Bài 1 :
a) =) \(\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{99}-\frac{1}{101}\)= \(1-\frac{1}{101}=\frac{100}{101}\)
b) =) \(\frac{5}{2}.\left(\frac{2}{1.3}+\frac{2}{3.5}+...+\frac{2}{99.101}\right)\)
=) \(\frac{5}{2}.\frac{100}{101}=\frac{250}{101}\)( theo phần a)
Bài 2 :
-Gọi d là UCLN \(\left(2n+1;3n+2\right)\)( d \(\in N\)* )
(=) \(2n+1⋮d\left(=\right)3.\left(2n+1\right)⋮d\)
(=) \(6n+3⋮d\)
và \(3n+2⋮d\left(=\right)2.\left(3n+2\right)⋮d\)
(=) \(6n+4⋮d\)
(=) \(\left(6n+4\right)-\left(6n+3\right)⋮d\)
(=) \(6n+4-6n-3⋮d\)
(=) \(1⋮d\left(=\right)d\in UC\left(1\right)\)(=) d = { 1;-1}
Vì d là UCLN\(\left(2n+1;3n+2\right)\)(=) \(d=1\)(=) \(\frac{2n+1}{3n+2}\)là phân số tối giản ( đpcm )
Bài 3 :
-Để A \(\in Z\)(=) \(n+2⋮n-5\)
Vì \(n-5⋮n-5\)
(=) \(\left(n+2\right)-\left(n-5\right)⋮n-5\)
(=) \(n+2-n+5⋮n-5\)
(=) \(7⋮n-5\)(=) \(n-5\in UC\left(7\right)\)= { 1;-1;7;-7}
(=) n = { 6;4;12;-2}
Vậy n = {6;4;12;-2} thì A \(\in Z\)
Bài 4:
A = \(10101.\left(\frac{5}{111111}+\frac{5}{222222}-\frac{4}{3.7.11.13.37}\right)\)
= \(10101.\left(\frac{5}{111111}+\frac{5}{222222}-\frac{4}{111111}\right)\)
= \(10101.\left(\frac{1}{111111}+\frac{5}{222222}\right)\)= \(10101.\left(\frac{2}{222222}+\frac{5}{222222}\right)\)
= \(10101.\frac{7}{222222}\)( không cần rút gọn \(\frac{7}{222222}\))
= \(\frac{7}{22}\)

Ôi, trang wed không tự nhận diện được công thức latex. Mình đăng lại bài giải:

a) Ta có

\(4T=\frac{4}{1+\sqrt{5}}+\frac{4}{\sqrt{5}+\sqrt{9}}+...+\frac{4}{\sqrt{2013}+\sqrt{2017}}\)

\(=\frac{\left(\sqrt{5}+1\right)\left(\sqrt{5}-1\right)}{\sqrt{5}+1}+...+\frac{\left(\sqrt{2017}+\sqrt{2013}\right)\left(\sqrt{2017}-\sqrt{2013}\right)}{\sqrt{2017}+\sqrt{2013}}\)

\(=\sqrt{5}-1+\sqrt{9}-\sqrt{5}+\sqrt{13}-\sqrt{9}+...+\sqrt{2017}-\sqrt{2013}\)

\(=\sqrt{2017}-1\)

\(\Rightarrow T=\frac{\sqrt{2017}-1}{4}\)

b) Ta có

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

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

\(=\frac{\sqrt{2}-\sqrt{1}}{\sqrt{2}\sqrt{1}}=\frac{1}{\sqrt{1}}-\frac{1}{\sqrt{2}}\)

Tương tự ta có

\(\frac{1}{3\sqrt{2}+2\sqrt{3}}=\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{3}}\)

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

\(\frac{1}{100\sqrt{99}+99\sqrt{100}}=\frac{1}{\sqrt{99}}-\frac{1}{\sqrt{100}}\)

Suy ra

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

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

a)\[\begin{array}{l}
4T = \frac{4}{{1 + \sqrt 5 }} + \frac{4}{{\sqrt 5  + \sqrt 9 }} + ... + \frac{4}{{\sqrt {2013}  + \sqrt {2017} }}\\
 = \frac{{(\sqrt 5  + 1)(\sqrt 5  - 1)}}{{1 + \sqrt 5 }} + ... + \frac{{(\sqrt {2017}  + \sqrt {2013} )(\sqrt {2017}  - \sqrt {2013} )}}{{\sqrt {2013}  + \sqrt {2017} }}\\
 = \sqrt 5  - 1 + \sqrt 9  - \sqrt 5  + ... + \sqrt {2017}  - \sqrt {2013} \\
 = 1 + \sqrt 5  - \sqrt 5  + \sqrt 9  - \sqrt 9  + ... + \sqrt {2013}  - \sqrt {2013}  + \sqrt {2017} \\
 = 1 + \sqrt {2017} \\
 \Rightarrow T = \frac{{1 + \sqrt {2017} }}{4}
\end{array}\]

