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25 tháng 4 2017
\(A=\frac{6}{5.8}+\frac{22}{8.19}+\frac{24}{19.31}+\frac{140}{31.101}+\frac{198}{101.200}\)
\(A=2.\left(\frac{1}{5}-\frac{1}{8}\right)+2.\left(\frac{1}{8}-\frac{1}{19}\right)+2.\left(\frac{1}{19}-\frac{1}{31}\right)+2.\left(\frac{1}{31}-\frac{1}{101}\right)+2.\left(\frac{1}{101}-\frac{1}{200}\right)\)
\(A=2.\left(\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{19}+\frac{1}{19}-\frac{1}{31}+\frac{1}{31}-\frac{1}{101}+\frac{1}{101}-\frac{1}{200}\right)\)
\(A=2.\left(\frac{1}{5}-\frac{1}{200}\right)\)
\(A=2.\frac{39}{200}\)
\(\Rightarrow A=\frac{39}{100}\)
HP
4 tháng 7 2016
\(\frac{5}{4.6}+\frac{5}{6.8}+\frac{5}{8.10}+....+\frac{5}{298.300}\)
\(=\frac{5}{2}.\left(\frac{2}{4.6}+\frac{2}{6.8}+\frac{2}{8.10}+....+\frac{2}{298.300}\right)\)
\(=\frac{5}{2}.\left(\frac{1}{4}-\frac{1}{6}+\frac{1}{6}-\frac{1}{8}+\frac{1}{8}-\frac{1}{10}+....+\frac{1}{298}-\frac{1}{300}\right)\)
\(=\frac{5}{2}.\left(\frac{1}{4}-\frac{1}{300}\right)=\frac{5}{2}.\frac{37}{150}=\frac{37}{60}\)
Vậy..........
19 tháng 4 2017
a)1.2.3.4...9-1.2.3.4...8-1.2.3.4...8.8
=1.2.3.4...8(9-1-8)
=1.2.3.4...8.0
=0
b)(3.4.216)2/11.123.411-169=(3.22.216)2/11.213.222-236=32.24.232/11.235-236=32.226/235.(11-2)
=32.236/235.9=32.236/235.32=2
c)70.(131313/565656+131313/727272+131313/909090
=70.(13/56+13/72+13/90)
=70.39/70=39
d)1/4.9+1/9.14+1/14.19+...+1/64.69
=4/4.9.4+4/9.4.14+4/14.19.4+...+4/64.69.4.
=1/4.(4/4.9+4/9.14+4/14.19+...+4/64.69)
=1/4.(1/4-1/9+1/9-1/14+1/14-1/19+...+1/64-1/69)
=1/4.(1/4-1/69)
=1/4.65/276=65/1104
~~~~~~~~Chúc bạn học giỏi nhé !~~~~~~~~
A
3
KT
26 tháng 7 2018
\(\frac{1}{4.6}+\frac{1}{6.8}+\frac{1}{8.10}+....+\frac{1}{2014.2016}\)
\(=\frac{1}{2}\left(\frac{2}{4.6}+\frac{2}{6.8}+\frac{2}{8.10}+....+\frac{2}{2014.2016}\right)\)
\(=\frac{1}{2}\left(\frac{1}{4}-\frac{1}{6}+\frac{1}{6}-\frac{1}{8}+\frac{1}{8}-\frac{1}{10}+....+\frac{1}{2014}-\frac{1}{2016}\right)\)
\(=\frac{1}{2}\left(\frac{1}{4}-\frac{1}{2016}\right)\)
\(=\frac{503}{4032}\)
28 tháng 9 2017
\(\frac{9^{14}.25^5.8^7}{18^{12}.625^3.24^3}\)
\(=\frac{\left(3^2\right)^{14}.\left(5^2\right)^5.\left(2^3\right)^7}{2^{12}.\left(3^2\right)^{12}.\left(5^4\right)^3.3^3.\left(2^3\right)^3}\)
\(=\frac{3^{28}.5^{10}.2^{21}}{2^{12}.3^{24}.5^{12}.3^3.2^9}\)
\(=\frac{3^{28}.5^{10}.2^{21}}{2^{21}.3^{27}.5^{12}}\)
\(=\frac{3}{5^2}\)
\(=\frac{3}{25}\)
26 tháng 3 2018
tìm y đúng hơn là tính :
\(\frac{1}{5\cdot8}+\frac{1}{8\cdot11}+\frac{1}{11\cdot14}+...+\frac{1}{y\left(y+3\right)}=\frac{98}{1545}\)
\(\Rightarrow\frac{1}{3}\left(\frac{3}{5\cdot8}+\frac{3}{8\cdot11}+\frac{3}{11\cdot14}+...+\frac{3}{y\left(y+3\right)}\right)=\frac{98}{1545}\)
\(\Rightarrow\frac{1}{3}\left(\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+\frac{1}{11}-\frac{1}{14}+...+\frac{1}{y}-\frac{1}{y+3}\right)=\frac{98}{1545}\)
\(\Rightarrow\frac{1}{3}\left(\frac{1}{5}-\frac{1}{y+3}\right)=\frac{98}{1545}\)
\(\Rightarrow\frac{1}{5}-\frac{1}{y+3}=\frac{98}{1545}\div\frac{1}{3}\)
\(\Rightarrow\frac{1}{5}-\frac{1}{y+3}=\frac{98}{515}\)
\(\Rightarrow\frac{1}{y+3}=\frac{1}{5}-\frac{98}{515}\)
\(\Rightarrow\frac{1}{y+3}=\frac{5}{515}=\frac{1}{103}\)
\(\Rightarrow y+3=103\)
\(\Rightarrow y=100\)
6 tháng 3 2017
Ta có: \(\left(2008-\dfrac{2}{135}+\dfrac{1}{50}\right)-\left(1-\dfrac{7}{135}+\dfrac{4}{150}\right)-\left(5+\dfrac{5}{135}+\dfrac{3}{50}\right)\)
= \(2008-\dfrac{2}{135}+\dfrac{1}{50}-1+\dfrac{7}{135}-\dfrac{4}{150}-5-\dfrac{5}{135}-\dfrac{3}{50}\)
= (2008-1-5) + \(\left(\dfrac{1}{50}-\dfrac{3}{50}\right)-\left(\dfrac{2}{135}-\dfrac{7}{135}\right)-\dfrac{4}{150}\)
=2002 \(-\dfrac{1}{25}\)+\(\dfrac{1}{27}\)\(-\dfrac{4}{150}\)
=2001,9(3)