a ) \(A=\frac{1}{2^2}+\frac{1}{4^2}+\frac{1}{6^2}+...+\frac{1}{\left(2n\right)^2}=\frac{1}{4}\left(\frac{1}{1^2}+\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{n^2}\right)\)

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

b )

\(B=\frac{1}{3^2}+\frac{1}{5^2}+...+\frac{1}{\left(2n+1\right)^2}< \frac{1}{3^2-1}+\frac{1}{5^2-1}+...+\frac{1}{\left(2n+1\right)^2-1}\)

\(=\frac{1}{2.4}+\frac{1}{4.6}+...+\frac{1}{2n\left(2n+2\right)}\)

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

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

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 \(A< \frac{1}{4}\)

\(a)\) Đặt \(A=\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{2010^2}\) ta có : 

\(A< \frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{2009.2010}\)

\(A< \frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2009}-\frac{1}{2010}\)

\(A< 1-\frac{1}{2010}=\frac{2009}{2010}< 1\)

\(\Rightarrow\)\(A< 1\) ( đpcm ) 

Vậy \(A< 1\)

Chúc bạn học tốt ~ 

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

\(< \frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{\left(2n-1\right)n}\)

\(=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2n-1}-\frac{1}{n}\)

\(=\frac{1}{2}-\frac{1}{n}< \frac{1}{2}^{\left(đpcm\right)}\)  (không chắc nha)

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

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

Ta có: \(\frac{1}{1}=\frac{1}{1},\frac{1}{2^2}< \frac{1}{1.2},\frac{1}{3^2}< \frac{1}{2.3},....,\frac{1}{n^2}< \frac{1}{\left(n-1\right).n}\)

=> \(A< \frac{1}{2^2}.\left[1+\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{\left(n-1\right)n}\right]\)

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

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

p/s: bài tớ ko bt đúng ko, nhưng tth bn làm vậy sẽ ko có quy luật, đoạn này

nếu cứ theo quy luật, tiếp tục sẽ ntn:\(\frac{1}{6^2}< \frac{1}{5.6};\frac{1}{8^2}< \frac{1}{6.7};\frac{1}{10^2}< \frac{1}{7.8}\)

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

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

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

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

Bạn tham khảo cách làm của bạn Alibabba nguyễn nha!!