đáp án là 150/101. cách giải hơi dài nên tớ không viết,thông cảm nha!

150/101

P=1+3/15+3/35+3/55+.............+3/9999

P=3/1*3+3/3*5+3/5*7+................+3/99*101

P=3/2(2/1*3+2/3*5+..............+2/99*101)

P=3/2(1-1/3+1/3-1/5+............+1/99-1/101)

P=3/2(1-1/101)

P=3/2*100/101

P=150/101

150/152

#)Giải :

\(P=1+\frac{9}{45}+\frac{9}{105}+\frac{9}{189}+...+\frac{9}{29997}\)

\(P=\frac{3}{1.3}+\frac{3}{3.5}+\frac{3}{5.7}+...+\frac{3}{99.101}\)

\(P=\frac{3}{2}\left(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{99.101} \right)\)

\(P=\frac{3}{2}\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{99}-\frac{1}{101}\right)\)

\(P=\frac{3}{2}\left(1-\frac{1}{101}\right)\)

\(P=\frac{3}{2}\times\frac{100}{101}\)

\(P=\frac{150}{101}\)

trả lời 

=150/101 

chúc bn 

hc tốt

trả lời 

=150/101

chúc bn 

hc tốt

=335510

335510

\(B=1+\frac{3}{5}+\frac{3}{35}+\frac{3}{55}+.....+\frac{3}{9999}\)

\(B=\frac{3}{1}.3+\frac{3}{3}.5+\frac{3}{5}.7+......+\frac{3}{99}.101\)

\(B=\frac{3}{2}\left(\frac{2}{1}.3+\frac{2}{3}.5+.......+\frac{3}{99}.101\right)\)

\(B=\frac{3}{2}\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+.........+\frac{1}{99}-\frac{1}{101}\right)\)

\(B=\frac{3}{2}\left(1-\frac{1}{101}\right)\)

\(B=\frac{3}{2}.\frac{100}{101}\)

\(B=\frac{150}{101}\)

\(\frac{4}{9}+\frac{4}{45}+\frac{4}{105}+\frac{4}{189}\)

\(=\frac{8}{15}+\frac{4}{105}+\frac{4}{189}\)

\(=\frac{4}{7}+\frac{4}{189}\)

\(=\frac{16}{27}\)

Học tốt #

\(S=\frac{1}{9}+\frac{1}{45}+\frac{1}{105}+\frac{1}{189}+\frac{1}{297}=\frac{1}{3}\left(\frac{1}{3}+\frac{1}{15}+\frac{1}{35}+\frac{1}{63}+\frac{1}{99}\right)\)

\(=\frac{1}{3}.\frac{1}{2}\left(\frac{2}{3}+\frac{2}{15}+\frac{2}{35}+\frac{2}{63}+\frac{2}{99}\right)=\frac{1}{6}\left(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+\frac{2}{9.11}\right)\)

\(=\frac{1}{6}\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11}\right)=\frac{1}{6}\left(1-\frac{1}{11}\right)=\frac{1}{6}.\frac{10}{11}\)

\(=\frac{5}{33}\)

Bài làm:

\(S=\frac{1}{9}+\frac{1}{45}+\frac{1}{105}+\frac{1}{189}+\frac{1}{297}\)

\(S=\frac{1}{6}\left(\frac{2}{3}+\frac{2}{15}+\frac{2}{35}+\frac{2}{63}+\frac{2}{99}\right)\)

\(S=\frac{1}{6}\left(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+\frac{2}{9.11}\right)\)

\(S=\frac{1}{6}\left(\frac{3-1}{1.3}+\frac{5-3}{3.5}+\frac{7-5}{5.7}+\frac{9-7}{7.9}+\frac{11-9}{9.11}\right)\)

\(S=\frac{1}{6}\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11}\right)\)

\(S=\frac{1}{6}\left(1-\frac{1}{11}\right)\)

\(S=\frac{1}{6}.\frac{10}{11}=\frac{5}{33}\)

Vậy \(S=\frac{5}{33}\)

Xin lỗi bạn Xyz nhé, mk ko có chép bài bạn đâu! với lại mk cx ko k sai bài bn đâu nhé!