Bạn tải photomath về là giải được ngay ! 

Nhớ k cho mình nhé !

Gọi biểu thức này là A

Ta có :

\(A=\frac{4}{45}+\frac{4}{105}+\frac{4}{189}+\frac{4}{297}+\frac{4}{929}\)

\(\frac{3}{2}A=\frac{3}{2}\times\left(\frac{4}{45}+\frac{4}{105}+\frac{4}{189}+\frac{4}{297}+\frac{4}{929}\right)\)

\(\frac{3}{2}A=\frac{6}{45}+\frac{6}{105}+\frac{6}{189}+\frac{6}{297}+\frac{6}{929}\)

\(\frac{3}{2}A=\frac{2}{15}+\frac{2}{35}+\frac{2}{63}+\frac{2}{99}+\frac{2}{???}\)

Bạn nên xem lại bài 

150/101 

làm cụ thể được không Yumi

mk đang cần gấp

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

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

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

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

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

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!

#)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