Đặt A=đã cho.

Ta thấy:

1/2^2<1/1*2(vì 2^2>1*2).

1/3^2<1/2*3(vì 3^2>2*3).

...

1/10^2<1/9*10(vì 10^2>9*10).

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

=>A<1-1/2+1/2-1/3+1/3-1/4+...+1/9-1/10.

=>A<1-1/10.

=>A<9/10.

Mà 9/10<1.

=>A<1.

Vậy A<1(đpcm).

khó quá mik trả lời ko được

Ta có : \(\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{10^2}< \frac{1}{1\cdot2}+\frac{1}{2\cdot3}+...+\frac{1}{9\cdot10}=1-\frac{1}{10}=\frac{9}{10}< 1\)

Vậy \(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{10^2}< 1\)     ( đpcm )

Ta có : \(\frac{1}{2^2}< \frac{1}{1.2}\)

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

....

\(\frac{1}{10^2}\)< \(\frac{1}{9.10}\)

=> \(\frac{1}{2^2}+....+\frac{1}{10^2}\)< \(\frac{1}{1.2}+\frac{1}{2.3}+.....+\frac{1}{9.10}\)

=> \(\frac{1}{2^2}+....+\frac{1}{10^2}\)< \(\frac{9}{10}\)< 1

=> \(\frac{1}{2^2}+....+\frac{1}{10^2}\)< 1 ( dpcm )

\(\frac{1}{4}\)+\(\frac{1}{9}\)+

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

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

\(A=1-\frac{1}{2^{10}}\)=> \(1-A=1-\left(1-\frac{1}{2^{10}}\right)=\frac{1}{2^{10}}>\frac{1}{2^{11}}\)=> đpcm

b) Đặt B = \(\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{100^2}\)

Vì \(\frac{1}{2^2}

Ta có \(D=\frac{1}{2^2}+\frac{1}{3^2}+....+\frac{1}{10^2}< \frac{1}{1.2}+\frac{1}{2.3}+....+\frac{1}{9.10}.\)

Mà \(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{9.10}=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+....+\frac{1}{9}-\frac{1}{10}\)

                                                                   \(=1-\frac{1}{10}=\frac{9}{10}< 1\)

\(\Rightarrow D< 1\)

Vậy \(D< 1\)

Ta có: 1/2² < 1/1.2

           1/3² <  1/2.3

          1/4² < 1/3.4

             ......

           1/10² < 1/9.10

=> D < 1/1.2 + 1/2.3 + 1/3.4 + ... + 1/9.10

=> D < 1 -1/2 + 1/2 - 1/3 + 1/3 - 1/4 + ... + 1/9 -1/10

=> D < 1 - 1/10

=> D < 9/10

=. D < 9/10 < 1

=> D < 1 ( đpcm )

Làm theo cách của Trắng nha , 

\(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{2019^2}< \frac{1}{2^2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{2018.2019}\)

\(\Rightarrow\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{2019^2}< \frac{1}{2^2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2018}-\frac{1}{2019}\)

\(\Rightarrow\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{2019^2}< \frac{1}{4}+\frac{1}{2}-\frac{1}{2019}\)

\(\Rightarrow\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{2019^2}< \frac{3}{4}-\frac{1}{2019}< \frac{3}{4}\left(Đpcm\right)\)

Ta có:  \(\frac{1}{2^2}=\frac{1}{2^2}\)

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

             ...................

             \(\frac{1}{2019^2}< \frac{1}{2018.2019}\)

\(\Rightarrow\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{2019^2}< \frac{1}{2^2}+\frac{1}{2.3}+...+\frac{1}{2018.2019}\)

\(=\frac{1}{2^2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{2018}-\frac{1}{2019}\)

\(=\frac{1}{4}+\frac{1}{2}-\frac{1}{2019}\)

\(=\frac{1}{4}+\frac{2}{4}-\frac{1}{2019}\)

\(=\frac{3}{4}-\frac{1}{2019}\)\(< \frac{3}{4}\)

\(\Rightarrow\)\(\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{2019^2}< \frac{3}{4}\)

                                              Điều phải chứng minh

Đặt \(A=\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{2019^2}\)

Ta có:

\(\frac{1}{2^2}=\frac{1}{2.2}< \frac{1}{1.2}\)

\(\frac{1}{3^2}=\frac{1}{3.3}< \frac{1}{2.3}\)

\(\frac{1}{4^2}=\frac{1}{4.4}< \frac{1}{3.4}\)

....

