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6 tháng 3 2016

\(\frac{3}{1.4}+\frac{3}{4.7}+..+\frac{3}{97.100}=\frac{0,33x}{2009}\)

\(1-\frac{1}{4}+\frac{1}{4}-...-\frac{1}{100}=\frac{0,33x}{2009}\)

\(1-\frac{1}{100}=\frac{0,33x}{2009}\)

\(\frac{99}{100}=\frac{0,33x}{20009}\Rightarrow2009.99=100.0,33x\)

x=6027

30 tháng 3 2018

\(\frac{1}{4}+\frac{1}{28}+\frac{1}{70}+...+\frac{1}{9700}=\frac{0,33x}{2009}\)

\(\frac{1}{1.4}+\frac{1}{4.7}+\frac{1}{7.10}+...+\frac{1}{97.100}=\frac{0,33x}{2009}\)

\(\frac{1}{1}-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{97}-\frac{1}{100}=\frac{0,33x}{2009}\)

\(\frac{1}{1}-\frac{1}{100}=\frac{0,33x}{2009}\)

\(\frac{100}{100}-\frac{1}{100}=\frac{0,33x}{2009}\)

\(\frac{99}{100}=\frac{0,33x}{2009}\)

\(\Rightarrow2009.99=100.0,33x\)

\(\Rightarrow2009.99=33x\)

\(\Rightarrow2009.99:33=x\)

\(\Rightarrow2009.3=x\)

\(\Rightarrow6027=x\)

Vậy \(x=6027\)(MK KO CHẮC NÓ ĐÚNG NHÉ )

7 tháng 6 2020

Ta có : \(\frac{1}{4}+\frac{1}{28}+....+\frac{1}{9700}=\frac{0,33x}{2009}\)

=> \(\frac{1}{1.4}+\frac{1}{4.7}+...+\frac{1}{97.100}=\frac{0.99x}{2009}\)

=> \(\frac{1}{3}\left(\frac{3}{1.4}+\frac{3}{4.7}+...+\frac{3}{97.100}\right)=\frac{0,33x}{2009}\)

=> \(\frac{1}{3}\left(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+...+\frac{1}{97}-\frac{1}{100}\right)=\frac{0,33x}{2009}\)

=> \(\frac{1}{3}\left(1-\frac{1}{100}\right)=\frac{0,33x}{2009}\)

=> \(\frac{33}{100}=\frac{0,33x}{2009}\Rightarrow33.2009=100.0,33x\)

=> 33.2009 = 33x

=> x = 2009

7 tháng 6 2020

Thanks bn nhìu nha, mình sẽ K cho bn ngay. Bn kb với mình nha.

4 tháng 9 2016

A = 1/4 + 1/28 + 1/70 +...+ 1/9700

A = 1/1.4 + 1/4.7 + 1/7.10 +...+ 1/97.100

3A = 3/1.4 + 3/4.7 + 3/7.10 +...+ 3/97.100

3A = 1 - 1/100

3A = 99/100

A=99/100:3=33/100

4 tháng 9 2016

\(=\frac{1}{1.4}+\frac{1}{4.7}+..+\frac{1}{97.100}\)

\(=\frac{1}{3}\left(\frac{3}{1.4}+\frac{3}{4.7}+...+\frac{3}{97.100}\right)\)

\(=\frac{1}{3}\left(\frac{1}{1}-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+...+\frac{1}{97}-\frac{1}{100}\right)\)

\(=\frac{1}{3}\left(\frac{1}{1}-\frac{1}{100}\right)\)

\(=\frac{1}{3}.\frac{99}{100}=\frac{33}{100}\)

31 tháng 8 2016

\(A=\frac{1}{4}+\frac{1}{28}+\frac{1}{70}+\frac{1}{130}+...+\frac{1}{9700}\)

\(A=\frac{1}{1.4}+\frac{1}{4.7}+\frac{1}{7.10}+\frac{1}{10.13}+...+\frac{1}{97.100}\)

