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15 tháng 7 2017

\(A=\dfrac{1}{1.2}+\dfrac{1}{3.4}+\dfrac{1}{5.6}+...+\dfrac{1}{99.100}\)

\(=1-\dfrac{1}{2}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{5}-\dfrac{1}{6}+...+\dfrac{1}{99}-\dfrac{1}{100}\)

\(=\left(1+\dfrac{1}{3}+\dfrac{1}{5}+...+\dfrac{1}{99}\right)-\left(\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{1}{6}+...+\dfrac{1}{100}\right)\)

\(=\left(1+\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+\dfrac{1}{5}+\dfrac{1}{6}+...+\dfrac{1}{99}+\dfrac{1}{100}\right)-2\left(\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{1}{6}+...+\dfrac{1}{100}\right)\)

\(=\left(1+\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{100}\right)-\left(1+\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{50}\right)\)

\(=\dfrac{1}{51}+\dfrac{1}{52}+...+\dfrac{1}{100}\)

\(=\left(\dfrac{1}{51}+\dfrac{1}{52}+...+\dfrac{1}{75}\right)+\left(\dfrac{1}{76}+\dfrac{1}{77}+...+\dfrac{1}{100}\right)\)

Ta có:

\(\dfrac{1}{51}+\dfrac{1}{52}+...+\dfrac{1}{75}>\dfrac{1}{75}+\dfrac{1}{75}+...+\dfrac{1}{75}=\dfrac{25}{75}=\dfrac{1}{3}\)

\(\dfrac{1}{76}+\dfrac{1}{77}+...+\dfrac{1}{100}>\dfrac{1}{100}+\dfrac{1}{100}+...+\dfrac{1}{100}=\dfrac{25}{100}=\dfrac{1}{4}\)

\(\Rightarrow A>\dfrac{1}{3}+\dfrac{1}{4}=\dfrac{7}{12}\) (1)

Lại có:

\(\dfrac{1}{51}+\dfrac{1}{52}+...+\dfrac{1}{75}< \dfrac{1}{50}+\dfrac{1}{50}+...+\dfrac{1}{50}=\dfrac{25}{50}=\dfrac{1}{2}\)

\(\dfrac{1}{76}+\dfrac{1}{77}+...+\dfrac{1}{100}< \dfrac{1}{75}+\dfrac{1}{75}+...+\dfrac{1}{75}=\dfrac{25}{75}=\dfrac{1}{3}\)

\(\Rightarrow A< \dfrac{1}{2}+\dfrac{1}{3}=\dfrac{5}{6}\) (2)

Từ (1) và (2) suy ra \(\dfrac{7}{12}< A< \dfrac{5}{6}\)

1 tháng 7 2015

\(A=\frac{1}{1.2}+\frac{1}{3.4}+...+\frac{1}{99.100}=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+..+\frac{1}{99}-\frac{1}{100}\)

\(=\left(1+\frac{1}{3}+...+\frac{1}{99}\right)-\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{100}\right)=\left(1+\frac{1}{2}+...+\frac{1}{100}\right)-2\cdot\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{100}\right)\)

\(=\left(1+\frac{1}{2}+...+\frac{1}{100}\right)-\left(1+\frac{1}{2}+...+\frac{1}{50}\right)=\frac{1}{51}+\frac{1}{52}+...+\frac{1}{100}\)

Do \(\frac{1}{51}>\frac{1}{52}>...>\frac{1}{100}\Rightarrow A=\frac{1}{51}+\frac{1}{52}+...+\frac{1}{100}>25\cdot\frac{1}{80}+25\cdot\frac{1}{100}=\frac{7}{12}\)

và \(A<10\cdot\frac{1}{50}+10\cdot\frac{1}{60}+...+10\cdot\frac{1}{90}=\frac{1}{5}+\frac{1}{6}+...+\frac{1}{9}=\frac{1879}{2520}<\frac{5}{6}\)