\(\frac{1}{2019^2}=\frac{1}{2019.2019}< \frac{1}{2018.2019}\)

\(\Rightarrow A< \frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{2018.2019}\)

\(\Rightarrow A< 1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2018}-\frac{1}{2019}\)

\(\Rightarrow A< 1-\frac{1}{2019}\)

\(\Rightarrow A< \frac{2018}{2019}\)

đến đây mới thấy mik sai ,xin lỗi

Đặt \(A=\frac{1}{2^2}+\frac{1}{3^2}+....+\frac{1}{2019^2}\)

\(\Rightarrow A=\frac{1}{2^2}+\left(\frac{1}{3^2}+\frac{1}{4^2}+\frac{1}{5^2}+...+\frac{1}{2019^2}\right)\)

\(\Rightarrow A< \frac{1}{4}+\left(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{2018.2019}\right)\)

\(\Rightarrow A< \frac{1}{4}+\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+..+\frac{1}{2018}-\frac{1}{2019}\right)\)

\(\Rightarrow A< \frac{1}{4}+\left(\frac{1}{2}-\frac{1}{2019}\right)\)

\(\Rightarrow A< \frac{1}{4}+\frac{1}{2}-\frac{1}{2019}=\frac{3}{4}-\frac{1}{2019}< \frac{3}{4}\)

\(\Rightarrow A< \frac{3}{4}\)

đặt A=1/2^2+....+1/2019^2

vì 1/2^2+....+1/2019^2<1/1.2+1/2.3+....+1/2018.2019

=> A<1/1-1/2+1/2-1/3+.....+1/2018-1/2019

=> A<1-1/2019=2018/2019<3/4.

=> A<3/4. 

vậy 1/2^2+....+1/2019^2<3/4

Ta có: \(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{2019^2}< \frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{2018.2019}\)

\(\Rightarrow\)\(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{2019^2}< 1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}\)\(+...+\frac{1}{2018}-\frac{1}{2019}\)

\(\Rightarrow\)\(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{2019^2}< 1-\frac{1}{2019}\)

\(\Rightarrow\)\(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{2019^2}< 1-\frac{1}{2019}\)

\(\Rightarrow\)\(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{2019^2}< \frac{2018}{2019}\)

Mà: \(\frac{3}{4}=\frac{2016}{2688}< \frac{2017}{2688}< \frac{2017}{2019}< \frac{2018}{2019}\)

\(\Rightarrow\frac{3}{4}< \frac{2018}{2019}\)

\(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{2011^2}\)

\(\text{Vì}\frac{1}{3^2}< \frac{1}{2.3};\frac{1}{4^2}< \frac{1}{3.4};...;\frac{1}{2011^2}< \frac{1}{2010.2011}\)

\(\Rightarrow\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{2011^2}< \frac{1}{2^2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{2010.2011}\)

\(\Rightarrow\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{2011^2}< \frac{1}{2^2}+\frac{1}{2}-\frac{1}{2011}\)

\(\Rightarrow\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{2011^2}< \frac{1}{4}+\frac{1}{2}-\frac{1}{2011}\)

\(\Rightarrow\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{2011^2}< \frac{3}{4}-\frac{1}{2011}< \frac{3}{4}\)

\(\Rightarrowđpcm\)

a) \(A=\frac{7}{10}+\frac{7}{10^2}+\frac{7}{10^3}+...\)

\(A=\frac{777...}{1000...}\)

b) 1/2+1/3+1/4+…+1/63=1/2+(1/3+1/4)+(1/5+1/6+…+1/10)+(1/11+1/12+….+1/20)+(1/21+1/22+….1/63).
Ta thấy:
1/3+1/4>1/4+1/4=1/2
1/5+1/6+…+1/10>5/10=1/2
1/11+1/12+….+1/20>10/20=1/2
Thêm.cái 1/2 sắn có là đủ >2 rồi nhể