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Bài 2:

n) Ta có: \(N=\dfrac{4}{2\cdot4}+\dfrac{4}{4\cdot6}+\dfrac{4}{6\cdot8}+...+\dfrac{4}{2014\cdot2016}\)

\(=2\left(\dfrac{2}{2\cdot4}+\dfrac{2}{4\cdot6}+\dfrac{2}{6\cdot8}+...+\dfrac{2}{2014\cdot2016}\right)\)

\(=2\left(\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{8}+...+\dfrac{1}{2014}-\dfrac{1}{2016}\right)\)

\(=2\cdot\left(\dfrac{1}{2}-\dfrac{1}{2016}\right)\)

\(=2\cdot\dfrac{1007}{2016}=\dfrac{1007}{1008}\)

o) Ta có: \(\dfrac{1}{18}+\dfrac{1}{54}+\dfrac{1}{108}+...+\dfrac{1}{990}\)

\(=\dfrac{1}{3\cdot6}+\dfrac{1}{6\cdot9}+\dfrac{1}{9\cdot12}+...+\dfrac{1}{30\cdot33}\)

\(=\dfrac{1}{3}\left(\dfrac{3}{3\cdot6}+\dfrac{3}{6\cdot9}+\dfrac{3}{9\cdot12}+...+\dfrac{3}{30\cdot33}\right)\)

\(=\dfrac{1}{3}\left(\dfrac{1}{3}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{12}+...+\dfrac{1}{30}-\dfrac{1}{33}\right)\)

\(=\dfrac{1}{3}\left(\dfrac{1}{3}-\dfrac{1}{33}\right)\)

\(=\dfrac{1}{3}\cdot\dfrac{10}{33}=\dfrac{10}{99}\)

a) Ta có: \(\left(6\dfrac{4}{9}+3\dfrac{7}{11}\right)-4\dfrac{4}{9}\)

\(=\dfrac{58}{9}+\dfrac{40}{11}-\dfrac{40}{9}\)

\(=2+\dfrac{40}{11}=\dfrac{62}{11}\)

Bài 2:

b) Ta có: \(10\dfrac{1}{5}-5\dfrac{1}{2}\cdot\dfrac{60}{11}+3:15\%\)

\(=\dfrac{51}{5}-\dfrac{11}{2}\cdot\dfrac{60}{11}+3:\dfrac{3}{20}\)

\(=\dfrac{51}{5}-30+20\)

\(=\dfrac{51}{5}-10=\dfrac{1}{5}\)

c) Ta có: \(\dfrac{2}{3\cdot5}+\dfrac{2}{5\cdot7}+\dfrac{2}{7\cdot9}+...+\dfrac{2}{97\cdot99}\)

\(=\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{9}+...+\dfrac{1}{97}-\dfrac{1}{99}\)

\(=\dfrac{1}{3}-\dfrac{1}{99}\)

\(=\dfrac{32}{99}\)

22 tháng 6 2016

Đặt Tử số là A ta có

\(2A=2+2^2+2^3+2^4+..+2^{2016}\)

\(A=2A-A=2^{2016}-1\)

\(\Rightarrow S=\frac{2^{2016}-1}{1-2^{2016}}=\frac{-\left(1-2^{2016}\right)}{1-2^{2016}}=-1\)

22 tháng 6 2016

\(S=\frac{1+2+2^2+2^3+...+2^{2015}}{1-2^{2016}}\)

\(\Rightarrow2S=\frac{2\left(1+2+2^2+2^3+...+2^{2015}\right)}{1-2^{2016}}\)

\(\Rightarrow2S=\frac{2+2^2+2^3+2^4+...+2^{2016}}{1-2^{2016}}\)

\(\Rightarrow2S-S=\frac{2+2^2+2^3+2^4+...+2^{2016}}{1-2^{2016}}-\frac{1+2+2^2+2^3+...+2^{2015}}{1-2^{2016}}\)

\(\Rightarrow S=\frac{2^{2016}-1}{1-2^{2016}}=-1\)

Khi nào có bài khó thì cứ đăng lên nhé, mình sẽ giúp ^.^

24 tháng 1 2017

em lớp 5, chị

5 tháng 11 2021

\(\Rightarrow5^2\cdot5^{x+3}=5^6\\ \Rightarrow5^{x+3}=5^4\\ \Rightarrow x+3=4\\ \Rightarrow x=1\)

5 tháng 11 2021

\(25.5^{x+3}=5^6\)

\(\Rightarrow5^2.5^{x+3}=5^6\)

\(\Rightarrow5^{x+3}=5^6:5^2\)

\(\Rightarrow5^{x+3}=5^4\)

\(\Rightarrow x+3=4\)

\(\Rightarrow x\)       \(=4-3\)

\(\Rightarrow x\)       \(=1\)

29 tháng 10 2017

Ta có :

\(M=4^2+4^4+4^6+...+4^{58}+4^{60}\)

\(=\left(4^2+4^4\right)+\left(4^6+4^8\right)+...+\left(4^{58}+4^{60}\right)\)

\(=4^2\left(1+4^2\right)+4^6\left(1+4^2\right)+...+4^{58}\left(1+4^2\right)\)

\(=\left(1+16\right)\left(4^2+4^6+...+4^{58}\right)\)

\(=\left(4^2+4^6+...+4^{58}\right).17⋮17\)

\(\Rightarrow M⋮17\)(đpcm)

Chúc bn hc giỏi!

29 tháng 10 2017

M = 4^2 + 4^4 + 4^6 + 4^8 +... +4^58+4^60

= (4^2+4^4)+...+(4^58+4^60)

=4^2.(1+4^2)+....+4^58.(1+4^2)

=4^2.17+....+4^58.17

= 17.(4^2+...+4^58)

Chia hết cho 17 

ĐPCM

có j không hiểu ib hỏi mình nhé

12 tháng 12 2021

\(\Rightarrow\)x+2\(\in\)Ư(9)

Ư(9)={\(\pm1\)\(\pm3\)\(\pm9\)}

\(\Rightarrow\)x+2\(\in\left\{\pm1;\pm3;\pm9\right\}\)

\(\Rightarrow\)x\(\in\left\{\pm1;-3;-5;-11;7\right\}\)

Vậy x\(\in\left\{\pm1;-3;-5;-11;7\right\}\)

4 tháng 9 2014

gõ nhầm nhé X+Y >=4
                    X+Y <=0

n.(n+2).(n+7)

=n.n.(2+7)

=2n.9

Vì \(9⋮3\Rightarrow2n.9⋮3\)

CHÚC BẠN HỌC TỐT !!!

26 tháng 5 2017

\(B=\left(\frac{1}{4}-1\right).\left(\frac{1}{9}-1\right)...\left(\frac{1}{100}-1\right)\)

\(B=\frac{-3}{4}.\frac{-8}{9}...\frac{-99}{100}\)

\(B=-\left(\frac{3}{4}.\frac{8}{9}...\frac{99}{100}\right)\)

\(B=-\left(\frac{1.3}{2.2}.\frac{2.4}{3.3}...\frac{9.11}{10.10}\right)\)

\(B=-\left(\frac{1.2...9}{2.3...10}.\frac{3.4...11}{2.3...10}\right)\)

\(B=-\left(\frac{1}{10}.\frac{11}{2}\right)\)

\(B=\frac{-11}{20}< \frac{-11}{21}\)

Vậy \(B< \frac{-11}{21}\)