Thuc hien phep tinh: A=\(\frac{2}{3.8}\)+ \(\frac{2}{8.13}\)+ ... + \(\frac{2}{48.53}\)
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S
5 tháng 3 2017
\(\frac{10}{3.8}+\frac{10}{8.13}+\frac{10}{13.18}+...+\frac{10}{48.53}\)
\(=\frac{10}{5}\left(\frac{1}{3}-\frac{1}{8}+\frac{1}{8}-\frac{1}{13}+\frac{1}{13}-\frac{1}{18}+...+\frac{1}{48}-\frac{1}{53}\right)\)
\(=2\left(\frac{1}{3}-\frac{1}{53}\right)\)
\(=2.\frac{50}{159}=\frac{100}{159}\)
ND
1
HN
3 tháng 12 2018
Ta có :\(\frac{4}{x+2}+\frac{2}{x-2}+\frac{5x-6}{4-x}\)
\(=\frac{4\left(x-2\right)\left(4-x\right)}{\left(x+2\right)\left(x-2\right)\left(4-x\right)}+\frac{2\left(x+2\right)\left(4-x\right)}{\left(x+2\right)\left(x-2\right)\left(4-x\right)}\)\(+\frac{\left(5x-6\right)\left(x+2\right)\left(x-2\right)}{\left(x+2\right)\left(x-2\right)\left(4-x\right)}\)
\(=\frac{24x-8x^2-32}{\left(x+2\right)\left(x-2\right)\left(4-x\right)}+\frac{4x-2x^2+16}{\left(x+2\right)\left(x-2\right)\left(4-x\right)}\)\(+\frac{5x^2+4x-12}{\left(x+2\right)\left(x-2\right)\left(4-x\right)}\)
\(=\frac{24x-8x^2-32+4x-2x^2+16+5x^2+4x-12}{\left(x+2\right)\left(x-2\right)\left(4-x\right)}\)
\(=\frac{32x-5x^2-28}{\left(x+2\right)\left(x-2\right)\left(4-x\right)}\)
BT
3
12 tháng 6 2020
-5/9+8/15+-2/11+-4/9/7/17
(-5/9+-4/9)+(8/15+7/15+-2/11
-1+1+-2/11
-2/11
(17/5+11/4)-22/5
123/20-22/5
123/20-88/20
35/20
PN
1
VV
3
M
27 tháng 7 2017
38.162/124.83
= 38.(24)2/(22.3)4.(23)3
= 38.28/28.34.29
=34/29
=81/512
27 tháng 7 2017
\(\frac{3^8\cdot16^2}{12^4\cdot8^3}\)
\(=\frac{3^8\cdot2^8}{3^4\cdot2^8\cdot2^9}\)
\(=\frac{81}{512}\)
Ta có : A=\(\frac{2}{3.8}+\frac{2}{8.13}+...+\frac{2}{48.53}\)
= \(2.\left(\frac{1}{3.8}+\frac{1}{8.13}+...+\frac{1}{48.53}\right)\)
= \(\frac{2}{5}.\left(\frac{5}{3.8}+\frac{5}{8.13}+...+\frac{5}{48.53}\right)\)
=\(\frac{2}{5}.\left(\frac{1}{3}-\frac{1}{8}+\frac{1}{8}-\frac{1}{15}+...+\frac{1}{48}-\frac{1}{53}\right)\)
=\(\frac{2}{5}\left(\frac{1}{3}-\frac{1}{53}\right)\)
=\(\frac{2}{5}.\frac{50}{159}\)
=\(\frac{20}{159}\)
Vậy A=\(\frac{20}{159}\)
\(A=\frac{2}{3.8}+\frac{2}{8.13}+...+\frac{2}{48.53}\)
\(=\frac{2}{5}\left(\frac{5}{3.8}+\frac{5}{8.13}+...+\frac{5}{48.53}\right)\)
\(=\frac{2}{5}\left(\frac{8-3}{3.8}+\frac{13-8}{8.13}+...+\frac{53-48}{48.53}\right)\)
\(=\frac{2}{5}\left(\frac{1}{3}-\frac{1}{8}+\frac{1}{8}-\frac{1}{13}+...+\frac{1}{48}-\frac{1}{53}\right)\)
\(=\frac{2}{5}\left(\frac{1}{3}-\frac{1}{53}\right)\)
\(=\frac{20}{159}\)