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NH
10 tháng 3 2023
\(A=\dfrac{3}{4\cdot9}+\dfrac{3}{9\cdot14}+\dfrac{3}{14\cdot19}+...+\dfrac{3}{154\cdot159}\) đề ntn phải ko ạ?
\(\dfrac{5}{3}A=\dfrac{5}{4\cdot9}+\dfrac{5}{9\cdot14}+\dfrac{5}{14\cdot19}+...+\dfrac{5}{154\cdot159}\)
\(\dfrac{5}{3}A=\dfrac{1}{4}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{14}+\dfrac{1}{14}-\dfrac{1}{19}+...+\dfrac{1}{154}-\dfrac{1}{159}\)
\(\dfrac{5}{3}A=\dfrac{1}{4}-\dfrac{1}{159}\)
\(\dfrac{5}{3}A=\dfrac{155}{636}\\ A=\dfrac{31}{212}\)
14 tháng 8 2018
-3/4. 2/11 + -3/4.9/11+11/4
= -3/4.(2/11+ 9/11)+11/4
= 3/4.1+11/4
= 7/2
14 tháng 8 2018
\(-\frac{3}{4}.\frac{2}{11}+\left(\frac{-3}{4}\right).\frac{9}{11}+\frac{11}{4}\)
=\(\frac{-3}{4}.\left(\frac{2}{11}+\frac{9}{11}\right)+\frac{11}{4}\)
=\(\frac{-3}{4}.1+\frac{11}{4}\)
=\(\frac{-3}{4}+\frac{11}{4}\) = \(\frac{8}{11}\)
Hok tốt !
2 tháng 10 2020
b) \(=\frac{1}{5}\left(\frac{1}{4}-\frac{1}{9}+\frac{1}{9}-\frac{1}{14}+\frac{1}{14}-\frac{1}{19}+...+\frac{1}{44}-\frac{1}{49}\right)\frac{2-\left(1+3+5+7+..+49\right)}{12}\)
\(=\frac{1}{5}\left(\frac{1}{4}-\frac{1}{49}\right)\frac{2-\left(12.50+25\right)}{89}=-\frac{5.9.7.89}{5.4.7.7.89}=\frac{-9}{28}\)
21 tháng 12 2022
\(=\dfrac{3^{14}\cdot5^4-5^4\cdot3^{12}}{3^{12}\cdot5^6+7\cdot3^{12}\cdot5^6}=\dfrac{3^{12}\cdot5^4\left(3^2-1\right)}{3^{12}\cdot5^6\left(1+7\right)}=\dfrac{1}{25}\)
ND
Nguyễn Đức Trí
VIP
6 tháng 8 2023
\(A=\dfrac{2}{4.7}-\dfrac{3}{5.9}+\dfrac{2}{7.10}-\dfrac{3}{9.13}+...+\dfrac{2}{301.304}-\dfrac{3}{401.405}\)
\(A=\dfrac{2}{4.7}+\dfrac{2}{7.10}+\dfrac{2}{301.304}...-\left(\dfrac{3}{5.9}+\dfrac{3}{9.13}+...+\dfrac{3}{401.405}\right)\)
\(A=2\left(\dfrac{1}{4.7}+\dfrac{1}{7.10}+\dfrac{1}{301.304}\right)...-3\left(\dfrac{1}{5.9}+\dfrac{1}{9.13}+...+\dfrac{1}{401.405}\right)\)
\(A=2\left(\dfrac{1}{4}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{10}+\dfrac{1}{301}-\dfrac{1}{304}\right)...-3\left(\dfrac{1}{5}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{13}+...+\dfrac{1}{401}-\dfrac{1}{405}\right)\)
\(A=2\left(\dfrac{1}{4}-\dfrac{1}{304}\right)-3\left(\dfrac{1}{5}-\dfrac{1}{405}\right)\)
\(A=2\left(\dfrac{76}{304}-\dfrac{1}{304}\right)-3\left(\dfrac{81}{5}-\dfrac{1}{405}\right)\)
\(A=2.\dfrac{75}{304}-3.\dfrac{80}{405}=\dfrac{75}{152}-\dfrac{80}{135}=\dfrac{10125-12160}{152.135}=-\dfrac{2035}{152.135}=-\dfrac{407}{4104}\)
PT
30 tháng 10 2017
\(\left|x+\dfrac{1}{1.5}\right|+\left|x+\dfrac{1}{5.9}\right|+\left|x+\dfrac{1}{9.14}\right|+...+\left|x+\dfrac{1}{397.401}\right|\ge0\)
