\(\dfrac{1}{3.4}\)+
K
Khách

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16 tháng 2 2023

\(\dfrac{1}{3.4}+\dfrac{1}{4.5}+\dfrac{1}{5.6}+...+\dfrac{1}{n.\left(n+1\right)}=\dfrac{3}{10}\)

\(\Rightarrow\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}+...+\dfrac{1}{n}-\dfrac{1}{n+1}=\dfrac{3}{10}\)

\(\Rightarrow\dfrac{1}{3}-\dfrac{1}{n+1}=\dfrac{3}{10}\)

\(\Rightarrow\dfrac{1}{n+1}=\dfrac{1}{30}\)

\(\Rightarrow n+1=30\)

\(\Rightarrow n=29\)

Vậy n = 29.

9 tháng 5 2017

a) Để phân số \(\dfrac{3}{n-2}\) là số nguyên thì n - 2 \(⋮\) 3

\(\Rightarrow\) n - 2 \(\in\) Ư(3)

\(\Rightarrow\) n - 2 \(\in\){3; -3; 1;-1}

n \(\in\){5; -1; 3; 2}

9 tháng 5 2017

c) \(\dfrac{1}{3.4}+\dfrac{1}{4.5}+\dfrac{1}{5.6}+......+\dfrac{1}{28.29}\)

\(=\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}+.....+\dfrac{1}{29}-\dfrac{1}{30}\)

\(=\dfrac{1}{3}-\dfrac{1}{30}\)

\(=\dfrac{10}{30}-\dfrac{1}{30}\)

\(=\dfrac{9}{30}\)

=\(\dfrac{3}{10}\)

22 tháng 3 2017

1,

B=\(\dfrac{1}{2}\)+\(\dfrac{1}{2^2}\)+\(\dfrac{1}{2^3}\)+\(\dfrac{1}{2^4}\)+.........+\(\dfrac{1}{2^{2017}}\)

2B=1+\(\dfrac{1}{2}\)+\(\dfrac{1}{2^2}\)+\(\dfrac{1}{2^3}\)+.......+\(\dfrac{1}{2^{2016}}\)

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

B=1-\(\dfrac{1}{2^{2017}}\)

Vậy B=1-\(\dfrac{1}{2^{2017}}\)

8 tháng 9 2017

Bài 1:

a, \(\left(x-2\right)^2=9\)

\(\Rightarrow x-2\in\left\{-3;3\right\}\Rightarrow x\in\left\{-1;5\right\}\)

b, \(\left(3x-1\right)^3=-8\)

\(\Rightarrow3x-1=-2\Rightarrow3x=-1\)

\(\Rightarrow x=-\dfrac{1}{3}\)

c, \(\left(x+\dfrac{1}{2}\right)^2=\dfrac{1}{16}\)

\(\Rightarrow x+\dfrac{1}{2}\in\left\{-\dfrac{1}{4};\dfrac{1}{4}\right\}\)

\(\Rightarrow x\in\left\{-\dfrac{3}{4};-\dfrac{1}{4}\right\}\)

d, \(\left(\dfrac{2}{3}\right)^x=\dfrac{4}{9}\)

\(\Rightarrow\left(\dfrac{2}{3}\right)^x=\left(\dfrac{2}{3}\right)^2\)

\(\dfrac{2}{3}\ne\pm1;\dfrac{2}{3}\ne0\) nên \(x=2\)

e, \(\left(\dfrac{1}{2}\right)^{x-1}=\dfrac{1}{16}\)

\(\Rightarrow\left(\dfrac{1}{2}\right)^{x-1}=\left(\dfrac{1}{2}\right)^4\)

\(\dfrac{1}{2}\ne\pm1;\dfrac{1}{2}\ne0\) nên \(x-1=4\Rightarrow x=5\)

f, \(\left(\dfrac{1}{2}\right)^{2x-1}=8\) \(\Rightarrow\left(\dfrac{1}{2}\right)^{2x-1}=\left(\dfrac{1}{2}\right)^{-3}\)\(\dfrac{1}{2}\ne\pm1;\dfrac{1}{2}\ne0\) nên \(2x-1=-3\) \(\Rightarrow2x=-2\Rightarrow x=-1\) Chúc bạn học tốt!!!
19 tháng 3 2018

\(\dfrac{1}{2.3}\) + \(\dfrac{1}{3.4}\) + \(\dfrac{1}{4.5}\) + \(\dfrac{1}{5.6}\) + \(\dfrac{1}{6.7}\) + \(\dfrac{1}{7.8}\)

= \(\dfrac{1}{2}\) - \(\dfrac{1}{3}\) + \(\dfrac{1}{3}\) - \(\dfrac{1}{4}\) + \(\dfrac{1}{4}\) - \(\dfrac{1}{5}\) + \(\dfrac{1}{5}\) - \(\dfrac{1}{6}\) + \(\dfrac{1}{6}\) - \(\dfrac{1}{7}\) + \(\dfrac{1}{7}\) - \(\dfrac{1}{8}\)

