7}{x+1} =\frac{x+3}{3x+7}\\
\Leftrightarrow 9x^{2} +42x+49=x^{2} +4x+3\\
\Leftrightarrow 8x^{2} +38x+46=0
\end{array}$
(vô nghiệm)
+ TH3: $\displaystyle u_{1} =x+3,\ u_{2} =x+1,\ u_{3} =3x+7$
$\displaystyle  \begin{array}{{>{\displaystyle}l}}
\Rightarrow \frac{x+1}{x+3} =\frac{3x+7}{x+1}\\
\Leftrightarrow x^{2} +2x+1=3x^{2} +16x+21\\
\Leftrightarrow 2x^{2} +14x+20=0\\
\Leftrightarrow x=-2,x=-5
\end{array}$
+ TH4: $\displaystyle u_{1} =x+3,u_{2} =3x+7,u_{3} =x+1$
$\displaystyle  \begin{array}{{>{\displaystyle}l}}
\Rightarrow \frac{3x+7}{x+3} =\frac{x+1}{3x+7}\\
\Leftrightarrow ( 3x+7)^{2} =( x+1)( x+3)
\end{array}$
vô nghiệm
+ TH5: $\displaystyle u_{1} =3x+7,u_{2} =x+1,u_{3} =x+3$
$\displaystyle  \begin{array}{{>{\displaystyle}l}}
\Rightarrow \frac{x+1}{3x+7} =\frac{x+3}{x+1}\\
\Rightarrow x=-2,x=-5
\end{array}$
+ TH6: $\displaystyle u_{1} =3x+7,u_{2} =x+3,u_{3} =x+1$
$\displaystyle  \begin{array}{{>{\displaystyle}l}}
\Rightarrow \frac{x+3}{3x+7} =\frac{x+1}{x+3}\\
\Rightarrow x=-1\pm \sqrt{2}
\end{array}$
Vậy có 4 nghiệm x thoả mãn