HT
1
K
Khách
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21 tháng 6 2021
a) \(A=\dfrac{3}{4}.\dfrac{8}{9}.\dfrac{15}{16}.\dfrac{24}{25}.....\dfrac{120}{121}.\dfrac{143}{144}\)
= \(\dfrac{1.3.2.4.3.5.4.6....10.12.11.13}{2^2.3^2.4^2.5^2...11^2.12^2}\)
= \(\dfrac{1.2.12.13}{2^2.12^2}=\dfrac{13}{2.12}=\dfrac{13}{24}\)
b) \(B=\dfrac{5}{9}.\dfrac{21}{25}.\dfrac{45}{49}.\dfrac{77}{81}....\dfrac{357}{361}.\dfrac{437}{441}\)
= \(\dfrac{1.5.3.7.5.9.7.11.....17.21.19.23}{3^2.5^2.7^2....19^2.21^2}=\dfrac{1.3.21.23}{3^2.21^2}\)
= \(\dfrac{23}{3.21}=\dfrac{23}{63}\)
NK
3
NK
Nguyễn Kim Tiến
VIP
24 tháng 7 2023
=2x2/1x3 x 3x3/2x4x4/3x5 x ... x 11x11/10x12
=2x2x3x3x4x4x...x11x11/1x3x2x4x3x5x...x10x12
= 2x3x4x...x11)x(2x3x4x...x11)/(1x2x3x4x...x11)/(1x2x3x...x10)x(2x3x4x...x11)
=11x2/1x12
=22/12
=11/6
24 tháng 7 2023
4/3x9/8x16/15x...x121/120=\(\dfrac{2^2}{1\cdot3}\)x\(\dfrac{3^2}{2\cdot4}\)x\(\dfrac{4^2}{3\cdot5}\)x...x\(\dfrac{11^2}{10\cdot12}\)=\(\dfrac{2}{3}\)x\(\dfrac{11}{10}\)=\(\dfrac{11}{15}\)
DN
3
9 tháng 4 2015
\(A=1+\frac{1}{3}+1+\frac{1}{8}+1+\frac{1}{15}+...+1+\frac{1}{120}=\left(1+1+...+1\right)+\left(\frac{1}{1\times3}+\frac{1}{2\times4}+\frac{1}{3\times5}+...+\frac{1}{10\times12}\right)\)\(A=10+\frac{1}{2}\times\left(\frac{2}{1\times3}+\frac{2}{2\times4}+\frac{2}{3\times5}+...+\frac{2}{10\times12}\right)=10+\frac{1}{2}\times\left(1-\frac{1}{3}+\frac{1}{2}-\frac{1}{4}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{10}-\frac{1}{12}\right)\)\(A=10+\frac{1}{2}\times\left(1+\frac{1}{2}+\left(-\frac{1}{3}+\frac{1}{3}\right)+\left(-\frac{1}{4}+\frac{1}{4}\right)+\left(-\frac{1}{5}+\frac{1}{5}\right)+...+\left(-\frac{1}{10}+\frac{1}{10}\right)-\frac{1}{11}-\frac{1}{12}\right)\)\(A=10+\frac{1}{2}\times\left(1+\frac{1}{2}-\frac{1}{11}-\frac{1}{12}\right)=10+\frac{175}{264}=\frac{2815}{264}\)
8 tháng 9 2016
\(\frac{3}{4}.\frac{8}{9}.\frac{15}{16}.....\frac{120}{121}\)
\(=\frac{2.3.4......10}{2.3.4.....11}.\frac{3.4.5.....12}{2.3.4.....11}\)
\(=\frac{1}{11}.6=\frac{6}{11}\)
C
1
8 tháng 7 2023
a: =6/12*7/14*8/16*9/18=1/2*1/2*1/2*1/2=1/16
b: =4/12*15/3*9/25*24/8
=1/3*3*5*9/25=9/5
LN
0
NT
Nguyễn Thị Thương Hoài
Giáo viên
VIP
19 tháng 8 2023
Bài 2:
a, \(\dfrac{5}{23}\) \(\times\) \(\dfrac{17}{26}\) + \(\dfrac{5}{23}\) \(\times\) \(\dfrac{9}{26}\)
= \(\dfrac{5}{23}\) \(\times\) ( \(\dfrac{17}{26}\) + \(\dfrac{9}{26}\))
= \(\dfrac{5}{23}\) \(\times\) \(\dfrac{26}{26}\)
= \(\dfrac{5}{23}\)
b, \(\dfrac{3}{4}\) \(\times\) \(\dfrac{7}{9}\) + \(\dfrac{7}{4}\) \(\times\) \(\dfrac{3}{9}\)
= \(\dfrac{7}{12}\) + \(\dfrac{7}{12}\)
= \(\dfrac{14}{12}\)
= \(\dfrac{7}{6}\)
PM
0
\(\frac{4}{3}.\frac{9}{8}.\frac{16}{15}.\frac{25}{24}...\frac{121}{120}.\frac{144}{143}\)
\(=\frac{2.2}{1.3}.\frac{3.3}{2.4}.\frac{4.4}{3.5}.\frac{5.5}{4.6}...\frac{11.11}{10.12}.\frac{12.12}{11.13}\)
\(=\frac{2.2.3.3.4.4.5.5...11.11.12.12}{1.3.2.4.3.5.4.6...10.12.11.13}\)
\(=\frac{\left(2.3.4.5...11.12\right).\left(2.3.4.5...11.12\right)}{\left(1.2.3.4...10.11\right).\left(3.4.5.6...12.13\right)}\)
\(=\frac{12.2}{1.13}\)
\(=\frac{24}{13}\)
( Dấu \(.\)là dấu \(\times\)nha )