A=2/3x5+3/5x8+11/8x9+13/9x32+25/32x57+30/57x87
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\(\frac{2}{3.5}+\frac{3}{5.8}+\frac{11}{8.19}+\frac{13}{19.32}+\frac{25}{32.57}+\frac{30}{57.87}\)
=\(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{19}+\frac{1}{19}-\frac{1}{32}+\frac{1}{32}-\frac{1}{57}+\frac{1}{57}-\frac{1}{87}\)
= 1/3 - 1/87
= 28/87
a,Ta có :
1/3 + 1/16 + 1/19 + 1/21 + 1/61 + 1/72 + 1/83 + 1/94 = 0,54
3/4 = 0,6
=>b < 3/4
a: \(\dfrac{1}{7}+\dfrac{1}{91}+\dfrac{1}{247}+\dfrac{1}{475}+\dfrac{1}{775}+\dfrac{1}{1147}\)
\(=\dfrac{1}{1\cdot7}+\dfrac{1}{7\cdot13}+\dfrac{1}{13\cdot19}+\dfrac{1}{19\cdot25}+\dfrac{1}{25\cdot31}+\dfrac{1}{31\cdot37}\)
\(=\dfrac{1}{6}\left(\dfrac{6}{1\cdot7}+\dfrac{6}{7\cdot13}+\dfrac{6}{13\cdot19}+\dfrac{6}{19\cdot25}+\dfrac{6}{25\cdot31}+\dfrac{6}{31\cdot37}\right)\)
\(=\dfrac{1}{6}\left(1-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{13}+...+\dfrac{1}{31}-\dfrac{1}{37}\right)\)
\(=\dfrac{1}{6}\left(1-\dfrac{1}{37}\right)=\dfrac{1}{6}\cdot\dfrac{36}{37}=\dfrac{6}{37}\)
b: Sửa đề:\(\dfrac{3}{5\cdot8}+\dfrac{11}{8\cdot19}+\dfrac{12}{19\cdot31}+\dfrac{80}{31\cdot101}+\dfrac{99}{101\cdot200}\)
\(=\dfrac{1}{5}-\dfrac{1}{8}+\dfrac{1}{8}-\dfrac{1}{19}+\dfrac{1}{19}-\dfrac{1}{31}+\dfrac{1}{31}-\dfrac{1}{101}+\dfrac{1}{101}-\dfrac{1}{200}\)
\(=\dfrac{1}{5}-\dfrac{1}{200}=\dfrac{39}{200}\)
b1
a) \(\dfrac{1}{5.6}+\dfrac{1}{6.7}+\dfrac{1}{7.8}+\dfrac{1}{8.9}+\dfrac{1}{9.10}\)
\(=\dfrac{1}{5}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{8}+\dfrac{1}{8}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{10}\)
\(=\dfrac{1}{5}-\dfrac{1}{10}\)
\(=\dfrac{2}{10}-\dfrac{1}{10}\)
\(=\dfrac{1}{10}\)
b) \(\dfrac{1}{1.2}+\dfrac{1}{2.3}+...+\dfrac{1}{99.100}\)
\(=\dfrac{1}{1}-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{99}-\dfrac{1}{100}\)
\(=\dfrac{1}{1}-\dfrac{1}{100}\)
\(=\dfrac{99}{100}\)
c) \(\dfrac{2}{3.5}+\dfrac{2}{5.7}+\dfrac{2}{7.9}+\dfrac{2}{9.11}\)
\(=\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{11}\)
\(=\dfrac{1}{3}-\dfrac{1}{11}\)
\(=\dfrac{8}{33}\)
d) \(\dfrac{2}{3.5}+\dfrac{2}{5.7}+...+\dfrac{2}{99.101}\)
\(=\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{99}-\dfrac{1}{101}\)
\(=\dfrac{1}{3}-\dfrac{1}{101}\)
\(=\dfrac{98}{303}\)
b)
S2=6/2x5+6/5x8+6/8x11+...+6/29x32
=2.(3/2.5+3/5.8+...+3/29.32)
=2.(1/2-1/5+1/5-1/8+...+1/29-1/32)
=2.(1/2-1/32)
=2.15/32
=15/16
a)
Ta có:
S1=2/3x5+2/5x7+2/7x9+...+2/97x99
=1/3-1/5+1/5-1/7+...+1/97-1/99
=1/3-1/99
=32/99
A=1/6+1/12+1/20+1/30+1/42+1/56+1/72
A=1/2*3+1/3*4+1/4*5+1/5*6+1/6*7+1/7*8+1/8*9
A=1/2-1/3+1/3-1/4+1/4-1/5+1/5-1/6+1/6-1/7+1/7-1/8+1/8-1/9
A=1/2-1/9
Câu B tương tự nha bạn :333
a) \(\dfrac{11}{3}-\dfrac{2}{3}\times\dfrac{5}{6}\)
\(=\dfrac{11}{3}-\dfrac{10}{18}\)
\(=\dfrac{11}{3}-\dfrac{5}{9}\)
\(=\dfrac{33}{9}-\dfrac{5}{9}\)
\(=\dfrac{28}{9}\)
b) \(\dfrac{13}{2}+\dfrac{7}{6}:\dfrac{2}{3}\)
\(=\dfrac{13}{2}+\dfrac{7}{6}\cdot\dfrac{3}{2}\)
\(=\dfrac{13}{2}+\dfrac{7}{4}\)
\(=\dfrac{26}{4}+\dfrac{7}{4}\)
\(=\dfrac{33}{4}\)
c) \(\dfrac{28}{9}-\dfrac{3}{4}+\dfrac{1}{3}\)
\(=\dfrac{28}{9}+\dfrac{3}{9}-\dfrac{3}{4}\)
\(=\dfrac{31}{9}-\dfrac{3}{4}\)
\(=\dfrac{124}{36}-\dfrac{27}{36}\)
\(=\dfrac{97}{36}\)
d) \(\dfrac{5}{8}:\dfrac{2}{3}\times\dfrac{1}{5}\)
\(=\dfrac{5}{8}\times\dfrac{3}{2}\times\dfrac{1}{5}\)
\(=\dfrac{15}{80}\)
\(=\dfrac{3}{16}\)
a) \(\dfrac{11}{3}-\dfrac{2}{3}\times\dfrac{5}{6}=\dfrac{11}{3}-\dfrac{5}{9}=\dfrac{28}{9}\)
b) \(\dfrac{13}{2}+\dfrac{7}{6}:\dfrac{2}{3}=\dfrac{13}{2}+\dfrac{7}{6}\times\dfrac{3}{2}=\dfrac{13}{2}+\dfrac{7}{4}=\dfrac{33}{4}\)
c) \(\dfrac{28}{9}-\dfrac{3}{4}+\dfrac{1}{3}=\dfrac{97}{36}\)
d) \(\dfrac{5}{8}:\dfrac{2}{3}\times\dfrac{1}{5}=\dfrac{5}{8}\times\dfrac{3}{2}\times\dfrac{1}{5}=\dfrac{3}{16}\)