= \(\dfrac{1}{2}\) + \(\dfrac{1}{8}\) MSC: 8

= \(\dfrac{4}{8}\) + \(\dfrac{1}{8}\)

= \(\dfrac{5}{8}\)

22 tháng 3 2018

\(\dfrac{1}{2.3}+\dfrac{1}{3.4}+\dfrac{1}{4.5}+\dfrac{1}{5.6}+\dfrac{1}{6.7}+\dfrac{1}{7.8}\)

= \(\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{8}\)

= \(\dfrac{1}{2}-\dfrac{1}{8}\)

=\(\dfrac{4}{8}-\dfrac{1}{8}\)

=\(\dfrac{3}{8}\)

6 tháng 4 2017

a) \(\left(1-\dfrac{1}{3}\right)\left(1-\dfrac{1}{6}\right)\left(1-\dfrac{1}{10}\right)...\left(1-\dfrac{1}{780}\right)\)

\(=\dfrac{2}{3}.\dfrac{5}{6}.\dfrac{9}{10}.....\dfrac{779}{780}\)\(=\)

5 tháng 4 2017

a, \(\dfrac{1}{n}-\dfrac{1}{n+a}=\dfrac{n+a}{n\left(n+a\right)}-\dfrac{n}{n\left(n+a\right)}=\dfrac{n+a-n}{n\left(n+a\right)}=\dfrac{a}{n\left(n+a\right)}\)

Vậy \(\dfrac{1}{n}-\dfrac{1}{n+a}=\dfrac{a}{n\left(n+a\right)}\)

b,

\(A=\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{99.100}\)

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

\(A=\dfrac{1}{2}-\dfrac{1}{100}=\dfrac{49}{100}\)

\(B=\dfrac{5}{1.4}+\dfrac{5}{4.7}+...+\dfrac{5}{100.103}\)

\(3B=\dfrac{5.3}{1.4}+\dfrac{5.3}{4.7}+...+\dfrac{5.3}{100.103}\)

\(3B=5\left(\dfrac{3}{1.4}+\dfrac{3}{4.7}+...+\dfrac{3}{100.103}\right)\)

\(3B=5\left(1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+...+\dfrac{1}{100}-\dfrac{1}{103}\right)\)

\(3B=5\left(1-\dfrac{1}{103}\right)=5\cdot\dfrac{102}{103}=\dfrac{510}{103}\)

\(B=\dfrac{510}{103}:3=\dfrac{170}{103}\)

\(C=\dfrac{1}{15}+\dfrac{1}{35}+...+\dfrac{1}{2499}\)

\(C=\dfrac{1}{3.5}+\dfrac{1}{5.7}+...+\dfrac{1}{49.51}\)

\(2C=\dfrac{2}{3.5}+\dfrac{2}{5.7}+...+\dfrac{2}{49.51}\)

\(2C=\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{49}-\dfrac{1}{51}\)

\(2C=\dfrac{1}{3}-\dfrac{1}{51}=\dfrac{16}{51}\)

\(C=\dfrac{16}{51}:2=\dfrac{8}{51}\)

17 tháng 4 2017

\(\dfrac{1}{2.3}+\dfrac{1}{3.4}+\dfrac{1}{4.5}+...+\dfrac{1}{x\left(x+1\right)}=\dfrac{299}{600}\)

\(\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+...+\dfrac{1}{x}-\dfrac{1}{x+1}=\dfrac{299}{600}\)

\(\dfrac{1}{2}-\dfrac{1}{x+1}=\dfrac{299}{600}\)

\(\dfrac{1}{x+1}=\dfrac{1}{2}-\dfrac{299}{600}\)

\(\dfrac{1}{x+1}=\dfrac{300}{600}-\dfrac{299}{600}\)

\(\dfrac{1}{x+1}=\dfrac{1}{600}\)

=> x + 1 = 600

x = 600 - 1

x = 599

Vậy x = 599

a: =>5x=3x-6

=>2x=-6

hay x=-3

b: \(\Leftrightarrow\left(x-3\right)^2=4\cdot5^2=100\)

=>x-3=10 hoặc x-3=-10

=>x=13 hoặc x=-7

c: \(\left|x^3+1\right|+2\ge2\forall x\)

