=1/2(2/3*5+2/5*7+...+2/97*99)

=1/2(1/3-1/5+1/5-1/7+...+1/97-1/99)

=1/2*32/99

=16/99

sao ở trên là \(\dfrac{1}{9.11}\)

mà ở dưới là 7.9 r ạ

a=78/35

b=22/12

c=1/1

d=40202090/4040090

e=1,24025667172...

f=871,82

ko biết đúng ko [0_0'] hihi

a. \(\dfrac{-3}{12}+\dfrac{1}{-4}=\dfrac{-3}{12}+\dfrac{-3}{12}=\dfrac{-3-3}{12}=\dfrac{-6}{12}=\dfrac{-1}{2}\)
 

b. \(\dfrac{5}{12}+\dfrac{-3}{28}=\dfrac{35}{84}+\dfrac{-9}{84}=\dfrac{35+\left(-9\right)}{84}=\dfrac{26}{84}=\dfrac{13}{42}\)
 

c. \(\dfrac{-7}{15}+\dfrac{3}{35}=\dfrac{-49}{105}+\dfrac{9}{105}=\dfrac{-49+9}{105}=\dfrac{-40}{105}=\dfrac{-8}{21}\)
 

d. \(\dfrac{-5}{7}+\dfrac{-3}{4}=\dfrac{-20}{28}+\dfrac{-21}{28}=\dfrac{-20+\left(-21\right)}{28}=\dfrac{-41}{28}\)

a: Ta có: \(A=\frac{-2}{9}+\frac{-3}{4}+\frac35+\frac{1}{15}+\frac{1}{57}+\frac13+\frac{-1}{36}\)

\(=\left(-\frac29-\frac34-\frac{1}{36}\right)+\left(\frac35+\frac{1}{15}+\frac13\right)+\frac{1}{57}\)

\(=\left(-\frac{8}{36}-\frac{27}{36}-\frac{1}{36}\right)+\left(\frac{9}{15}+\frac{1}{15}+\frac{5}{15}\right)+\frac{1}{57}\)

\(=-1+1+\frac{1}{57}=\frac{1}{57}\)

b: \(B=\frac12+\frac{-1}{5}+\frac{-5}{7}+\frac16+\frac{-3}{35}+\frac13+\frac{1}{41}\)

\(=\left(\frac12+\frac13+\frac16\right)-\left(\frac15+\frac57+\frac{3}{35}\right)+\frac{1}{41}\)

\(=\left(\frac36+\frac26+\frac16\right)-\left(\frac{7}{35}+\frac{25}{35}+\frac{3}{35}\right)+\frac{1}{41}\)

\(=1-1+\frac{1}{41}=\frac{1}{41}\)

c: \(C=\frac{-1}{2}+\frac35+\frac{-1}{9}+\frac{1}{127}+\frac{-7}{18}+\frac{4}{35}+\frac27\)

\(=\left(-\frac12-\frac19-\frac{7}{18}\right)+\left(\frac35+\frac27+\frac{4}{35}\right)+\frac{1}{127}\)

\(=\left(-\frac{9}{18}-\frac{2}{18}-\frac{7}{18}\right)+\left(\frac{21}{35}+\frac{10}{35}+\frac{4}{35}\right)+\frac{1}{127}\)

\(=-1+1+\frac{1}{127}=\frac{1}{127}\)

133/99

quy đòng r tính nha ra \(\dfrac{199}{33}\)

\(M=\dfrac{6}{5}+\dfrac{3}{2}\left(\dfrac{2}{35}+\dfrac{2}{63}+...+\dfrac{2}{9603}+\dfrac{2}{9999}\right)\)

\(=\dfrac{6}{5}+\dfrac{3}{2}\left(\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{9}+...+\dfrac{1}{97}-\dfrac{1}{99}+\dfrac{1}{99}-\dfrac{1}{101}\right)\)

\(=\dfrac{6}{5}+\dfrac{3}{2}\cdot\dfrac{96}{505}=\dfrac{150}{101}\)

\(=\dfrac{4}{3}\cdot\dfrac{9}{8}\cdot\dfrac{16}{15}\cdot\dfrac{25}{24}\cdot\dfrac{36}{35}=\dfrac{12}{7}\)

= 4/3*9/8*16/15*25/24*36/35

=2*2/1*3 * 3*3/2*4 *4*4/3*5 *5*5/4*6 * 6*6/5*7

= (2*3*4*5*6 / 1*2*3*4*5) * ( 2*3*4*5*6 / 3*4*5*6*7)

=6/1* 2/7 

= 12/7

a: Ta có: \(\dfrac{8}{9}-\left(\dfrac{1}{2}+\dfrac{1}{6}+\dfrac{1}{12}+...+\dfrac{1}{72}\right)\)

\(=\dfrac{8}{9}-\left(1-\dfrac{1}{2}+\dfrac{1}{2}+\dfrac{1}{3}-...+\dfrac{1}{8}-\dfrac{1}{9}\right)\)

=0

câu b, đâu ạ?

Bài 1:

1: \(B=\frac{1}{15}+\frac{1}{35}+\frac{1}{63}+\frac{1}{99}+\frac{1}{143}\)

\(=\frac{1}{3\cdot5}+\frac{1}{5\cdot7}+\cdots+\frac{1}{11\cdot13}\)

\(=\frac12\left(\frac{2}{3\cdot5}+\frac{2}{5\cdot7}+\cdots+\frac{2}{11\cdot13}\right)\)

\(=\frac12\left(\frac13-\frac15+\frac15-\frac17+\cdots+\frac{1}{11}-\frac{1}{13}\right)\)

\(=\frac12\left(\frac13-\frac{1}{13}\right)=\frac12\cdot\frac{10}{39}=\frac{5}{39}\)

2: \(C=\frac12+\frac{1}{14}+\frac{1}{35}+\frac{1}{65}+\frac{1}{104}+\frac{1}{152}\)

\(=\frac24+\frac{2}{28}+\frac{2}{70}+\frac{2}{130}+\frac{2}{208}+\frac{2}{304}\)

\(=\frac{2}{1\cdot4}+\frac{2}{4\cdot7}+\frac{2}{7\cdot10}+\frac{2}{10\cdot13}+\frac{2}{13\cdot16}+\frac{2}{16\cdot19}\)

\(=\frac23\left(\frac{3}{1\cdot4}+\frac{3}{4\cdot7}+\frac{3}{7\cdot10}+\frac{3}{10\cdot13}+\frac{3}{13\cdot16}+\frac{3}{16\cdot19}\right)\)

\(=\frac23\left(1-\frac14+\frac14-\frac17+\frac17-\frac{1}{10}+\frac{1}{10}-\frac{1}{13}+\frac{1}{13}-\frac{1}{16}+\frac{1}{16}-\frac{1}{19}\right)\)

\(=\frac23\left(1-\frac{1}{19}\right)=\frac23\cdot\frac{18}{19}=\frac{2\cdot6}{19}=\frac{12}{19}\)

Bài 2:

\(\frac{1}{101}>\frac{1}{300};\frac{1}{102}>\frac{1}{300};\ldots;\frac{1}{299}>\frac{1}{300};\frac{1}{300}=\frac{1}{300}\)

Do đó: \(\frac{1}{101}+\frac{1}{102}+\cdots+\frac{1}{300}>\frac{1}{300}+\frac{1}{300}+\cdots+\frac{1}{300}=\frac{200}{300}=\frac23\) (ĐPCM)