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

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

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

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

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

...............

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

\(\Rightarrow A>\frac{1}{4}+\left(\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{10.11}\right)=\frac{1}{4}+\left(\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{10}-\frac{1}{11}\right)=\frac{1}{4}+\left(\frac{1}{3}-\frac{1}{100}\right)=\frac{1}{4}+\frac{8}{33}=\frac{65}{132}\)

Vậy A > 65/132

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!

Câu hỏi của gam vu - Toán lớp 6 | Học trực tuyến

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

a>\(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+.....+\frac{1}{10.11}\)

  =\(\frac{1}{2}-\frac{1}{11}\)