vì tất cả các mẫu số cộng vào thì sẽ lớn hơn mẫu số 132 nên A>65/132

Ta có : \(A=\frac{1}{4}+\frac{1}{9}+\frac{1}{16}+...+\frac{1}{81}+\frac{1}{100}\)

\(A=\frac{1}{2.2}+\frac{1}{3.3}+\frac{1}{4.4}+....+\frac{1}{9.9}+\frac{1}{10.10}\)

\(=>A>\frac{1}{2.2}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{9.10}+\frac{1}{10.11}\)

\(=>A>\frac{1}{4}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+....+\frac{1}{9}-\frac{1}{10}+\frac{1}{10}-\frac{1}{11}\)

\(=>A>\frac{1}{4}+\frac{1}{3}-\frac{1}{11}=\frac{65}{132}\)

\(=>A>\frac{65}{132}\left(đpcm\right)\)

A=1/2*2+1/3*3+1/4*4+...+1/10*10.

A>1/1*2+1/2*3+1/3*4+...+1/9*10.

A>1-1/2+1/2-1/3+...+1/9-1/10.

A>1-1/10.

A>9/10.

=>A>1/2.

Mà 1/2=66/132>65/132.

=>A>65/132.

Vậy A>65/132.

A=1/2^2+1/3^2+1/4^2+......+1/9^2+1/10^2

=1/4+1/3×3+1/4×4+.....+1/9×9+1/10×10

=>A>1/4+(1/3×4+1/4×5+...+1/9×10+1/10×11)

=>A>1/4+(1/3-1/11)

=>A>1/4+8/33

=>A>65/132( đpcm)

A=1/201+1/202+...+1/300 ( Ta xét mẫu thì thấy có 100 số => ta xếp cặp sao cho cặp là Ư(100) và cặp đó là 2(cặp)=> mỗi cặp 50 số)
Cặp 1: (1) Vì 1/201<1/200
              1/202<1/200
               ...
              1/250<1/200        (Vì từ 201->250 có 50 số nên mới có số 250)
Cặp 2: (2)
     Vì 1/251<1/250
         1/252<1/250
         ...
          1/300<1/250
Từ (1) và (2),ta cộng cặp 1 với cặp 2 ta có:
A=(1/201+1/202+...+1/250)+(1/251+1/252+...+1/300) < (1/200+1/200+1/200+...+1/200)+(1/250+1/250+...+1/250)
                                                                                             50 phân số 1/200            50 phân số 1/250
=> A=(1/201+1/202+...+1/250)+(1/251+1/252+...+1/300) <   50/200 + 50/250
=> A < 1/4+1/5
=> A < 9/20 (đpcm)
* Chú ý : +Nhớ k mình nhé :)
+Mình làm hơi khó hiểu nên hãy hỏi mình chỗ nào bạn không hiểu ^_^

hay quá

Đề sai nha:

Sửa lại:

Cho \(A=\dfrac{1}{4}+\dfrac{1}{9}+\dfrac{1}{16}+...+\dfrac{1}{81}+\dfrac{1}{100}\). Chứng tỏ rằng \(A>\dfrac{65}{132}\)

Giải:

Có:

\(A=\dfrac{1}{4}+\dfrac{1}{9}+\dfrac{1}{16}+...+\dfrac{1}{81}+\dfrac{1}{100}\)

\(A=\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+...+\dfrac{1}{9^2}+\dfrac{1}{10^2}\)

Mà: \(\dfrac{1}{3^2}>\dfrac{1}{3.4}\);

\(\dfrac{1}{4^2}>\dfrac{1}{4.5}\);

...

\(\dfrac{1}{9^2}>\dfrac{1}{9.10}\);

\(\dfrac{1}{10^2}>\dfrac{1}{10.11}\).

\(\Rightarrow A>\dfrac{1}{2^2}+\dfrac{1}{3.4}+\dfrac{1}{4.5}+...+\dfrac{1}{9.10}+\dfrac{1}{10.11}\)

\(A>\dfrac{1}{2^2}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+...+\dfrac{1}{9}-\dfrac{1}{10}+\dfrac{1}{10}-\dfrac{1}{11}\)

\(\Rightarrow A>\dfrac{1}{2^2}+\dfrac{1}{3}-0-0-...-0-\dfrac{1}{11}\)

\(\Rightarrow A>\dfrac{1}{2^2}+\dfrac{1}{3}-\dfrac{1}{11}\)

\(\Rightarrow A>\dfrac{65}{132}\)

Chúc bạn học tốt!

A = \(\frac{1}{4}+\frac{1}{9}+\frac{1}{16}+...+\frac{1}{100}\)

= \(\frac{1}{4}+\left(\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{10^2}\right)\)

Ta có: \(\frac{1}{3^2}>\frac{1}{3.4}\)

\(\frac{1}{4^2}>\frac{1}{4.5}\)

.........

\(\frac{1}{10^2}>\frac{1}{10.11}\)

\(\Rightarrow A>\frac{1}{4}+\left(\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{10.11}\right)\)

\(\Rightarrow A>\frac{1}{4}+\left(\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{10}-\frac{1}{11}\right)\)

\(\Rightarrow A>\frac{1}{4}+\left(\frac{1}{3}-\frac{1}{11}\right)=\frac{1}{4}+\frac{8}{33}=\frac{65}{132}\)

Vậy A > 65/132

Ta có:
\(A=\frac{1}{4}+\frac{1}{9}+\frac{1}{16}+...+\frac{1}{81}+\frac{1}{100}\)

\(\Leftrightarrow A=\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{9^2}+\frac{1}{10^2}\)

\(\Leftrightarrow A>\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+\frac{1}{4\cdot5}+...+\frac{1}{9\cdot10}+\frac{1}{10\cdot11}\)

\(\Leftrightarrow A>\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{9}-\frac{1}{10}+\frac{1}{10}-\frac{1}{11}\)

\(\Leftrightarrow A>\frac{1}{2}-\frac{1}{11}\)

\(\Leftrightarrow A>\frac{9}{22}\)

Ta lại có:

\(\frac{9}{22}=\frac{9.11}{22\cdot11}=\frac{99}{132}\)

Ta thấy: 99>65

\(\Rightarrow\frac{99}{132}>\frac{65}{132}\)

\(\Rightarrow A>\frac{65}{132}\)

Vậy \(A>\frac{65}{132}\left(đpcm\right)\)

\(A=\frac{1}{4}+\frac{1}{9}+\frac{1}{16}+...+\frac{1}{81}+\frac{1}{100}\)

\(A=\frac{1}{4}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{9^2}+\frac{1}{10^2}\)

\(A>\frac{1}{4}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{9.10}+\frac{1}{10.11}\)

\(A>\frac{1}{4}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{10}-\frac{1}{11}\)

\(A>\frac{1}{4}+\frac{1}{3}-\frac{1}{11}\)

\(A>\frac{33}{132}+\frac{44}{132}-\frac{12}{132}\)

\(A>\frac{65}{132}\)