Cảm ơn Tran Tu, kết bn đê

OK. Tối nhớ giải hộ mik nha

Mik hứa sẽ lik-e cho bạn

mình ko biết

\(\left(1\right)\frac{1}{2}-\frac{1}{3}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{98}-\frac{1}{99}>\frac{1}{5}\)

\(=\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{4}-\frac{1}{5}\right)+\left(\frac{1}{6}-\frac{1}{7}\right)+\left(\frac{1}{8}-\frac{1}{9}\right)+...+\left(\frac{1}{98}-\frac{1}{99}\right)\)

\(=\frac{13}{60}+\left(\frac{1}{6}-\frac{1}{7}\right)+\left(\frac{1}{8}-\frac{1}{9}\right)+...\left(\frac{1}{98}-\frac{1}{99}\right)\)

Ta thấy        \(\frac{13}{60}>\frac{12}{60}=\frac{1}{5}\)

          \(\frac{1}{6}-\frac{1}{7}>0\)

          \(\frac{1}{8}-\frac{1}{9}>0\)

\(...\)\(\frac{1}{98}-\frac{1}{99}>0\)

\(\Rightarrow\frac{1}{2}-\frac{1}{3}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{98}-\frac{1}{99}>\frac{1}{5}\)

\(\left(2\right)\frac{1}{2}-\frac{1}{3}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{98}-\frac{1}{99}< \frac{2}{5}\)

\(=\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{4}-\frac{1}{5}+\frac{1}{6}\right)-\left(\frac{1}{7}-\frac{1}{8}\right)-\left(\frac{1}{9}-\frac{1}{10}\right)-...-\left(\frac{1}{97}-\frac{1}{98}\right)-\frac{1}{99}\)

\(=\frac{23}{60}-\left(\frac{1}{7}-\frac{1}{8}\right)-\left(\frac{1}{9}-\frac{1}{10}\right)-...-\left(\frac{1}{97}-\frac{1}{98}\right)-\frac{1}{99}\)

Ta thấy   \(\frac{23}{60}< \frac{24}{60}=\frac{2}{5}\)

      \(\frac{1}{7}-\frac{1}{8}>0\)

     \(\frac{1}{9}-\frac{1}{10}>0\)

\(...\frac{1}{97}-\frac{1}{98}>0\)

                \(\frac{1}{99}>0\)

\(\Rightarrow\frac{1}{2}-\frac{1}{3}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{98}-\frac{1}{99}< \frac{2}{5}\)

2S=\(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+\frac{2}{9.11}+\frac{2}{11.13}+\frac{2}{13.15}\)

      = \(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{13}-\frac{1}{15}\)

      =\(1-\frac{1}{15}=\frac{14}{15}\)

\(\Rightarrow S=\frac{7}{15}\)

a. Ta có:A= 1/1.3+1/3.5+1/5.7+1/7.9+1/9.11+1/11.13+1/13.15

A=1/2(1/1.3+1/3.5+1/5.7+1/7.9+1/9.11+1/11.13+1/13.15)

A=1/2(1/1-1/3+1/3-1/5+1/5-1/7+1/7-1/9+1/9-1/11+1/11-1/13+1/13-1/15)

A=2(1-1/15)

A=1/2.14/15

A=7/15

phần b nè

pt \(\Rightarrow90-6ab=3a\)\(\Leftrightarrow3a\left(b+2\right)=90\)vì b>0 \(\Leftrightarrow a=\frac{30}{b+2}\)mà a,b \(\inℕ^∗\)

\(\Rightarrow\)b+2\(\inƯ\left(30\right)\)MÀb\(\inℕ^∗\)\(b+2\in\left\{3;5;6;10;15;30\right\}\)khi đó tìm đc b \(\rightarrow\)thau vào tìm a . nhớ thử lại vào pt ban đầu nhé 

k cho mk nha mn ^.^

\(A=\frac{2015+2013+2011+...+5+3+1}{2015-2013+2011-2009+...+7-5+3-1}\)

