tớ ko biết

S = 3 - \(\frac{3}{100}\)= \(\frac{300}{100}-\frac{3}{100}=\frac{297}{100}\)

\(B=\frac{3}{1.4}+\frac{3}{4.7}+\frac{3}{7.10}+..+\frac{3}{97.100}\)

\(=\frac{1}{1}-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-....+\frac{1}{94}-\frac{1}{97}+\frac{1}{97}-\frac{1}{100}\)

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

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

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

\(A=\frac{3^2}{1.4}+\frac{3^2}{4.7}+\frac{3^2}{7.10}+.....+\frac{3^2}{97.100}\)

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

Ta thấy :

 \(\frac{3}{1.4}=\frac{4-1}{1.4}=1-\frac{1}{4}\)

\(\frac{3}{4.7}=\frac{7-4}{4.7}=\frac{1}{4}-\frac{1}{7}\)

\(.........\)

\(\frac{3}{97.100}=\frac{100-97}{97.100}=\frac{1}{97}-\frac{1}{100}\)

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

\(=3\left(1-\frac{1}{100}\right)=3\cdot\frac{99}{100}=\frac{297}{100}\)

đáp án = \(\frac{297}{100}\)

đúng không?

kết bạn với mh nha

A:3=\(\frac{3}{1.4}+\frac{3}{4.7}\)\(+.....+\frac{3}{97.100}\)

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

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

A:3=\(\frac{99}{100}\)

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

A=\(\frac{297}{100}\)

\(A:3=\frac{3}{1.4}+\frac{3}{4.7}+\frac{3}{7.10}+....+\frac{3}{97.100}\)

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

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

\(A:3=\frac{99}{100}\)

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

\(A=\frac{297}{100}\)

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

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

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

Trả lời

\(A=\frac{3}{1.4}+\frac{3}{4.7}+\frac{3}{4.7}+...+\frac{3}{97.100}\)

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

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

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

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

=\(\frac{1}{3}\left(\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+........+\frac{1}{97}-\frac{1}{100}\right)\)

=\(\frac{1}{3}\left(\frac{1}{4}-\frac{1}{100}\right)\)

=\(\frac{1}{3}x\frac{6}{25}\)=\(\frac{2}{25}\)

vậy biểu thức trên có giá trị bằng\(\frac{2}{25}\)

dễ lắm pn ạ

\(\frac{3^2}{1.4}+\frac{3^2}{4.7}+\frac{3^2}{7.10}+\frac{3^2}{10.13}+...+\frac{3^2}{97.100}\)

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

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

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

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

a) 1/1 - 1/3 +1/3 - 1/5 +........+1/49 - 1/51

=1/1 - 1/51 (các số liền kề nhau cộng lại bằng 0)

=50/51

còn câu b bạn tự giải

nhớ thank mik nha!!!!!

b,khoảng cách của nó là 3 mà tử của nó bằng 3 chứng  tỏ nó là dạng đủ 

1/1-1/4+1/4-1/7+...+1/97-1/100

1-1/100=99/100

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

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

= \(3\left(1-\frac{1}{100}\right)\)

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

= \(3.\frac{99}{100}\)

= \(\frac{297}{100}\)

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

\(A=3\left(1-\frac{1}{100}\right)=3.\frac{99}{100}=\frac{297}{100}\)