Question

\(\frac{\left(\frac{-2}{11}\right)^{n+1}}{\left(\frac{-2}{11}\right)^n}\left(n\ge1\right)\)

Answer 1

\(\frac{\left(\frac{-2}{11}\right)^{n+1}}{\left(\frac{-2}{11}\right)^n}=\frac{\left(\frac{-2}{11}\right)^n.\left(\frac{-2}{11}\right)}{\left(\frac{-2}{11}\right)^n}=\frac{\left(\frac{-2}{11}\right)^n}{\left(\frac{-2}{11}\right)^n}.\frac{\left(\frac{-2}{11}\right)}{\left(\frac{-2}{11}\right)^n}=1.\left(\frac{-2}{11}\right).\frac{1}{\left(\frac{-2}{11}\right)^n}\) \(=\frac{1}{1^n}\)

Nếu n =1 thì biểu thức sẽ bằng 1

Answer 2

Làm biếng quá sai bét nhè chè đỗ đen mà vẫn k đúng, khó hỉu :))