Ta có : 2015 + 2013 + 2011 + ... + 5 + 3 + 1  

= [(2015 - 1) : 2 + 1].(2015 + 1) : 2

= 1008.2016 : 2 = 1016064

Lại có :  2015 - 2013 + 2011 - 2009 + ... + 7 - 5 + 3 - 1 (1008 số hạng

= (2015 - 2013) + (2011 - 2009) + ... + (7 - 5) + (3 - 1) (504 cặp)

= 2 + 2 + ... + 2 + 2 (504 số hạng 2)

= 2 x 504 = 1008 

Khi đó A = \(\frac{1016064}{1008}=1008\)

b) tTa có : B = \(\frac{1+\frac{1}{3}+\frac{1}{5}+...+\frac{1}{97}+\frac{1}{99}}{\frac{1}{1.99}+\frac{1}{3.97}+\frac{1}{5.95}+...+\frac{1}{97.3}+\frac{1}{99.1}}\)

=> \(\frac{B}{100}\) = \(\frac{1+\frac{1}{3}+\frac{1}{5}+...+\frac{1}{97}+\frac{1}{99}}{\frac{100}{1.99}+\frac{100}{3.97}+\frac{100}{5.95}+...+\frac{100}{97.3}+\frac{100}{99.1}}\)

\(=\frac{1+\frac{1}{3}+\frac{1}{5}+...+\frac{1}{97}+\frac{1}{99}}{1+\frac{1}{99}+\frac{1}{3}+\frac{1}{97}+\frac{1}{5}+\frac{1}{95}+..+\frac{1}{97}+\frac{1}{3}+\frac{1}{99}+1}=\frac{1+\frac{1}{3}+\frac{1}{5}+...+\frac{1}{97}+\frac{1}{99}}{2\left(1+\frac{1}{3}+\frac{1}{5}+...+\frac{1}{97}+\frac{1}{99}\right)}=\frac{1}{2}\)

Khi đó : B/100 = 1/2

=> B = 50 

Vậy B = 50

giỏi ghê vậy Hân

chứ sao

2/

S = 2 + 2² + 2³ +...+ 2⁹⁹

= (2+2²+2³) +...+ (2⁹⁷+2⁹⁸+2⁹⁹)

= 2(1+2+2²)+...+2⁹⁷(1+2+2²)

= 2.7 +...+ 2⁹⁷.7

= 7(2+...+2⁹⁷) chia hết cho 7

S = 2+2²+2³+...+2⁹⁹

= (2+2²+2³+2⁴+2⁵)+...+(2⁹⁵+2⁹⁶+2⁹⁷+2⁹⁸+2⁹⁹)

= 2(1+2+2²+2³+2⁴)+...+2⁹⁵(1+2+2²+2³+2⁴)

= 2.31+...+2⁹⁵.31

= 31(2+...+2⁹⁵) chia hết cho 31

3/

A = 1+5+5²+....+5¹⁰⁰ (1)

5A = 5+5²+5³+...+5101 (2)

Lấy (2) - (1) ta được

4A = 5¹⁰¹ - 1

A = \(\frac{5^{101}-1}{4}\)

4/

Đặt A là tên của biểu thức trên

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

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

........

\(\frac{1}{8^2}< \frac{1}{7.8}=\frac{1}{7}-\frac{1}{8}\)

\(\Rightarrow A< \frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{7}-\frac{1}{8}=\frac{1}{1}-\frac{1}{8}=\frac{7}{8}< 1\)

Vậy...

5/

a, Gọi UCLN(n+1,2n+3) = d

Ta có : n+1 chia hết cho d => 2(n+1) chia hết cho d => 2n+2 chia hết cho d

           2n+3 chia hết cho d

=> 2n+2 - (2n+3) chia hết cho d

=> -1 chia hết cho d => d = {-1;1}

Vậy...

b, Gọi UCLN(2n+3,4n+8) = d

Ta có: 2n+3 chia hết cho d => 2(2n+3) chia hết cho d => 4n+6 chia hết cho d

          4n+8 chia hết cho d 

=> 4n+6 - (4n+8) chia hết cho d

=> -2 chia hết cho d => d = {1;-1;2;-2}

Mà 2n+3 lẻ => d lẻ => d khác 2;-2 => d = {1;-1}

Vậy...

6/

a,Vì B > 1

\(\Rightarrow B=\frac{2016^{2016}}{2016^{2016}-3}>\frac{2016^{2016}+2}{2016^{2016}-3+2}=\frac{2016^{2016}+2}{2016^{2016}-1}=A\) 

Vậy A < B

b, C = \(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+...+\frac{1}{9900}\)

\(=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{99.100}\)

\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}\)

\(=1-\frac{1}{100}=\frac{99}{100}\)