Bài 1:

\(\frac{4}{3.5}+\frac{4}{5.7}+...+\frac{4}{97.99}\)

\(=2\left(\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{97.99}\right)\)

\(=2\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{97}-\frac{1}{99}\right)\)

\(=2\left(\frac{1}{3}-\frac{1}{99}\right)\)

\(=2.\frac{32}{99}=\frac{64}{99}\)

Bài 2:

a) \(2.4^x-18=110\)

\(\Leftrightarrow2.4^x=128\)

\(\Leftrightarrow4^x=64\)

\(\Leftrightarrow4^x=4^3\Leftrightarrow x=3\)

Vậy x = 3

b) \(\left(\frac{3}{2}x-1\right)^5=1\)

\(\Leftrightarrow\frac{3}{2}x-1=1\)

\(\Leftrightarrow\frac{3}{2}x=2\)

\(\Leftrightarrow x=\frac{4}{3}\)

Vậy \(x=\frac{4}{3}\)

a) 4/3.5 + 3/5.7 + .... + 4/97.99

= 4( 1/3.5 +1/5.7 + ... + 1/97.99 )

= 4 . 1/2 . 2 ( 1/3.5 +1/5.7 + ... + 1/97.99 )

= 4/2 ( 2/3.5 + 2/5.7 + .... + 2/97.99 )

= 2 ( 5-3/3.5 + 7-5/5.7 + ..... + 99-97/97.99 )

= 2 (5/3.5 - 3/3.5 + 7/5.7 - 5/5.7 + .... + 99/97.99 - 97/97.99 )

= 2 ( 1/3 - 1/5 + 1/5 - 1/7 + ..... + 1/97 - 1/99 )

= 2 ( 1/3 -1/99 )

= 2 (33/99 - 1/99 )

= 2 . 32/99

= 32.2/99

=64/99

b) 2. 4^x  - 18 = 110

    2. 4^x = 110 + 18

    2. 4^x = 128 

    4^x = 128 : 2

    4^x = 64

    4^x = 4³

=> x = 3

Ta có : 

\(\frac{4}{3.5}+\frac{4}{5.7}+\frac{4}{7.9}+...+\frac{4}{97.99}\)

\(=\)\(2\left(\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+...+\frac{2}{97.99}\right)\)

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

\(=\)\(2\left(\frac{1}{3}-\frac{1}{99}\right)\)

\(=\)\(2.\frac{32}{99}\)

\(=\)\(\frac{64}{99}\)

Chúc bạn học tốt ~ 

de y \(\frac{1}{3}\)-\(\frac{1}{5}\)=\(\frac{2}{3.5}\)

tuong tu suy ra

M=\(\frac{1}{3}\)-\(\frac{1}{5}\)+\(\frac{1}{5}\)-\(\frac{1}{7}\)+......+\(\frac{1}{97}\)-\(\frac{1}{99}\)

M=\(\frac{1}{3}\)-\(\frac{1}{99}\)

M=\(\frac{32}{99}\)

N ở đâu hả bạn

N là số tự nhiên đó

a. \(A=\frac{4}{3.5}+\frac{4}{5.7}+...+\frac{4}{97.99}\)

\(\Rightarrow A=2\left(\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{97.99}\right)\)

\(\Rightarrow A=2\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{97}-\frac{1}{99}\right)\)

\(\Rightarrow A=2\left(\frac{1}{3}-\frac{1}{99}\right)\)

\(\Rightarrow A=2.\frac{32}{99}\)

\(\Rightarrow A=\frac{64}{99}\)

b. \(M=\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{2020^2}\notin N\)

Ta cần chứng minh M < 1 và M khác 0

Dễ thấy M khác 0 ( 1 )

Ta có : \(M=\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{2020^2}< \frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{2019.2020}\)

\(M< 1-\frac{1}{2020}=\frac{2019}{2020}\)

=> M < 1 ( 2 )

Từ ( 1 ) và ( 2 ) => Đpcm

Bài 1 : 

Ta có :

\(A=\frac{10^{17}+1}{10^{18}+1}=\frac{\left(10^{17}+1\right).10}{\left(10^{18}+1\right).10}=\frac{10^{18}+10}{10^{19}+10}\)

Mà : \(\frac{10^{18}+10}{10^{19}+10}>\frac{10^{18}+1}{10^{19}+1}\)

Mà \(A=\frac{10^{18}+10}{10^{19}+10}\)nên \(A>B\)

Vậy \(A>B\)

Bài 2 :

Ta có :

\(S=\frac{2013}{2014}+\frac{2014}{2015}+\frac{2015}{2016}+\frac{2016}{2013}\)

\(\Rightarrow S=\frac{2014-1}{2014}+\frac{2015-1}{2015}+\frac{2016-1}{2016}+\frac{2013+3}{2013}\)

\(\Rightarrow S=1-\frac{1}{2014}+1-\frac{1}{2015}+1-\frac{1}{2016}+1+\frac{3}{2013}\)

\(\Rightarrow S=4+\frac{3}{2013}-\left(\frac{1}{2014}+\frac{1}{2015}+\frac{1}{2016}\right)\)

Vì \(\frac{1}{2013}>\frac{1}{2014}>\frac{1}{2015}>\frac{1}{2016}\)nên  \(\frac{3}{2013}-\left(\frac{1}{2014}+\frac{1}{2015}+\frac{1}{2016}\right)>0\)

