Sai rồi bn ơi!!! 

     \(k^3=8\)

\(\Leftrightarrow k=2\)

nhé. Còn bn lm

     \(k^3=8\)

\(\Leftrightarrow k=\pm2\)

là sai r.

Để \(M< 1\), ta phải có điều kiện: \(n\in\) R*. Nếu \(n=0\) thì \(M\) không xác định.
\(M=\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+...+\dfrac{1}{n^2}< \dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{\left(n-1\right)n}\)
                                                 \(=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{n-1}-\dfrac{1}{n}\)
                                                 \(=1-\dfrac{1}{n}< 1\)
Vậy \(M< 1\) với \(n\in\) R*.

Nhầm rồi, kết quả là -(584/495) nhé.

    \(\left(-\dfrac{5}{9}\right)+\dfrac{8}{18}+\left(-\dfrac{12}{11}\right)+\left(-\dfrac{4}{9}\right)+\dfrac{7}{15}\)
\(=\left[\dfrac{4}{9}+\left(-\dfrac{4}{9}\right)\right]+\left(-\dfrac{5}{9}\right)+\left(-\dfrac{12}{11}\right)+\dfrac{7}{15}\)
\(=-\dfrac{224}{495}\)

    \(\dfrac{4}{3}.\dfrac{9}{8}.\dfrac{16}{15}.\dfrac{25}{24}...\dfrac{100}{99}\)
\(=\dfrac{2^2}{1.3}.\dfrac{3^2}{2.4}.\dfrac{4^2}{3.5}.\dfrac{5^2}{4.6}...\dfrac{10^2}{9.11}\)
\(=\dfrac{2^2.3^2.4^2.5^2...10^2}{1.2.3^2.4^2.5^2...9^2.10.11}\)
\(=\dfrac{2.10}{1.11}=\dfrac{20}{11}\)

Tôi thik nhất Toán, Lí, Hóa, Công nghệ.  :)