Vậy 7/12<A<5/6

1 tháng 7 2015

olm lag kinh đang làm lag thoát ra mất tiêu

-------đề đúng------------

28 tháng 1 2018

\(\dfrac{1}{5^2}+\dfrac{1}{6^2}+\dfrac{1}{7^2}+........+\dfrac{1}{100^2}\)

Ta có :

\(\dfrac{1}{5^2}< \dfrac{1}{4.5}\)

\(\dfrac{1}{6^2}< \dfrac{1}{5.6}\)

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

\(\dfrac{1}{100^2}< \dfrac{1}{99.100}\)

\(\Leftrightarrow\dfrac{1}{5^2}+\dfrac{1}{6^2}+....+\dfrac{1}{100^2}< \dfrac{1}{4.5}+\dfrac{1}{5.6}+.......+\dfrac{1}{99.100}=\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}+......+\dfrac{1}{99}-\dfrac{1}{100}=\dfrac{1}{4}-\dfrac{1}{100}=\dfrac{6}{25}\)

\(\dfrac{1}{6}< \dfrac{5}{26}< \dfrac{1}{4}\)

\(\dfrac{1}{5^2}+\dfrac{1}{6^2}+.........+\dfrac{1}{100^2}< \dfrac{6}{25}\)

\(\Leftrightarrow\dfrac{1}{6}< \dfrac{1}{5^2}+\dfrac{1}{6^2}+.......+\dfrac{1}{100^2}< \dfrac{1}{4}\left(đpcm\right)\) \(\left(1\right)\)

a: x>-3/5 nên x+3/5>0

x<1/7 nên x-1/7<0

A=1/7-x-x-3/5+4/5=-2x+12/35

b: B=|x-1/7|+|x+3/5|-1/3

x>-3/5 nên x+3/5>0

x<1/7 nên x-1/7<0

B=1/7-x+3/5+x-1/3=43/105

19 tháng 1 2018

\(S=\dfrac{1}{5^2}+\dfrac{1}{5^4}+\dfrac{1}{5^6}+...+\dfrac{1}{5^{2018}}\\ 25S=25\left(\dfrac{1}{5^2}+\dfrac{1}{5^4}+\dfrac{1}{5^6}+...+\dfrac{1}{5^{2018}}\right)\\ 25S=1+\dfrac{1}{5^2}+\dfrac{1}{5^4}+...+\dfrac{1}{5^{2016}}\\ 25S-S=\left(1+\dfrac{1}{5^2}+\dfrac{1}{5^4}+...+\dfrac{1}{5^{2016}}\right)-\left(\dfrac{1}{5^2}+\dfrac{1}{5^4}+\dfrac{1}{5^6}+...+\dfrac{1}{5^{2018}}\right)\\ 24S=1-\dfrac{1}{5^{2018}}< 1\\ \Rightarrow S< \dfrac{1}{24}\)

22 tháng 7 2017

1. Tính:

a. \(\dfrac{\text{−1 }}{\text{4 }}+\dfrac{\text{5 }}{\text{6 }}=\dfrac{-3}{12}+\dfrac{10}{12}=\dfrac{7}{12}\)

b. \(\dfrac{\text{5 }}{\text{12 }}+\dfrac{\text{-7 }}{8}=\dfrac{10}{24}+\dfrac{-21}{24}=\dfrac{-11}{24}\)

c. \(\dfrac{-7}{6}+\dfrac{-3}{10}=\dfrac{-35}{30}+\dfrac{-9}{30}=\dfrac{-44}{30}=\dfrac{-22}{15}\)

d.\(\dfrac{-3}{7}+\dfrac{5}{6}=\dfrac{-18}{42}+\dfrac{35}{42}=\dfrac{17}{42}\)

2. Tính :

a. \(\dfrac{2}{14}-\dfrac{5}{2}=\dfrac{2}{14}-\dfrac{35}{14}=\dfrac{-33}{14}\)

b.\(\dfrac{-13}{12}-\dfrac{5}{18}=\dfrac{-39}{36}-\dfrac{10}{36}=\dfrac{49}{36}\)

c.\(\dfrac{-2}{5}-\dfrac{-3}{11}=\dfrac{-2}{5}+\dfrac{3}{11}=\dfrac{-22}{55}+\dfrac{15}{55}=\dfrac{-7}{55}\)

d. \(0,6--1\dfrac{2}{3}=\dfrac{6}{10}--\dfrac{5}{3}=\dfrac{3}{5}+\dfrac{5}{3}=\dfrac{9}{15}+\dfrac{25}{15}=\dfrac{34}{15}\)