Nên : \(M>4\)

Vậy \(M>4\)

Bài 3 : 

Ta có :

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

Suy ra : \(A< \frac{1}{1.3}+\frac{1}{2.4}+\frac{1}{3.5}+....+\frac{1}{99.101}\)

\(\Rightarrow A< \frac{1}{2}.\left(\frac{2}{1.3}+\frac{2}{2.4}+......+\frac{2}{99.101}\right)\)

\(\Rightarrow A< \frac{1}{2}.\left(1-\frac{1}{3}+\frac{1}{2}-\frac{1}{4}+\frac{1}{3}-......-\frac{1}{101}\right)\)

\(\Rightarrow A< \frac{1}{2}.\left[\left(1+\frac{1}{2}+\frac{1}{3}+.....+\frac{1}{99}\right)-\left(\frac{1}{3}+\frac{1}{4}+......+\frac{1}{101}\right)\right]\)

\(\Rightarrow A< \frac{1}{2}.\left(1+\frac{1}{2}-\frac{1}{100}-\frac{1}{101}\right)\)

\(\Rightarrow A< \frac{1}{2}.\left(1+\frac{1}{2}\right)\)

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

Vậy \(A< \frac{3}{4}\)

Bài 4 :

\(a)A=\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+....+\frac{1}{2015.2017}\)

\(\Rightarrow A=\frac{1}{2}.\left(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+.....+\frac{1}{2015.2017}\right)\)

\(\Rightarrow A=\frac{1}{2}.\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+.....+\frac{1}{2015}-\frac{1}{2017}\right)\)

\(\Rightarrow A=\frac{1}{2}.\left(1-\frac{1}{2017}\right)\)

\(\Rightarrow A=\frac{1}{2}.\frac{2016}{2017}\)

\(\Rightarrow A=\frac{1008}{2017}\)

Vậy \(A=\frac{1008}{2017}\)

\(b)\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+......+\frac{1}{x\left(x+2\right)}=\frac{1008}{2017}\)

\(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+......+\frac{2}{x.\left(x+2\right)}=\frac{2016}{2017}\)

\(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+.....+\frac{1}{x}-\frac{1}{x+2}=\frac{2016}{2017}\)

\(1-\frac{1}{x+2}=\frac{2016}{2017}\)

\(\Rightarrow\frac{1}{x+2}=1-\frac{2016}{2017}\)

\(\Rightarrow\frac{1}{x+2}=\frac{1}{2017}\)

\(\Rightarrow x+2=2017\)

\(\Rightarrow x=2017-2=2015\)

Vậy \(x=2015\)

mk làm đc,vs điều kiện bạn phải tick cho mk khi mk giải xog, ko đc tick cho ai

4/58678

tich minh cho minh len thu 8 tren bang sep hang cai

ta co : 65%=0,65

goi A= 4.(1/3.5+1/5.7+1/7.9+............+1/97.99)

2A=4.( 2/3.5+2/5.7+2/7.9+...............+2/97.99)

2A=4.(1/3-1/5+1/5-1/7+1/7-1/9+...+1/97-1/99)

2A=4.(1/3-1/99)

2A=4.(33/=99+1/99)

2A=4.34/99

2A=136/99

A=136/99:2

A=68/99=0,69=0,68

Vi A=0,68 > 0,65

=> A > 65%

a) \(\frac{5}{1.3}+\frac{5}{3.5}+\frac{5}{5.7}+...+\frac{5}{99.101}\)

\(=5.\left(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{99.101}\right)\)

\(=5.\left(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{99.101}\right):2\)

\(=5.\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{99}-\frac{1}{101}\right):2\)

\(=5.\left(1-\frac{1}{101}\right):2=5.\frac{100}{101}:2=\frac{500}{101}.\frac{1}{2}\)\(=\frac{250}{101}\)

b) \(\frac{1}{18}+\frac{1}{54}+\frac{1}{108}+...+\frac{1}{990}\)

\(=\frac{1}{3.6}+\frac{1}{6.9}+\frac{1}{9.12}+...+\frac{1}{30.33}\)

\(=3\left(\frac{1}{3.6}+\frac{1}{6.9}+...+\frac{1}{30.33}\right)\)\(.\frac{1}{3}\)

\(=(\frac{3}{3.6}+\frac{3}{6.9}+...+\frac{3}{30.33}).\frac{1}{3}\)

\(=(\frac{1}{3}-\frac{1}{6}+\frac{1}{6}-\frac{1}{9}+...+\frac{1}{30}-\frac{1}{33}).\frac{1}{3}\)

\(=(\frac{1}{3}-\frac{1}{33}).\frac{1}{3}=\frac{10}{33}.\frac{1}{3}=\frac{10}{99}\)

câu c bạn có thể viết rõ được ko

Làm nốt câu c hộ các bạn:

\(\frac{\frac{3}{41}-\frac{12}{47}+\frac{27}{53}}{\frac{4}{41}-\frac{16}{47}+\frac{36}{53}}=\frac{3\left(\frac{1}{41}-\frac{4}{47}+\frac{9}{53}\right)}{4\left(\frac{1}{41}-\frac{4}{47}+\frac{9}{53}\right)}=\frac{3}{4}\)