CHO A = \(\frac{2\times9\times8+3\times12\times10+4\times15\times12+...+98\times297\times200}{2\times3\times4+3\times4\times5+4\times5\times6+...+98\times99\times100}\)
TÍNH A\(^2\)
K
Khách
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HT
18 tháng 8 2015
C = \(\frac{3}{2.3.4}+\frac{3}{3.4.5}+.....+\frac{3}{98.99.100}\)
C = \(3.\left(\frac{1}{2.3.4}+\frac{1}{3.4.5}+...+\frac{1}{98.99.100}\right)\)
C = \(3.\frac{1}{2}.\left(\frac{2}{2.3.4}+\frac{2}{3.4.5}+...+\frac{2}{98.99.100}\right)\)
C = \(\frac{3}{2}.\left(\frac{4-2}{2.3.4}+\frac{5-3}{3.4.5}+...+\frac{100-98}{98.99.100}\right)\)
C = \(\frac{3}{2}.\left(\frac{4}{2.3.4}-\frac{2}{2.3.4}+\frac{5}{3.4.5}-\frac{3}{3.4.5}+...+\frac{100}{98.99.100}-\frac{99}{98.99.100}\right)\)
C = \(\frac{3}{2}.\left(\frac{1}{2.3}-\frac{1}{3.4}+\frac{1}{3.4}-\frac{1}{4.5}+...+\frac{1}{98.99}-\frac{1}{99.100}\right)\)
C = \(\frac{3}{2}.\left(\frac{1}{2.3}-\frac{1}{99.100}\right)\)
C = \(\frac{3}{2}.\frac{1649}{9900}\)
C = \(\frac{1649}{6600}\)
1 tháng 5 2015
\(\frac{1}{1.2.3}+\frac{1}{2.3.4}+...+\frac{1}{8.9.10}=\frac{1}{2}.\left(\frac{1}{1.2}-\frac{1}{2.3}\right)+\frac{1}{2}.\left(\frac{1}{2.3}-\frac{1}{3.4}\right)+...+\frac{1}{2}.\left(\frac{1}{8.9}-\frac{1}{9.10}\right)\)
\(=\frac{1}{2}.\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+...+\frac{1}{8.9}-\frac{1}{9.10}\right)\)
\(\frac{1}{2}.\left(\frac{1}{1.2}-\frac{1}{9.10}\right)=\frac{1}{2}.\frac{22}{45}=\frac{11}{45}\)
LT
Lương Thị Vân Anh
CTVHS
21 tháng 2 2023
a) \(\dfrac{30\times25\times7\times8}{75\times8\times12\times14}=\dfrac{3\times2\times5\times25\times7\times8}{25\times3\times8\times3\times4\times2\times7}=\dfrac{5}{3\times4}=\dfrac{5}{12}\)
b) \(\dfrac{8\times3\times4}{16\times3}=\dfrac{8\times3\times2\times2}{8\times2\times3}=2\)
c) \(\dfrac{4\times5\times6}{3\times10\times8}=\dfrac{4\times5\times3\times2}{3\times5\times2\times4\times2}=\dfrac{1}{2}\)
7 tháng 5 2017
\(=\frac{1.2}{99.100}\)
\(=\frac{2}{9900}=\frac{1}{4950}\)
24 tháng 5 2019
\(A=\frac{1}{1\times6\times6}+\frac{1}{2\times9\times8}+\frac{1}{3\times12\times10}+...+\frac{1}{98\times297\times200}\)
\(A=\frac{1}{1\times\left(2\times3\right)\times\left(2\times3\right)}+\frac{1}{2\times\left(3\times3\right)\times\left(2\times4\right)}+...+\frac{1}{98\times\left(99\times3\right)\times\left(100\times2\right)}\)
\(A=\frac{1}{6}\times\left(\frac{1}{1\times2\times3}+\frac{1}{2\times3\times4}+...+\frac{1}{98\times99\times100}\right)\)
\(12\times A=\frac{2}{1\times2\times3}+\frac{2}{2\times3\times4}+...+\frac{2}{98\times99\times100}\)
\(12\times A=\left(\frac{1}{1\times2}-\frac{1}{2\times3}\right)+\left(\frac{1}{2\times3}-\frac{1}{3\times4}\right)+...+\left(\frac{1}{98\times99}-\frac{1}{99\times100}\right)\)
\(12\times A=\frac{1}{1\times2}-\frac{1}{99\times100}=\frac{4949}{9900}\)
\(A=\frac{4949}{118800}\)
7 tháng 8 2016
\(C=\frac{1.5.6+2.10.12+24.8.10}{1.3.5+2.6.10+8.6.20}\)
\(C=\frac{1.5.6.\left(1^3+2^3+8^2\right)}{1.3.5.\left(1^3+2^3+8^2\right)}=\frac{6}{3}=2\)
AD
23 tháng 7 2023
\(1,\\ =\dfrac{2-1}{1\times2}+\dfrac{3-2}{2\times3}+\dfrac{4-3}{3\times4}+\dfrac{5-4}{4\times5}+.....+\dfrac{99-98}{98\times99}+\dfrac{100-99}{99\times100}\\ =1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+....+\dfrac{1}{98}-\dfrac{1}{99}+\dfrac{1}{99}-\dfrac{1}{100}\\ =1-\dfrac{1}{100}=\dfrac{100-1}{100}=\dfrac{99}{100}\)
\(2,=\dfrac{13-11}{11\times13}+\dfrac{15-13}{13\times15}+....+\dfrac{21-19}{19\times21}+\dfrac{23-21}{21\times23}\\ =\dfrac{1}{11}-\dfrac{1}{13}+\dfrac{1}{13}-\dfrac{1}{15}+....+\dfrac{1}{19}-\dfrac{1}{21}+\dfrac{1}{21}-\dfrac{1}{23}\\ =\dfrac{1}{11}-\dfrac{1}{23}\\ =\dfrac{23-11}{11\times23}=\dfrac{12}{253}\)
@seven
23 tháng 7 2023
a: 1/1*2+1/2*3+...+1/99*100
=1-1/2+1/2-1/3+...+1/99-1/100
=1-1/100
=99/100
b: 2/11*13+2/13*15+...+2/21*23
=1/11-1/13+1/13-1/15+...+1/21-1/23
=1/11-1/23
=12/253
\(A=\frac{2\cdot9\cdot8+3\cdot12\cdot10+4\cdot15\cdot12+...+98\cdot297\cdot200}{2\cdot3\cdot4+3\cdot4\cdot5+4\cdot5\cdot6+...+98\cdot99\cdot100}\)
\(=\frac{2\cdot1\cdot3\cdot3\cdot4\cdot2+3\cdot1\cdot4\cdot3\cdot5\cdot2+...+98\cdot1+99\cdot3+100\cdot2}{2\cdot3\cdot4+3\cdot4\cdot5+...+98\cdot99\cdot100}\)
\(=\frac{1\cdot3\cdot2\cdot\left(2\cdot3\cdot4+3\cdot4\cdot5+...+98\cdot99\cdot100\right)}{2\cdot3\cdot4+3\cdot4\cdot5+...+98\cdot99\cdot100}\)
\(=1\cdot3\cdot2\)
\(=6\)
\(A^2=6^2=36\)