3. Tính :

a.\(\dfrac{-1}{39}+\dfrac{-1}{52}=\dfrac{-4}{156}+\dfrac{-3}{156}=\dfrac{-7}{156}\)

b.\(\dfrac{-6}{9}-\dfrac{12}{16}=\dfrac{2}{3}-\dfrac{3}{4}=\dfrac{8}{12}-\dfrac{9}{12}=\dfrac{-17}{12}\)

c. \(\dfrac{-3}{7}-\dfrac{-2}{11}=\dfrac{-3}{7}+\dfrac{2}{11}=\dfrac{-33}{77}+\dfrac{14}{77}=\dfrac{-19}{77}\)

d.\(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...\dfrac{1}{8.9}+\dfrac{1}{9.10}\)

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

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

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

Chế Kazuto Kirikaya thử tham khảo thử đi !!!

23 tháng 7 2017

Mấy câu trên kia dễ rồi mình chữa mình câu \(c\) bài \(3\) thôi nhé Kazuto Kirikaya

d) \(\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+\dfrac{1}{3\cdot4}+...+\dfrac{1}{9\cdot10}\)

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

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

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

a: \(\Leftrightarrow-\dfrac{23}{5}\cdot\dfrac{50}{23}< x< \dfrac{-13}{5}:\dfrac{21}{15}=\dfrac{-13}{5}\cdot\dfrac{5}{7}=\dfrac{-13}{7}\)

=>-10<x<-13/7

hay \(x\in\left\{-9;-8;-7;-6;-5;-4;-3;-2\right\}\)

b: \(\Leftrightarrow-\dfrac{13}{3}\cdot\dfrac{1}{3}< x< \dfrac{-2}{3}\cdot\dfrac{4-3-9}{12}\)

\(\Leftrightarrow-\dfrac{13}{9}< x< \dfrac{4}{9}\)

mà x là số nguyên

nên \(x\in\left\{-1;0\right\}\)

20 tháng 8 2017

1.Tính

a.\(\dfrac{7}{23}\left[(-\dfrac{8}{6})-\dfrac{45}{18}\right]=\dfrac{7}{23}.-\dfrac{12}{6}=-\dfrac{7}{6}\)

b.\(\dfrac{1}{5}\div\dfrac{1}{10}-\dfrac{1}{3}(\dfrac{6}{5}-\dfrac{9}{4})=2-(-\dfrac{7}{20})=\dfrac{47}{20}\)

c.\(\dfrac{3}{5}.(-\dfrac{8}{3})-\dfrac{3}{5}\div(-6)=-\dfrac{3}{2}\)

d.\(\dfrac{1}{2}.(\dfrac{4}{3}+\dfrac{2}{5})-\dfrac{3}{4}.(\dfrac{8}{9}+\dfrac{16}{3})=-\dfrac{19}{5}\)

e.\(\dfrac{6}{7}\div(\dfrac{3}{26}-\dfrac{3}{13})+\dfrac{6}{7}.(\dfrac{1}{10}-\dfrac{8}{5})=-\dfrac{61}{7}\)

Bài 2

a.\(1^2_5x+\dfrac{3}{7}=\dfrac{4}{5}\)

\(x=\dfrac{13}{49}\)

b.\(\left|x-1,5\right|=2\)

Xảy ra 2 trường hợp

TH1

\(x-1,5=2\)

\(x=3,5\)

TH2

\(x-1,5=-2\)

\(x=-0,5\)

Vậy \(x=3,5\) hoặc \(x=-0,5\) .

Ngại làm quá trời ơi,lần sau bn tách ra nhá làm vậy mỏi tay quá.

20 tháng 8 2017

Ths bn nhé