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 )

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

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

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

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

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

A = \(\dfrac{1}{2^2}\).(\(\dfrac{1}{1^2}\) + \(\dfrac{1}{2^2}\) + \(\dfrac{1}{3^2}\) + ... + \(\dfrac{1}{n^2}\)) 

A = \(\dfrac{1}{4}\).(\(\dfrac{1}{1}\) + \(\dfrac{1}{2.2}\) + \(\dfrac{1}{3.3}\) + ... + \(\dfrac{1}{n.n}\))

Vì \(\dfrac{1}{2.2}\) < \(\dfrac{1}{1.2}\); \(\dfrac{1}{3.3}\) < \(\dfrac{1}{2.3}\); ...; \(\dfrac{1}{n.n}\) < \(\dfrac{1}{\left(n-1\right)n}\)

nên A < \(\dfrac{1}{4}\).(\(\dfrac{1}{1}\) + \(\dfrac{1}{1.2}\) + \(\dfrac{1}{2.3}\) + ... + \(\dfrac{1}{\left(n-1\right)n}\))

A < \(\dfrac{1}{4.}\)(1 + \(\dfrac{1}{1}\) - \(\dfrac{1}{2}\) + \(\dfrac{1}{2}\) - \(\dfrac{1}{3}\) + \(\dfrac{1}{n-1}\) - \(\dfrac{1}{n}\))

A < \(\dfrac{1}{4}\).(1 + 1 - \(\dfrac{1}{n}\))

A < \(\dfrac{1}{4}\).(2 - \(\dfrac{1}{n}\))

A < \(\dfrac{1}{2}\) - \(\dfrac{1}{4n}\) < \(\dfrac{1}{2}\) (đpcm)

Từ "lạc trôi" có nghĩa là gì trong câu:

"Mây bềnh bồng lạc trôi/mượt mà như tuổi ngọc."

Bạn MAI XUÂN LONG trả lời lộn câu hỏi rồi à

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

\(=\frac{-\left(2n-2\right)^2-1+4n^2+1}{\left(4n^2+1\right)\left[\left(2n-2\right)^2+1\right]}=\frac{4\left(2n-1\right)}{\left(4n^2-4n+1+4n\right)\left(4n^2-4n+1-6n+4\right)}\)
\(=\frac{4\left(2n-1\right)}{\left(4n^2-4n+1\right)^2+4\left(4n^2-4n+1\right)-16n^2+16n}=\frac{4\left(2n-1\right)}{\left(2n-1\right)^4+4}\)
\(\Rightarrow\frac{n^2}{4n^2+1}-\frac{\left(n-1\right)^2}{4\left(n-1\right)^2+1}=\frac{2n-1}{4+\left(2n-1\right)^4}\)
-> đpcm theo phương pháp quy nạp

uk e ms hk lop 6 thuj oy

Câu 1 :
 A = (2012+2) . [ ( 2012-2) : 3+1 ] : 2 = 2014 . 671 : 2 = 675697
 B = \(\frac{1}{2}\).  \(\frac{2}{3}\).  \(\frac{3}{4}\)+...+  \(\frac{2010}{2011}\).  \(\frac{2011}{2012}\)= \(\frac{1.2.3.....2010.2011}{2.3.4.....2011.2012}\)=  \(\frac{1}{2012}\)
Câu 2 :
 a) \(2x.\left(3y-2\right)+\left(3y-2\right)=-55\)
=> \(\left(3y-2\right).\left(2x+1\right)=-55\)
=>  \(3y-2;2x+1\in\: UC\left(-55\right)\)
=>  \(3y-2;2x+1=\left\{1;-1;5;-5;11;-11;55;-55\right\}\)
- Vậy ta có bảng 

\(\Leftrightarrow\)Những cặp (x;y) tìm được là : 
(-1;19)  ;   (2;-3)   ;    (5;-1)    ;    (-28;1)
b) Ta đặt vế đó là A
Ta xét A :   \(\frac{1}{4^2}\)<  \(\frac{1}{2.4}\)
                  \(\frac{1}{6^2}\)<  \(\frac{1}{4.6}\)
                  \(\frac{1}{8^2}\)<  \(\frac{1}{6.8}\)
                          ...
                 \(\frac{1}{\left(2n\right)^2}\)<  \(\frac{1}{\left(2n-2\right).2n}\)

  \(\Leftrightarrow\)A < \(\frac{1}{2.4}\)+  \(\frac{1}{4.6}\)+...+  \(\frac{1}{\left(2n-2\right).2n}\)
  \(\Leftrightarrow\)A < \(\frac{1}{2}\). ( \(\frac{2}{2.4}\)+  \(\frac{2}{4.6}\)+...+  \(\frac{2}{\left(2n-2\right).2n}\))
  \(\Leftrightarrow\)A < \(\frac{1}{2}\). ( \(\frac{1}{2}\)-  \(\frac{1}{4}\)+  \(\frac{1}{4}\)-  \(\frac{1}{6}\)+...+  \(\frac{1}{2n-2}\)-  \(\frac{1}{2n}\))
  \(\Leftrightarrow\)A < \(\frac{1}{2}\). ( \(\frac{1}{2}\)-  \(\frac{1}{2n}\)) = \(\frac{1}{2}\).  \(\frac{1}{2}\)-  \(\frac{1}{2}\).  \(\frac{1}{2n}\)
  \(\Leftrightarrow\)A < \(\frac{1}{4}\)-  \(\frac{1}{4n}\)<  \(\frac{1}{4}\) ( Vì n \(\in\)N )
  \(\Leftrightarrow\)A <  \(\frac{1}{4}\)( đpcm ) .

Bạn Phùng Quang Thịnh làm đúng hết rồi