Ta có : 

\(A=\frac{1}{4^2}+\frac{1}{6^2}+\frac{1}{8^2}+...+\frac{1}{\left(2n\right)^2}\)

\(A=\frac{1}{2^2}\left(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{n^2}\right)< \frac{1}{2^2}\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{\left(n-1\right)n}\right)\)

\(A< \frac{1}{4}\left(\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{n-1}-\frac{1}{n}\right)=\frac{1}{4}\left(1-\frac{1}{n}\right)\)

\(A< \frac{1}{4}-\frac{1}{4n}\)

Lại có \(n>0\) nên \(\frac{1}{4n}>0\)

\(\Rightarrow\)\(\frac{1}{4}-\frac{1}{4n}< \frac{1}{4}\)

Vậy \(A< \frac{1}{4}\)

Đặt \(A=\frac{1}{4^2}+\frac{1}{6^2}+...+\frac{1}{\left(2n\right)^2}\)

\(=\frac{1}{2^2}.\left(\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{n^2}\right)\)

Có:

\(\frac{1}{2^2}< \frac{1}{1.2}\)

\(\frac{1}{3^2}< \frac{1}{2.3}\)

\(...\)

\(\frac{1}{n^2}< \frac{1}{\left(n-1\right)n}\)

\(\Rightarrow\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{n^2}< \frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{\left(n-1\right)n}\)

\(\Rightarrow\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{n^2}< 1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{n-1}-\frac{1}{n}\)

\(\Rightarrow\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{n^2}< 1-\frac{1}{n}< 1\)

\(\Rightarrow A< \frac{1}{2^2}.1=\frac{1}{4}\)

\(\begin{equation} x = a_0 + \cfrac{1}{740_1 + \cfrac{1}{897654_2 + \cfrac{1}{672_3 + \cfrac{1}{100_4} } } } \end{equation}\)

Đề là chứng minh N < 1/4 sẽ đúng hơn

Ta có :

\(N=\frac{1}{4^2}+\frac{1}{6^2}+\frac{1}{8^2}+...+\frac{1}{\left(2n\right)^2}\)

\(\Rightarrow2^2.N=\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{n^2}\)

Ta lại có :

\(4N=\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{n^2}< \frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{\left(n-1\right)n}=1-\frac{1}{n}\)

\(\Rightarrow N< \left(1-\frac{1}{n}\right):4=\frac{1}{4}\left(1-\frac{1}{n}\right)\)

Mà \(n\in N;n\ge2\)=> 1 -\(\frac{1}{n}\)< 1

=> \(N< \frac{1}{4}\left(1-\frac{1}{n}\right)< \frac{1}{4}\)

=> \(N< \frac{1}{4}\)( đpcm )

Thank you very much

1/ Ta có:

\(a^5-a^3+a=2\)

Dễ thấy a = 0 không phải là nghiệm từ đó ta có:

\(a^6-a^4+a^2=2a\)

\(\Rightarrow2a=a^6+a^2-a^4\ge2a^4-a^4\ge a^4\)

\(\Rightarrow\hept{\begin{cases}2a\ge a^4\\a>0\end{cases}}\)

\(\Leftrightarrow\hept{\begin{cases}2\ge a^3\\a>0\end{cases}}\)

\(\Leftrightarrow\hept{\begin{cases}4\ge a^6\\a>0\end{cases}}\)

Dấu = không xảy ra 

Vậy \(a^6< 4\)

Câu 2/

Câu hỏi của XPer Miner - Toán lớp 9 - Học toán với OnlineMath

\(N=\frac{1}{4^2}+\frac{1}{6^2}+\frac{1}{8^2}+...+\frac{1}{\left(2n\right)^2}\)

\(N=\frac{1}{2^2}.\left(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{n^2}\right)< \frac{1}{2^2}.\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{...1}{\left(n-1\right).n}\right)\)

\(N< \frac{1}{4}.\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{n-1}-\frac{1}{n}\right)\)

\(N< \frac{1}{4}.\left(1-\frac{1}{n}\right)< \frac{1}{4}.1=\frac{1}{4}\)

=> \(N< \frac{1}{4}\)(đpcm)

bài này tui bít làm nhưng dài lắm

Cố gắng làm hộ mình với !