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

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

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

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

x=6027

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

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

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

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

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

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

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

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

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

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

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

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

= \(\frac{1}{3}.\frac{99}{100}\)

=\(\frac{33}{100}\)

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

bằng \(\frac{539}{1080}\)

= 539/1080

câu trả lời là \(\frac{539}{1080}\)

pt đã cho có dạng \(\frac{1}{\left(x+1\right)\left(x+4\right)}+\frac{1}{\left(x+4\right)\left(x+7\right)}+\frac{1}{\left(x+7\right)\left(x+10\right)}=\frac{4}{13}\)
\(\Leftrightarrow\frac{1}{x+1}-\frac{1}{x+4}+\frac{1}{x+4}-\frac{1}{x+7}+\frac{1}{x+7}-\frac{1}{x+10}=\frac{4}{13}\)
\(\Leftrightarrow\frac{1}{x+1}-\frac{1}{x+10}=\frac{4}{13}\Leftrightarrow....\)

bạn tuấn mình thấy vậy nè

Gỉa sử cho x=1 ta thấy \(\frac{1}{1\times4}\ne\frac{1}{1}-\frac{1}{4}\)

Bạn bấm máy tính thử xem dấu bằng chỉ áp dụng với 2 số tự nhiên liên tiếp thôi còn cái này cách 3 lận

giải thích giúp mình với

à 

ý là nhân hai về pt cho 3 đúg kk 

mình hiểu rồi nha

cảm ơn nha

Mình lười chép lại đề nên làm lun nha

3 * (1/4 + 1/28 + 1/70 +...+1/8554) * x = 31/94 * 3

(3/4 + 3/28 + 3/70 + ... +3/8554) * x = 93/94

(3/1*4 + 3/4*7 + 3/7*10 + ... + 3/ 91*94) * x = 93/94

(1/1 - 1/94) * x = 93/94

93/94 * x = 93/94

=> x = 93/94 : 93/94 = 1

Vậy x=1

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