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7 tháng 3 2017

Ta có: \(A=\dfrac{1}{5^2}+\dfrac{2}{5^3}+...+\dfrac{11}{5^{12}}\)

\(\Rightarrow5A=\dfrac{1}{5}+\dfrac{2}{5^2}+...+\dfrac{11}{5^{11}}\)

\(\Rightarrow5A-A=\dfrac{1}{5}+\dfrac{1}{5^2}+...+\dfrac{1}{5^{11}}-\dfrac{11}{5^{12}}\)

\(\Rightarrow4A=\dfrac{1}{5}+\dfrac{1}{5^2}+...+\dfrac{1}{5^{11}}-\dfrac{11}{5^{12}}\)

\(\Rightarrow20A=1+\dfrac{1}{5}+...+\dfrac{1}{5^{10}}-\dfrac{11}{5^{11}}\)

\(\Rightarrow20A-4A=\left(1+\dfrac{1}{5}+...+\dfrac{1}{5^{10}}-\dfrac{11}{5^{11}}\right)-\left(\dfrac{1}{5}+\dfrac{1}{5^2}+...+\dfrac{1}{5^{11}}-\dfrac{11}{5^{12}}\right)\)

\(\Rightarrow16A=1-\dfrac{12}{5^{11}}+\dfrac{11}{5^{12}}< 1\)

\(\Rightarrow A< \dfrac{1}{16}\)

22 tháng 1 2018
Ta có: A=152+253+...+11512A=152+253+...+11512

⇒5A=15+252+...+11511⇒5A=15+252+...+11511

⇒5A−A=15+152+...+1511−11512⇒5A−A=15+152+...+1511−11512

⇒4A=15+152+...+1511−11512⇒4A=15+152+...+1511−11512

⇒20A=1+15+...+1510−11511⇒20A=1+15+...+1510−11511

⇒20A−4A=(1+15+...+1510−11511)−(15+152+...+1511−11512)⇒20A−4A=(1+15+...+1510−11511)−(15+152+...+1511−11512)

⇒16A=1−12511+11512<1⇒16A=1−12511+11512<1

⇒A<116⇒A<116

leuleu

Bài 2: 

a) Ta có: \(A=\dfrac{4}{n-1}+\dfrac{6}{n-1}-\dfrac{3}{n-1}\)

\(=\dfrac{4+6-3}{n-1}\)

\(=\dfrac{7}{n-1}\)

Để A là số tự nhiên thì \(7⋮n-1\)

\(\Leftrightarrow n-1\inƯ\left(7\right)\)

\(\Leftrightarrow n-1\in\left\{1;7\right\}\)

hay \(n\in\left\{2;8\right\}\)

Vậy: \(n\in\left\{2;8\right\}\)

27 tháng 3 2021

ta có B=2n+9/n+2-3n+5n+1/n+2=4n+10/n+2                                                   Để B là STN thì 4n+10⋮n+2                          4n+8+2⋮n+2                                  4n+8⋮n+2                                                      ⇒2⋮n+2                                     n+2∈Ư(2)                                                        Ư(2)={1;2}                                  Vậy n=0                                                                                  

16 tháng 4 2017

a) Giải:

Ta có: \(4n-5=4\left(n-3\right)+7\)

Để \(\left(4n-5\right)⋮\left(n-3\right)\Leftrightarrow7⋮n-3\)

\(\Rightarrow n-3\inƯ\left(7\right)\)

\(Ư\left(7\right)\in\left\{\pm1;\pm7\right\}\)

Nên ta có bảng sau:

\(n-3\) \(n\)
\(1\) \(4\)
\(-1\) \(2\)
\(-7\) \(-4\)
\(7\) \(10\)

Vậy \(n=\left\{2;4;-4;10\right\}\)

b) Ta có:

\(S=\dfrac{1}{5}+\dfrac{1}{13}+\dfrac{1}{14}+\dfrac{1}{15}+\dfrac{1}{61}+\dfrac{1}{62}+\dfrac{1}{63}\)

\(=\dfrac{1}{5}+\left(\dfrac{1}{13}+\dfrac{1}{14}+\dfrac{1}{15}\right)+\left(\dfrac{1}{61}+\dfrac{1}{62}+\dfrac{1}{63}\right)\)

Nhận xét:

\(\dfrac{1}{13}+\dfrac{1}{14}+\dfrac{1}{15}< \dfrac{1}{12}+\dfrac{1}{12}+\dfrac{1}{12}=\dfrac{1}{4}\)

\(\dfrac{1}{61}+\dfrac{1}{62}+\dfrac{1}{63}< \dfrac{1}{60}+\dfrac{1}{60}+\dfrac{1}{60}=\dfrac{1}{20}\)

\(\Rightarrow S< \dfrac{1}{5}+\dfrac{1}{4}+\dfrac{1}{20}=\dfrac{1}{2}\)

