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TP
2
2 tháng 3 2018
\(A=\dfrac{18}{2.5}+\dfrac{18}{5.8}+...+\dfrac{18}{203.206}\)
\(A=\dfrac{6.3}{2.5}+\dfrac{6.3}{5.8}+...+\dfrac{6.3}{203.206}\)
\(A=6\left(\dfrac{3}{2.5}+\dfrac{3}{5.8}+...+\dfrac{3}{203.206}\right)\)
\(A=6\left(\dfrac{1}{2}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{8}+...+\dfrac{1}{203}-\dfrac{1}{206}\right)\)
\(A=6\left(\dfrac{1}{2}-\dfrac{1}{206}\right)\)
\(A=6.\dfrac{51}{103}\)
\(A=\dfrac{306}{103}\)
30 tháng 3 2019
a) \(\frac{4}{3.5}+\frac{4}{5.7}+...+\frac{4}{97.99}\)
\(=4.\left(\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{97.99}\right)\)
\(=4.\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{97}-\frac{1}{99}\right)\)
\(=4.\left(\frac{1}{3}-\frac{1}{99}\right)\)
\(=4.\frac{32}{99}\)
\(=\frac{128}{99}\)
KN
30 tháng 3 2019
\(\frac{4}{3.5}+\frac{4}{5.7}+...+\frac{4}{97.99}\)
\(=2\left(\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{97.99}\right)\)
\(=2\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{97}-\frac{1}{99}\right)\)
\(=2\left(\frac{1}{3}-\frac{1}{99}\right)\)
\(=2.\frac{32}{99}\)
\(=\frac{64}{99}\)
HT
17 tháng 7 2015
\(\frac{18}{2.5}+\frac{18}{5.8}+....+\frac{18}{103.106}\)
=\(6\left(\frac{3}{2.5}+\frac{3}{5.8}+....+\frac{3}{103.106}\right)\)
=\(6\left(\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+....+\frac{1}{103}-\frac{1}{106}\right)\)
=\(6\left(\frac{1}{2}-\frac{1}{106}\right)\)
=\(6.\frac{26}{53}\)
=\(\frac{156}{53}\)
NT
2
22 tháng 6 2023
`@` `\text {Ans}`
`\downarrow`
\(\dfrac{7}{5}\cdot\dfrac{8}{19}+\dfrac{7}{5}\cdot\dfrac{12}{19}-\dfrac{7}{5}\cdot\dfrac{1}{18}\)
`=`\(\dfrac{7}{5}\cdot\left(\dfrac{8}{19}+\dfrac{12}{19}-\dfrac{1}{18}\right)\)
`=`\(\dfrac{7}{5}\cdot\left(\dfrac{20}{19}-\dfrac{1}{18}\right)\)
`=`\(\dfrac{7}{5}\cdot\dfrac{341}{342}=\dfrac{2387}{1710}\)
T
0
TD
18 tháng 2 2017
\(\frac{9^{14}.25^5.8^7}{18^{12}.625^3.24^3}\)
\(=\frac{\left(3^2\right)^{14}.\left(5^2\right)^5.\left(2^3\right)^7}{\left(2.3^2\right)^{12}.\left(5^4\right)^3.\left(2^3.3\right)^3}\)
\(=\frac{3^{28}.5^{10}.2^{21}}{2^{12}.3^{24}.5^{12}.2^9.3}\)
\(=\frac{3^{28}.5^{10}.2^{21}}{2^{21}.3^{25}.5^{12}}\)
\(=\frac{3^3.1.1}{1.1.5^2}\)
\(=\frac{27}{25}\)
=6.3/2.5 +6.3/5.8+...+6.3/203.206
=6(3/2.5+3/5.8+...+3/203.206)
=6(1/2-1/5+1/5-1/8+...+1/203-1/206)
=6[(1/2-1/206)+(1/5-1/5)+(1/8-1/8)+...+(1/203-1/203)]
=6(1/2-1/206)=6(103/206-1/206)=6. 102/206=6. 51/103=306/103
A=6.( 3/2.5+3/5.8+...+3/203.206)
=6.(1/2-1/5+1/3-1/8+...+1/202-1/206)
=6.(1/2-1/206)=306/103