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14 tháng 6 2016

Hỏi đáp Toán

14 tháng 6 2016

a) S1 = \(-\frac{1}{1.2}-\frac{1}{2.3}-...-\frac{1}{99.100}\)

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

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

          = \(-\frac{101}{100}\)

12 tháng 9 2019

1 Tính : 

a) \(A=\frac{1}{1.2}-\frac{1}{2.3}-\frac{1}{3.4}-...-\frac{1}{\left(n-1\right).n}\)

\(=\frac{1}{1.2}-\left(\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{\left(n-1\right).n}\right)\)

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

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

\(=\frac{1}{2}-\frac{1}{2}+\frac{1}{n}\)

\(=\frac{1}{n}\)

b) \(B=\frac{4}{1.5}-\frac{4}{5.9}-\frac{4}{9.13}-...-\frac{4}{\left(n-4\right).n}\)

\(=\frac{4}{1.5}-\left(\frac{4}{5.9}+\frac{4}{9.13}+...+\frac{4}{\left(n-4\right).n}\right)\)

\(=\frac{4}{5}-\left(\frac{1}{5.9}+\frac{1}{9.13}+...+\frac{1}{\left(n-4\right).n}\right)\)

\(=\frac{4}{5}-\left(\frac{1}{5}-\frac{1}{9}+\frac{1}{9}-\frac{1}{13}+...+\frac{1}{n-4}-\frac{1}{n}\right)\)

\(=\frac{4}{5}-\left(\frac{1}{5}-\frac{1}{n}\right)\)

\(=\frac{4}{5}-\frac{1}{5}+\frac{1}{n}\)

\(=\frac{3}{5}+\frac{1}{n}\)

c) \(C=1-\frac{1}{2}-\frac{1}{2^2}-\frac{1}{2^3}-...-\frac{1}{2^{10}}\)

\(=1-\left(\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{10}}\right)\)

Đặt \(B=\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{10}}\)

\(\Rightarrow C=1-B\left(1\right)\)

\(\Rightarrow2B=1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^9}\)

Lấy 2B trừ B ta có : 

\(2B-B=\left(1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^9}\right)-\left(\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{10}}\right)\)

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

Thay (2) vào (1) ta có :

\(C=1-\left(1-\frac{1}{10}\right)\)

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

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

Vậy \(C=\frac{1}{10}\)

7 tháng 12 2016

B= 333300

C=328350

D=(n+1) /( n nhân 2)

E=(1/3 trừ 1/3^100):2

7 tháng 12 2016

1)=>3B=1.2.3+2.3.3+3.4.3+...+99.100.3

3B=1.2.3+2.3.(4-1)+3.4.(5-2)+...+99.100.(101-98)

3B=1.2.3+2.3.4-1.2.3+3.4.5-2.3.4+...+99.100.101-98.99.100

3B=99.100.101

=>B=333300

 

31 tháng 5 2016

1.

a.

\(\frac{1}{3}+\left(\frac{1}{5}-\frac{1}{7}\right)\)

\(=\frac{1}{3}+\frac{1}{5}-\frac{1}{7}\)

\(=\frac{35-21-15}{105}\)

\(=-\frac{1}{105}\)

b.

\(\frac{3}{5}-\left(\frac{3}{4}-\frac{1}{2}\right)\)

\(=\frac{3}{5}-\frac{3}{4}+\frac{1}{2}\)

\(=\frac{12-15+10}{20}\)

\(=\frac{7}{20}\)

c.

\(\frac{4}{7}-\left(\frac{2}{5}+\frac{1}{3}\right)\)

\(=\frac{4}{7}-\frac{2}{5}-\frac{1}{3}\)

\(=\frac{60-42-35}{105}\)

\(=-\frac{17}{105}\)

2.

a.

\(S=-\frac{1}{1\times2}-\frac{1}{2\times3}-\frac{1}{3\times4}-...-\frac{1}{\left(n-1\right)\times n}\)

\(S=-\left(\frac{1}{1\times2}+\frac{1}{2\times3}+\frac{1}{3\times4}+...+\frac{1}{\left(n-1\right)\times n}\right)\)

\(S=-\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{n-1}-\frac{1}{n}\right)\)

\(S=-\left(1-\frac{1}{n}\right)\)

\(S=-1+\frac{1}{n}\)

b.

