\(\forall n\in N;n\ne0\) Ta có : \(\frac{1}{n}-\frac{1}{n+1}-\frac{1}{n\left(n+1\right)}=\frac{\left(n+1\right)-n-1}{n\left(n+1\right)}=\frac{0}{\left(n+1\right)n}=0\)

\(\Rightarrow\sqrt{1+\frac{1}{n^2}+\frac{1}{\left(n+1\right)^2}}=\sqrt{1+\frac{1}{n^2}+\frac{1}{\left(n+1\right)^2}+2\left[\frac{1}{n}-\frac{1}{n+1}-\frac{1}{n\left(n+1\right)}\right]}\)

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

Áp dụng ta được :

\(A=1+\frac{1}{2}-\frac{1}{3}+1+\frac{1}{3}-\frac{1}{4}+.....+1+\frac{1}{1100}-\frac{1}{1101}\)

\(=1099+\left(\frac{1}{2}+\frac{1}{3}+....+\frac{1}{1100}\right)-\left(\frac{1}{3}+\frac{1}{4}+....+\frac{1}{1101}\right)\)

\(=1099+\frac{1}{2}-\frac{1}{1101}=\frac{2421097}{2202}\)

Ta có : \(\frac{x-1}{2}=\frac{x+1}{3}\)

<=> \(3\left(x-1\right)=2\left(x+1\right)\)

<=> \(3x-3=2x+2\)

<=> \(3x-2x=2+3\)

<=> x = 5 

a, \(\frac{x-1}{2}=\frac{x+1}{3}\)

=> (x-1)3 = 2(x+1)

=> 3x - 3 = 2x + 2

=> 3x - 2x = 2 + 3

=> x = 5

b, \(A=\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{99.100}\)

\(A=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{99}-\frac{1}{100}\)

\(A=1-\frac{1}{100}=\frac{99}{100}< 1\) (ĐPCM)

= \(\frac{17}{8}:\frac{25}{14}-\left(15-\frac{40}{3}\right):\frac{25}{6}\)

= \(\frac{17}{8}.\frac{14}{25}-\left(\frac{45}{3}-\frac{40}{3}\right).\frac{6}{25}\)

= \(\frac{119}{100}-\frac{5}{3}.\frac{6}{25}\)  =  \(\frac{119}{100}-\frac{2}{5}\)

=  \(\frac{119}{100}-\frac{40}{100}=\frac{79}{100}\)

Chúc bạn Hk tốt!!!!!

=\(\frac{79}{100}\)

Sorry bạn nha , mình bấm nhầm nút

\(A=\frac{5}{4}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{100^2}\)

\(A< \frac{5}{4}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{99.100}\)

\(A< \frac{5}{4}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}\)

\(A< \frac{5}{4}+\frac{1}{2}-\frac{1}{100}< \frac{5}{4}+\frac{1}{2}=\frac{7}{4}\)

\(\Rightarrow\)\(A< \frac{7}{4}\)

Vậy , \(\frac{5}{4}< A< \frac{7}{4}\left(ĐPCM\right)\)

BÀI KHÓ CỦA TRƯỜNG MÌNH ĐÓ THI HK2

GIÚP MÌNH NHÉ!!!!!!THANKS!!!!!!

A=\(\left(\frac{1}{2^2}-1\right)\)\(\left(\frac{1}{3^2}-1\right)\)\(\left(\frac{1}{4^2}-1\right)\)...\(\left(\frac{1}{98^2}-1\right)\)\(\left(\frac{1}{99^2}-1\right)\)

Do tích A có(99-2)+1=98 thừa số nguyên âm nên tích A dương

A=\(\frac{3}{4}\).\(\frac{8}{9}\).\(\frac{15}{16}\)...\(\frac{97.99}{98^2}\).\(\frac{98.100}{99^2}\)=\(\frac{1.2.3.4.5...97.98.99.100}{2^2.3^3.4^2...98^2.99^2}\)

=\(\frac{1.2.3.4...98}{2.3.4...98.99}.\frac{3.4.5...99.100}{2.3.4...98.99}=\frac{1}{99}.\frac{100}{2}=\frac{50}{99}\)

Đề gõ sai, xin sửa lại:
Chứng minh:

\({1 \over {11}^2} + {1 \over {12}^2} + {1 \over {13}^2} + {1 \over {14}^2} + ... + {1 \over {100}^2}<{1 \over {10}}\)

Cảm ơn

Đặt biểu thức là A     ta có:

1/11^2 < 1/10.11 = 1/10 - 1/11

1/12^2 < 1/11.12 = 1/11 - 1/12

1/13^2 < 1/12.13 = 1/12 - 1/13

. . . . . . . . . 

1/100^2 < 1/99.100 = 1/99 - 1/100

 => A < 1/10 - 1/11 + 1/11 - 1/12 + 1/12 - 1/13 + . . . .+ 1/99 - 1/100

  => A < 1/10 -  1/100

 => A < 1/10

                Bạn nhớ k cho mình nha

Toán vui mỗi tuần có lời giải rồi bạn ơi

Vào đó mà đọc.

\(\frac{20}{11}=2-\frac{2}{n}\Rightarrow\frac{2}{n}=2-\frac{20}{11}=\frac{2}{11}\Rightarrow n=11\)

\(A=1+\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{99}}+\frac{1}{2^{100}}+\frac{1}{2^{100}}\)

=>\(2A=2+1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{98}}+\frac{1}{2^{99}}+\frac{1}{2^{99}}\)

=>\(A=2A-A=2+\frac{1}{2^{99}}+\frac{1}{2^{99}}\)

\(A=2+\frac{1}{2^{98}}\)

Vậy: \(A=2+\frac{1}{2^{98}}\)

Gọi \(B=1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{100}}\)

\(\Rightarrow2B=2+1+\frac{1}{2}+...+\frac{1}{2^{99}}\)

\(\Rightarrow2B-B=\left(2+1+\frac{1}{2}+...+\frac{1}{2^{99}}\right)-\left(1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{100}}\right)\)

\(\Rightarrow B=2-\frac{1}{2^{100}}\)

\(\Rightarrow A=2\)

Vậy A = 2