\(\Rightarrow101x\ge0\)
\(\Rightarrow x\ge0\)
\(\Rightarrow x+\dfrac{1}{1.5}+x+\dfrac{1}{5.9}+...+x+\dfrac{1}{397.401}=101x\)
\(\Rightarrow101x+\left(\dfrac{1}{1.5}+\dfrac{1}{5.9}+...+\dfrac{1}{397.401}\right)=x\)
\(\Rightarrow\dfrac{1}{4}\left(\dfrac{4}{1.5}+\dfrac{4}{5.9}+...+\dfrac{4}{397.401}\right)=x\)
\(\Rightarrow x=\dfrac{1}{4}\left(1-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{9}+....+\dfrac{1}{397}-\dfrac{1}{401}\right)\)
\(\Rightarrow x=\dfrac{1}{4}\left(1-\dfrac{1}{401}\right)\)
\(\Rightarrow x=\dfrac{1}{4}.\dfrac{400}{401}\)
\(\Rightarrow x=\dfrac{100}{401}\)
11 tháng 7 2023
`@` `\text {Ans}`
`\downarrow`
\(\dfrac{2^8-2^3}{2^5-1}=\dfrac{2^3\left(2^5-1\right)}{2^5-1}=\dfrac{2^3}{1}=2^3=8\)
_____
\(\dfrac{4^8\cdot9^4}{6^6\cdot8^3}\)
`=`\(\dfrac{\left(2^2\right)^8\cdot\left(3^2\right)^4}{2^6\cdot3^6\cdot\left(2^3\right)^3}\)
`=`\(\dfrac{2^{16}\cdot3^8}{2^6\cdot3^6\cdot2^9}\)
`=`\(\dfrac{2^{16}\cdot3^8}{2^{15}\cdot3^6}\)
`=`\(\dfrac{3^2}{2}\) `=`\(\dfrac{9}{2}\)
______
\(\dfrac{27^4\cdot2^3-3^{10}\cdot4^3}{6^4\cdot9^3}\)
`=`\(\dfrac{\left(3^3\right)^4\cdot2^3-3^{10}\cdot\left(2^2\right)^3}{2^4\cdot3^4\cdot\left(3^2\right)^3}\)
`=`\(\dfrac{3^{12}\cdot2^3-3^{10}\cdot2^6}{2^4\cdot3^4\cdot3^6}\)
`=`\(\dfrac{3^{10}\cdot\left(3^2\cdot2^3-2^6\right)}{3^{10}\cdot2^4}\)
`=`\(\dfrac{72-2^6}{2^4}=\dfrac{8}{16}=\dfrac{1}{2}\)
ND
Nguyễn Đức Trí
VIP
11 tháng 7 2023
\(\dfrac{2^8-2^3}{2^5-1}=\dfrac{2^3\left(2^5-1\right)}{2^5-1}=2^3=8\)
\(\dfrac{4^8.9^4}{6^6.8^3}=\dfrac{2^{16}.3^8}{2^6.3^6.2^9}=2.3^2=18\)
\(\dfrac{27^4.2^3-3^{10}.4^3}{6^4.9^3}=\dfrac{3^{12}.2^3-3^{10}.2^6}{2^4.3^4.3^6}=\dfrac{2^3.3^{10}.\left(3^2-2^3\right)}{2^4.3^{10}}=\dfrac{9-8}{2}=\dfrac{1}{2}\)
NM
1
5 tháng 8 2015
\(\frac{4}{1.5}+\frac{4}{5.9}+\frac{4}{9.13}+...+\frac{4}{177.181}+\frac{4}{181.185}\)
\(=\left(\frac{1}{1}-\frac{1}{5}\right)+\left(\frac{1}{5}-\frac{1}{9}\right)+\left(\frac{1}{9}-\frac{1}{13}\right)+...+\left(\frac{1}{177}-\frac{1}{181}\right)+\left(\frac{1}{181}-\frac{1}{185}\right)\)
\(=\frac{1}{1}-\frac{1}{185}\)
\(=\frac{184}{185}\)
\(\frac{3}{1.5}+\frac{3}{5.9}+...+\frac{3}{94.99}=3.\left(\frac{1}{1.5}+\frac{1}{5.9}+...+\frac{1}{94.99}\right)\)
\(=\frac{3}{4}.\left(\frac{4}{1.5}+\frac{4}{5.9}+...+\frac{4}{94.99}\right)\)
\(=\frac{3}{4}.\left(1-\frac{1}{5}+\frac{1}{5}-\frac{1}{9}+...+\frac{1}{94}-\frac{1}{99}\right)\)
\(=\frac{3}{4}.\left(1-\frac{1}{99}\right)\)
\(=\frac{3}{4}.\frac{98}{99}\)
\(=\frac{49}{66}\)