tớ ko biết

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

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

S=1/1-1/4+1/4-1/7+1/7-1/10+.....+1/97-1/100

S=1-1/100

S=99/100

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

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

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

Trả lời

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

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

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

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

#)Giải :

\(A=\frac{3}{1.4}+\frac{3}{4.7}+\frac{3}{7.11}+...+\frac{3}{97.100}\)

\(A=1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{11}+...+\frac{1}{97}-\frac{1}{100}\)

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

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

=1/1-1/4+1/4-1/7+1/7-1/10+...+1/40-1/41

=1-1/41=40/41

= 3(1/1.4+1/4.7+1/7.10+.......+1/40.43)

=3(1-1/4+1/4-1/7+1/7-1/10+....+1/40-1/43)

=3(1-1/43)

=3.42/43

=126/43

3/1.4+3/4.7+3/7.10+.....+3/40.43

=1/1-1/4+1/4-1/7+1/7-1/10+1/10-.....+1/40-1/43

=1+(-1/4+1/4)+(-1/7+1/7)+(-1/10+1/10)+.....-1/43

=1+0+0+0+....-1/43

1-1/43=43/43-1/43=42/43

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

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

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

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

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

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

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

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

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

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

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

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

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

\(B=3.\left(\frac{1}{1.4}+\frac{1}{4.7}+\frac{1}{7.10}+........+\frac{1}{27.30}\right)\)  

\(B=3.\frac{1}{3}.\left(\frac{1}{1}-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-.......-\frac{1}{27}+\frac{1}{27}-\frac{1}{30}\right)\) 

\(B=1.\left(\frac{1}{1}-\frac{1}{30}\right)\) 

\(B=\frac{29}{30}\)

B =\(\frac{3}{1.4}+\frac{3}{4.7}+\frac{3}{7.10}+...+\frac{3}{27.30}\)

B = \(\frac{1}{1}-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{27}-\frac{1}{30}\)

B =\(\frac{1}{1}-\frac{1}{30}\)

B =\(\frac{29}{30}\)

Ta có:

\(B=\frac{3}{1.4}+\frac{3}{4.7}+\frac{3}{7.10}+...+\frac{3}{27.30}\)

\(=1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+...+\frac{1}{27}-\frac{1}{30}\)

\(=1-\frac{1}{30}=\frac{29}{30}\)

Vậy \(B=\frac{29}{30}\)

\(\frac{5}{1.4}+\frac{5}{4.7}+\frac{5}{7.10}+...+\frac{5}{100.103}\)

\(=\frac{5}{3}\left(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{100}-\frac{1}{103}\right)\)

\(=\frac{5}{3}\left(1-\frac{1}{103}\right)\)

\(=\frac{5}{3}.\frac{102}{103}=\frac{170}{103}\)

\(\frac{5}{1.4}+\frac{5}{4.7}+\frac{5}{7.10}+...+\frac{5}{100.103}=\frac{5}{3}.\left(\frac{3}{1.4}+\frac{3}{4.7}+\frac{3}{7.10}+...+\frac{3}{100.103}\right)=\frac{5}{3}.\left(\frac{1}{1}-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{100}-\frac{1}{103}\right)=\frac{5}{3}.\left(\frac{1}{1}-\frac{1}{103}\right)=\frac{5}{3}.\frac{102}{103}=\frac{170}{103}\)

=5/3.(3/1.4+3/4.7+3/7.10+...+3/100.103)

=5/3.(1-1/4+1/4-1/7+1/7-1/10+...+1/100-1/103)

=5/3.(1-1/103)=5/3.102/103=170/103

                                       đáp số : 170/103