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31 tháng 8 2016

C = 1/100 - ( 1/2.1 + 1/3.2 + ... + 1/98.97 + 1/99.98 + 1/100.99

C = 1/100 - (  1- 1/2+ 1/2 - 1/3 + ... + 1/97 - 1/98 + 1/98 - 1/99 + 1/99 - 1/100 )

C = 1/100 - ( 1 - 1/100 )

C = 1/100 - 99/100

C = \(\frac{-49}{50}\)

24 tháng 8 2015

A= 1/3- 3/4+ 3/5+ 1/72- 2/9- 1/36+ 1/15
A= ( 1/3- 3/5+ 1/15) - (3/4- 1/72+ 2/9+ 1/36)
A= (5/15- 9/15+ 1/15) - (54/72- 1/72+ 16/72+ 2/36)
A= 1- 71/72
A= 1/72
 
 

17 tháng 7 2017

A=1/72

1 tháng 8 2016

a)

\(A=\left(\frac{1}{9}-\frac{1}{10}\right)-\left(\frac{1}{8}-\frac{1}{9}\right)-....-\left(1-\frac{1}{2}\right)=\frac{1}{9}-\frac{1}{10}-\frac{1}{8}+\frac{1}{9}-....-1+\frac{1}{2}\)

\(A=-\left(\frac{1}{10}+1\right)=-\frac{11}{10}\)

21 tháng 2 2017

a)\(A=\frac{1}{90}-\frac{1}{72}-\frac{1}{56}-\frac{1}{42}-\frac{1}{30}-\frac{1}{20}-\frac{1}{12}-\frac{1}{6}-\frac{1}{2}\\ \Rightarrow A=-\frac{1}{2}-\frac{1}{6}-\frac{1}{12}-\frac{1}{20}-\frac{1}{30}-\frac{1}{42}-\frac{1}{56}-\frac{1}{72}-\frac{1}{90}\\ \Rightarrow A=-\left(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+\frac{1}{42}+\frac{1}{56}+\frac{1}{72}+\frac{1}{90}\right)\)Đặt \(B=\frac{1}{2}+\frac{1}{6}+...+\frac{1}{72}+\frac{1}{90}\)

\(\Rightarrow B=\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+...+\frac{1}{8\cdot9}+\frac{1}{9\cdot10}\)

\(\Rightarrow B=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{8}-\frac{1}{9}+\frac{1}{9}-\frac{1}{10}\)

\(\Rightarrow B=1-\frac{1}{10}=\frac{9}{10}\)

Ta có : \(A=-B\)

\(\Rightarrow A=-\frac{9}{10}\)

\(\left(\frac{1}{4}-x\right)\left(x+\frac{2}{5}\right)=0\)

Ta xét 2 trường hợp 

\(\begin{cases}\frac{1}{4}-x=0\\x+\frac{2}{5}=0\end{cases}\Rightarrow\orbr{\begin{cases}x=\frac{1}{4}\\x=-\frac{2}{5}\end{cases}}\)

tớ mới làm bài 1 thôi bài 2 3 tớ ko có thời gian 

25 tháng 3 2020

1, =\(\frac{2\left(\frac{1}{5}+\frac{1}{7}-\frac{1}{9}-\frac{1}{11}\right)}{4\left(\frac{1}{5}+\frac{1}{7}-\frac{1}{9}-\frac{1}{11}\right)}=\frac{1}{2}\)

2, A=\(\frac{1}{2}\cdot\frac{2}{3}\cdot\frac{3}{4}\cdot...\cdot\frac{99}{100}\)

\(\frac{1\cdot2\cdot3\cdot....\cdot99}{2\cdot3\cdot4\cdot...\cdot100}=\frac{1}{100}\)

Vậy ......

hok tốt

\(B=\frac{1}{3}-\frac{3}{4}+0,6+\frac{1}{64}-\frac{2}{9}-\frac{1}{36}+\frac{1}{15}\)

\(\Rightarrow B=\frac{3}{15}-\frac{48}{64}+\frac{9}{15}+\frac{1}{64}-\frac{8}{36}-\frac{1}{36}+\frac{1}{15}\)

\(\Rightarrow B=\frac{3}{15}+\frac{9}{15}+\frac{1}{15}+\left(-\frac{48}{64}+\frac{1}{64}\right)+\left(-\frac{8}{36}-\frac{1}{36}\right)\)

\(\Rightarrow B=\frac{13}{15}-\frac{47}{64}-\frac{1}{4}\)

\(\Rightarrow B=-\frac{113}{960}\)

\(C=0\)

\(D=\frac{1}{99}-\frac{1}{99.98}-...-\frac{1}{3.2}-\frac{1}{2.1}\)

\(\Rightarrow D=\frac{1}{99}-\frac{1}{99}+\frac{1}{98}-\frac{1}{98}+...-\frac{1}{3}+\frac{1}{2}-\frac{1}{2}+1\)

\(\Rightarrow D=1\)

11 tháng 8 2019

D= \(\frac{1}{99}-\frac{1}{99.98}-\frac{1}{98.97}......-\frac{1}{3.2}-\frac{1}{2.1}\)

=\(\frac{1}{99}-\left(\frac{1}{1.2}+\frac{1}{2.3}+.......+\frac{1}{97.98}+\frac{1}{98.99}\right)\)

=\(\frac{1}{99}-\left(1-\frac{1}{2}+\frac{1}{2}-.....-\frac{1}{98}-\frac{1}{99}\right)\)

=\(\frac{1}{99}-\left[1-(\frac{1}{2}-\frac{1}{2}+......+\frac{1}{98}-\frac{1}{99})\right]\)

=\(\frac{1}{99}-\left(1-0-0-.....-0-\frac{1}{99}\right)\)

=\(\frac{1}{99}-1-\frac{1}{99}\)

=1