\(A=\frac{3}{3}\left(\frac{1}{1.4}+\frac{1}{4.7}+\frac{1}{7.10}+\frac{1}{10.13}+...+\frac{1}{97.100}\right)\)

\(A=\frac{1}{3}\left(\frac{3}{1.4}+\frac{3}{4.7}+\frac{3}{7.10}+\frac{3}{10.13}+...+\frac{3}{97.100}\right)\)

\(A=\frac{1}{3}\left(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+\frac{1}{10}-\frac{1}{13}+...+\frac{1}{97}-\frac{1}{100}\right)\)

\(A=\frac{1}{3}\left(1-\frac{1}{100}\right)\)

\(A=\frac{1}{3}.\frac{99}{100}=\frac{33}{100}\)

27 tháng 3 2018

\(\frac{1}{4}+\frac{1}{28}+\frac{1}{70}+\frac{1}{130}+...+\frac{1}{9700}\)

\(=\frac{1}{1.4}+\frac{1}{4.7}+\frac{1}{7.10}+...+\frac{1}{97.100}\)

\(=\frac{1}{1}-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{97}-\frac{1}{100}\)

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

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

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

16 tháng 8 2016

a) số số x là 4 nên ta có:

(x.4)+1/2+1/4+1/8+1/16=1 mà 1/2+1/4+1/8+1/16=15/16 nên x=1-15/16=1/16:4=1/64

8 tháng 8 2015

\(\frac{2}{6}+\frac{2}{12}+...+\frac{2}{x\left(x-1\right)}=\frac{2007}{2009}\)

\(2\cdot\left(\frac{1}{6}+\frac{1}{12}+..+\frac{1}{x\left(x-1\right)}\right)=\frac{2007}{2009}\)

\(\frac{1}{6}+\frac{1}{12}+..+\frac{1}{x\left(x-1\right)}=\frac{2007}{2009}:2\)

\(\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{\left(x-1\right)x}=\frac{2007}{4018}\)

\(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+..+\frac{1}{x-1}-\frac{1}{x}=\frac{2007}{4018}\)

\(\frac{1}{2}-\frac{1}{x}=\frac{2007}{4018}\)

\(\frac{1}{x}=\frac{1}{2}-\frac{2007}{4018}\)

\(\frac{1}{x}=\frac{1}{2009}\)

=> x = 2009 

13 tháng 8 2017

\(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{1}{x.\left(x+1\right):2}=\frac{2009}{2011}\)

\(\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{x.\left(x+1\right)}=\frac{2009}{4022}\)(nhân mỗi vế với 1/2)

\(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{x.\left(x+1\right)}=\frac{2009}{4022}\)

\(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{x}-\frac{1}{x+1}=\frac{2009}{4022}\)

\(\frac{1}{2}-\frac{1}{x+1}=\frac{2009}{4022}\)\(\Rightarrow\frac{1}{x+1}=\frac{1}{2}-\frac{2009}{4022}=\frac{1}{2011}\)

\(\Rightarrow x+1=2011\Rightarrow x=2010\)

13 tháng 8 2017

\(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{1}{x\left(x+1\right):2}=\frac{2009}{2011}\)

\(\Rightarrow\frac{1}{2}\left(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{1}{x\left(x+1\right):2}\right)=\frac{2009}{4022}\)

\(\Rightarrow\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{x\left(x+1\right)}=\frac{2009}{4022}\)

\(\Rightarrow\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{x\left(x+1\right)}\)\(=\frac{2009}{4022}\)

\(\Rightarrow\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{x}-\frac{1}{x+1}\)\(=\frac{2009}{4022}\)

\(\Rightarrow\frac{1}{2}-\frac{1}{x+1}=\frac{2009}{4022}\)

\(\Rightarrow\frac{1}{x+1}=\frac{1}{2}-\frac{2009}{4022}\)

\(\Rightarrow\frac{1}{x+1}=\frac{1}{2011}\)

\(\Rightarrow x+1=2011\)

\(\Rightarrow x=2010\)

29 tháng 6 2018

mk lỡ lm lộn bài của bn huỳnh kim đạt ở bài dưới nha 

mk xin lỗi !