Dấu '=' xảy ra khi x=-1

Bài 1: a, Chứng tỏ rằng với n thuộc N, n khác 0 thì: \(\dfrac{1}{n\left(n+1\right)}\)=\(\dfrac{1}{n}\) - \(\dfrac{1}{n+1}\) b, Áp dụng kết quả ở câu a để tính nhanh: A=\(\dfrac{1}{1.2}\)+\(\dfrac{1}{2.3}\)+\(\dfrac{1}{3.4}\)+.....+\(\dfrac{1}{9.10}\) Bài 2: Tính nhanh: C=\(\dfrac{1}{2}\)+\(\dfrac{1}{14}\)+\(\dfrac{1}{35}\)+\(\dfrac{1}{65}\)+\(\dfrac{1}{104}\)+\(\dfrac{1}{152}\) Bài 3: a, Cho 2 phân số...
Đọc tiếp

Bài 1:

a, Chứng tỏ rằng với n thuộc N, n khác 0 thì:

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

b, Áp dụng kết quả ở câu a để tính nhanh:

A=\(\dfrac{1}{1.2}\)+\(\dfrac{1}{2.3}\)+\(\dfrac{1}{3.4}\)+.....+\(\dfrac{1}{9.10}\)

Bài 2: Tính nhanh:

C=\(\dfrac{1}{2}\)+\(\dfrac{1}{14}\)+\(\dfrac{1}{35}\)+\(\dfrac{1}{65}\)+\(\dfrac{1}{104}\)+\(\dfrac{1}{152}\)

Bài 3:

a, Cho 2 phân số \(\dfrac{1}{n}\)\(\dfrac{1}{n+1}\) (n thuộc Z, n > 0). Chứng tỏ rằng tích của 2 phân số này bằng hiệu của chúng.

b, Áp dụng kết quả trên để tính giá trị các biểu thức sau:

A=\(\dfrac{1}{2}\) . \(\dfrac{1}{3}\) + \(\dfrac{1}{3}\) . \(\dfrac{1}{4}\) + \(\dfrac{1}{4}\) . \(\dfrac{1}{5}\) + \(\dfrac{1}{5}\) . \(\dfrac{1}{6}\) + \(\dfrac{1}{6}\) . \(\dfrac{1}{7}\) + \(\dfrac{1}{7}\) . \(\dfrac{1}{8}\) + \(\dfrac{1}{8}\) . \(\dfrac{1}{9}\)

B=\(\dfrac{1}{30}\)+\(\dfrac{1}{42}\)+\(\dfrac{1}{56}\)+\(\dfrac{1}{72}\)+\(\dfrac{1}{90}\)+\(\dfrac{1}{110}\)+\(\dfrac{1}{132}\)

Các bạn giúp mk với nha!vui

4
18 tháng 3 2017

Bài 1:

a) \(\dfrac{1}{n\left(n+1\right)}=\dfrac{1}{n}-\dfrac{1}{n+1}\)

Quy đồng \(VP\) ta được:

\(VP=\dfrac{1}{n}-\dfrac{1}{n+1}\)

\(\Rightarrow VP=\dfrac{n+1}{n\left(n+1\right)}-\dfrac{n}{n\left(n+1\right)}\)

\(\Rightarrow VP=\dfrac{n+1-n}{n\left(n+1\right)}=\dfrac{1}{n\left(n+1\right)}\)

\(\Rightarrow VP=VT\)

Vậy \(\forall n\in Z,n>0\Rightarrow\dfrac{1}{n\left(n+1\right)}=\dfrac{1}{n}-\dfrac{1}{n+1}\) (Đpcm)

b) \(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{9.10}\)

\(=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{9}-\dfrac{1}{10}\)

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

\(=\dfrac{9}{10}\)

18 tháng 3 2017

Bài 3:

a) \(\dfrac{1}{n}-\dfrac{1}{n+1}=\dfrac{1+1}{n\left(n+1\right)}-\dfrac{n}{n\left(n+1\right)}=\dfrac{1}{n\left(n+1\right)}\)

b) A=\(\dfrac{1}{2}.\dfrac{1}{3}+\dfrac{1}{3}.\dfrac{1}{4}+\dfrac{1}{4}.\dfrac{1}{5}+\dfrac{1}{5}.\dfrac{1}{6}+\dfrac{1}{6}.\dfrac{1}{7}+\dfrac{1}{7}.\dfrac{1}{8}+\dfrac{1}{8}.\dfrac{1}{9}\)

\(=\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{8}+\dfrac{1}{8}-\dfrac{1}{9}\)

\(=\dfrac{1}{2}-\dfrac{1}{9}\)

\(=\dfrac{7}{18}\)

B=\(\dfrac{1}{30}+\dfrac{1}{42}+\dfrac{1}{56}+\dfrac{1}{72}+\dfrac{1}{90}+\dfrac{1}{110}+\dfrac{1}{132}\)

\(=\dfrac{1}{5.6}+\dfrac{1}{6.7}+\dfrac{1}{7.8}+\dfrac{1}{8.9}+\dfrac{1}{9.10}+\dfrac{1}{10.11}+\dfrac{1}{11.12}\)

\(=\dfrac{1}{5}-\dfrac{1}{12}\)

\(=\dfrac{7}{60}\)