Vậy \(S=\dfrac{1}{5}+\dfrac{1}{13}+\dfrac{1}{14}+\dfrac{1}{15}+\dfrac{1}{61}+\dfrac{1}{62}+\dfrac{1}{63}\) \(< \dfrac{1}{2}\) (Đpcm)

18 tháng 5 2017

a)Ta có:\(\dfrac{1}{b}-\dfrac{1}{b+1}=\dfrac{b+1-b}{b\left(b+1\right)}=\dfrac{1}{b^2+b}< \dfrac{1}{b^2}\)(do b>1)

\(\dfrac{1}{b-1}-\dfrac{1}{b}=\dfrac{b-b+1}{\left(b-1\right)b}=\dfrac{1}{b^2-b}>\dfrac{1}{b^2}\)(do b>1)

b)Áp dụng từ câu a

=>\(\dfrac{1}{2}-\dfrac{1}{3}< \dfrac{1}{2^2}< \dfrac{1}{1}-\dfrac{1}{2}\)

\(\dfrac{1}{3}-\dfrac{1}{4}< \dfrac{1}{3^2}< \dfrac{1}{2}-\dfrac{1}{3}\)

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

\(\dfrac{1}{9}-\dfrac{1}{10}< \dfrac{1}{9^2}< \dfrac{1}{8}-\dfrac{1}{9}\)

=>\(\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{9}-\dfrac{1}{10}< S< 1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{8}-\dfrac{1}{9}\)

=>\(\dfrac{1}{2}-\dfrac{1}{10}< S< 1-\dfrac{1}{9}\)

=>\(\dfrac{2}{5}< S< \dfrac{8}{9}\)(đpcm)

18 tháng 5 2017

thanks bn nhìu

25 tháng 1 2021

\(\left(1\dfrac{3}{4}-\dfrac{4}{6}\right):\left(1\dfrac{1}{5}+2\dfrac{2}{5}+\dfrac{1}{5}\right)< x< 1\dfrac{1}{5}.1\dfrac{1}{4}+3\dfrac{2}{11}:2\dfrac{3}{121}\)

\(\Leftrightarrow\left(\dfrac{7}{4}-\dfrac{4}{6}\right):\left(\dfrac{6}{5}+\dfrac{12}{5}+\dfrac{1}{5}\right)< x< \dfrac{6}{5}.\dfrac{5}{4}+\dfrac{35}{11}:\dfrac{245}{121}\) \(\Leftrightarrow\left(\dfrac{21}{12}-\dfrac{8}{12}\right):\dfrac{19}{5}< x< \dfrac{3}{2}+\dfrac{35}{11}.\dfrac{121}{245}\) \(\Leftrightarrow\dfrac{13}{12}.\dfrac{5}{19}< x< \dfrac{3}{2}+\dfrac{2}{7}\) \(\Leftrightarrow\dfrac{65}{228}< x< \dfrac{21}{14}+\dfrac{4}{14}\) \(\Leftrightarrow\dfrac{65}{228}< x< \dfrac{25}{14}\) \(\Leftrightarrow x=1\)
19 tháng 4 2022

a) \(2\left(\dfrac{2}{3.5}+\dfrac{4}{5.9}+...+\dfrac{16}{n\left(n+16\right)}\right)=\dfrac{16}{25}\)

\(\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{9}+...+\dfrac{1}{n}-\dfrac{1}{n+16}=\dfrac{8}{25}\)

\(\dfrac{1}{3}-\dfrac{1}{n+16}=\dfrac{8}{25}\)

\(\dfrac{n+13}{3\left(n+16\right)}=\dfrac{8}{25}\)

\(24n+384=25n+325\)

\(25n-24n=384-325\)

\(n=59\)

19 tháng 4 2022

b) Sai đề nha

\(\left\{{}\begin{matrix}\dfrac{2018}{2019}< 1\\\dfrac{2019}{2020}< 1\\\dfrac{2020}{2021}< 1\\\dfrac{2021}{2022}< 1\end{matrix}\right.\)

\(\Rightarrow\dfrac{2018}{2019}+\dfrac{2019}{2020}+\dfrac{2020}{2021}+\dfrac{2021}{2022}< 4\)

Câu 1 : Thực hiện phép tính 1 cách hợp lý : a) \(\dfrac{-12}{7}.\dfrac{4}{35}+\dfrac{12}{7}.\dfrac{\left(-31\right)}{35}-\dfrac{2}{7}\) b) \(1+2-3-4+5+5-7-8+...+97+98-99-100\) c) \(A=157.\left(-37\right)-\left(41.53-37.157\right)+51.53\) d) \(B=\left(\dfrac{1}{11}+\dfrac{1}{21}+\dfrac{1}{31}+\dfrac{1}{41}+\dfrac{1}{51}\right)\left(\dfrac{-41}{123}+\dfrac{31}{-186}-\dfrac{-51}{102}\right)\) Câu 2 : a) 12 ( x - 5 ) = 7x - 5 b) Tìm x \(\in\) Z sao cho : ( 2x - 3 ) 2010 = ( 2x - 3 )...
Đọc tiếp

