Áp dụng tính chất dãy tỉ số bằng nhau ta có :0
\(\dfrac{a_1-1}{9}=\dfrac{a_2-2}{8}=..............=\dfrac{a_9-9}{1}=\dfrac{\left(a_1+a_2+......+a_9\right)-\left(1+2+....+9\right)}{9+8+..+1}\)
\(=\dfrac{90-45}{45}=1\)
+) \(\dfrac{a_1-1}{9}=1\Leftrightarrow a_1=10\)
+) \(\dfrac{a_2-1}{8}=1\Leftrightarrow a_2=10\)
........................
+) \(\dfrac{a_9-9}{1}=1\Leftrightarrow a_9=10\)
Vậy \(a_1=a_2=..........=a_9=10\)
Áp dụng tính chất dãy tỉ số bằng nhau ta có:
\(\dfrac{a_1-1}{9}=\dfrac{a_2-2}{8}=\dfrac{a_3-3}{7}=...=\dfrac{a_9-9}{1}\)
\(=\dfrac{a_1+a_2+...+a_9-\left(1+2+...+9\right)}{9+8+7+...+1}\)\(=\dfrac{90-45}{45}=1\)
\(\Rightarrow\left\{{}\begin{matrix}\dfrac{a_1-1}{9}=1\\\dfrac{a_2-2}{8}=1\\.................\\\dfrac{a_9-9}{1}=1\end{matrix}\right.\)\(\Rightarrow\left\{{}\begin{matrix}a_1-1=9\\a_2-2=8\\.................\\a_9-9=1\end{matrix}\right.\)\(\Rightarrow a_1=a_2=...=a_9=10\)