\(S=-\frac{4}{1\times5}-\frac{4}{5\times9}-\frac{4}{9\times13}-...-\frac{4}{\left(n-4\right)\times n}\)

\(S=-\left(\frac{4}{1\times5}+\frac{4}{5\times9}+\frac{4}{9\times13}+...+\frac{4}{\left(n-4\right)\times n}\right)\)

\(S=-\left(1-\frac{1}{5}+\frac{1}{5}-\frac{1}{9}+\frac{1}{9}-\frac{1}{13}+...+\frac{1}{n-4}-\frac{1}{n}\right)\)

\(S=-\left(1-\frac{1}{n}\right)\)

\(S=-1+\frac{1}{n}\)

Chúc bạn học tốtok

 

10 tháng 7 2018

a, \(\left(\frac{1}{2}\right)^x+\left(\frac{1}{2}\right)^{x+4}=17\)

\(\Rightarrow\frac{1}{2^x}+\frac{1}{2^x}\cdot\frac{1}{16}=17\)

\(\Rightarrow\frac{1}{2^x}\left(1+\frac{1}{16}\right)=17\)

\(\Rightarrow\frac{1}{2^x}\cdot\frac{17}{16}=17\)

\(\Rightarrow\frac{1}{2^x}=17:\frac{17}{16}=\frac{1}{16}=\frac{1}{2^4}\)

=> x = 4

b, Ta có: \(\left|x+\frac{1}{1.2}\right|\ge0;\left|x+\frac{1}{2.3}\right|\ge0;....;\left|x+\frac{1}{99.100}\right|\ge0\)

\(\Rightarrow\left|x+\frac{1}{1.2}\right|+\left|x+\frac{1}{2.3}\right|+...+\left|x+\frac{1}{99.100}\right|\ge0\)

\(\Rightarrow100x\ge0\Rightarrow x\ge0\)

\(\Rightarrow x+\frac{1}{1.2}+x+\frac{1}{2.3}+...+x+\frac{1}{99.100}=100x\)

\(\Rightarrow\left(x+x+...+x\right)+\left(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{99.100}\right)=100x\)

\(\Rightarrow99x+1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{99}-\frac{1}{100}=100x\)

\(\Rightarrow100x-99x=1-\frac{1}{100}\)

\(\Rightarrow x=\frac{99}{100}\)

25 tháng 12 2018

M = - ( 4/1.5 + 4/5.9 + ..................+ 4/(n-4).n

M = - (1-1/5 + 1/5 - 1/9 +..............+1/(n-4) - 1/n

M = -(1-1/n)

M = -1 + 1/n 

M = -n + 1

27 tháng 12 2018

xie xie bn nha

21 tháng 9 2015

\(S=\frac{-4}{1.5}-\frac{4}{5.9}-\frac{4}{9.13}-...-\frac{4}{\left(n-4\right).n}\)

\(=-\left(\frac{1}{1}-\frac{1}{5}\right)-\left(\frac{1}{5}-\frac{1}{9}\right)-\left(\frac{1}{9}-\frac{1}{13}\right)-...-\left(\frac{1}{n-4}-\frac{1}{n}\right)\)

\(=-\frac{1}{1}+\frac{1}{5}-\frac{1}{5}+\frac{1}{9}-\frac{1}{9}+\frac{1}{13}-...-\frac{1}{n-4}+\frac{1}{n}\)

\(=-\frac{1}{1}+\frac{1}{n}=\frac{1}{n}+1\)

4 tháng 3 2017

Ta có : \(-\frac{4}{1.5}-\frac{4}{5.9}-\frac{4}{9.13}-.....-\frac{4}{\left(n+4\right)n}\)

\(=-\left(\frac{4}{1.5}+\frac{4}{5.9}+\frac{4}{9.13}+......+\frac{4}{n\left(4+n\right)}\right)\)

\(=-\left(1-\frac{1}{5}+\frac{1}{5}-\frac{1}{9}+......+\frac{1}{n}-\frac{1}{n+4}\right)\)

\(=-\left(1-\frac{1}{n+4}\right)\)

\(=-\left(\frac{n+4}{n+4}-\frac{1}{n+4}\right)\)

\(=-\frac{n+3}{n+4}\)

19 tháng 7 2016

ai kb voi mk ko !!!

mk cho