Câu 1 : Thực hiện phép tính 1 cách hợp lý :

a) \(\dfrac{-12}{7}.\dfrac{4}{35}+\dfrac{12}{7}.\dfrac{\left(-31\right)}{35}-\dfrac{2}{7}\)

b) \(1+2-3-4+5+5-7-8+...+97+98-99-100\)

c) \(A=157.\left(-37\right)-\left(41.53-37.157\right)+51.53\)

d) \(B=\left(\dfrac{1}{11}+\dfrac{1}{21}+\dfrac{1}{31}+\dfrac{1}{41}+\dfrac{1}{51}\right)\left(\dfrac{-41}{123}+\dfrac{31}{-186}-\dfrac{-51}{102}\right)\)

Câu 2 :

a) 12 ( x - 5 ) = 7x - 5

b) Tìm x \(\in\) Z sao cho : ( 2x - 3 ) 2010 = ( 2x - 3 ) 2012

Câu 3 :

1) Cho biểu thức S = 1 + 3 + 32 + 33 +...+ 3202 + 3 203

a) chứng tỏ rằng tổng S chia hết cho 52 .

b) Tìm Chữ số tận cùng trong tổng S .

2 ) Cho biểu thức A= \(\dfrac{2n+1}{2n+5}\) . Chứng tỏ rằng với mọi số tự nhiên n thì A là phân số tối giản .

Câu 4 : So sánh tổng gồm 1006 số hạng :

\(S=\dfrac{1}{1.1.3}+\dfrac{1}{2.3.5}+\dfrac{1}{3.5.7}+...+\dfrac{1}{1006.2011.2013}\) với \(\dfrac{2}{3}\)

1
10 tháng 12 2022

Câu 2:

a: \(\Leftrightarrow12x-60=7x-5\)

=>5x=55

=>x=11

b: \(\Leftrightarrow\left(2x-3\right)^{2010}\left[\left(2x-3\right)^2-1\right]=0\)

=>(2x-3)(2x-2)(2x-4)=0

hay \(x\in\left\{\dfrac{3}{2};1;2\right\}\)

23 tháng 3 2017

1,

đặt A= \(\dfrac{1}{2}\)+\(\dfrac{1}{3}\)+\(\dfrac{1}{4}\)+....+\(\dfrac{1}{2016}\)+\(\dfrac{1}{2017}\)

2A=1+\(\dfrac{1}{2}\)+\(\dfrac{1}{3}\)+....+\(\dfrac{1}{2015}\)+\(\dfrac{1}{2016}\)

2A-A=(1+\(\dfrac{1}{2}\)+\(\dfrac{1}{3}\)+....+\(\dfrac{1}{2015}\)+\(\dfrac{1}{2016}\))-(\(\dfrac{1}{2}\)+\(\dfrac{1}{3}\)+\(\dfrac{1}{4}\)+....+\(\dfrac{1}{2016}\)+\(\dfrac{1}{2017}\))

A=1-\(\dfrac{1}{2017}\)

A=\(\dfrac{2016}{2017}\)

vậy A=\(\dfrac{2016}{2017}\)

23 tháng 3 2017

Bạn ơi hnhf như đề bài phải là tính \(^{\dfrac{a}{b}}\)chứ k thì làm sao mak tính đc phần b

bài 3:

\(A=\dfrac{9}{1\cdot2}+\dfrac{9}{2\cdot3}+\dfrac{9}{3\cdot4}+...+\dfrac{9}{2021\cdot2022}\)

\(=9\left(\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+...+\dfrac{1}{2021\cdot2022}\right)\)

\(=9\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{2021}-\dfrac{1}{2022}\right)\)

\(=9\cdot\dfrac{2021}{2022}=\dfrac{6063}{674}\)

Bài 1:

a: \(\left(\dfrac{1}{2}+\dfrac{16}{30}\right)-\left(1+\dfrac{1}{30}\right)\)

\(=\dfrac{15+16}{30}-1-\dfrac{1}{30}\)

\(=\dfrac{30}{30}-1\)

=1-1

=0

b: \(\dfrac{-5}{11}\cdot\dfrac{4}{13}+\dfrac{-5}{11}\cdot\dfrac{9}{13}+3\dfrac{5}{11}\)

\(=-\dfrac{5}{11}\left(\dfrac{4}{13}+\dfrac{9}{13}\right)+3+\dfrac{5}{11}\)

\(=-\dfrac{5}{11}+3+\dfrac{5}{11}\)

=3

c: \(3^2-12\left(\dfrac{3}{4}-\dfrac{2}{3}\right)\)

\(=9-12\cdot\dfrac{9-8}{12}\)

=9-1

=8