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H=\(\frac{1\cdot2\cdot3+2\cdot4\cdot6+3\cdot6\cdot9+5\cdot10\cdot15}{1\cdot3\cdot6+2\cdot6\cdot12+3\cdot9\cdot18+5\cdot15\cdot30}=\frac{1.2.3+2^3.\left(1.2.3\right)+3^3.\left(1.2.3\right)+5^3.\left(1.2.3\right)}{1.3.6+2^3.\left(1.3.6\right)+3^3.\left(1.3.6\right)+5^3.\left(1.3.6\right)}=\frac{1.2.3.\left(1+2^3+3^3+5^3\right)}{1.3.6.\left(1+2^3+3^3+5^3\right)}=\frac{2}{6}=\frac{1}{3}\)
Tách phần lử trên ra sao cho có thể rút gọn với phần ơn dưới
S = 1/5.6 + 1/10.9+....+ 1/3350.2013
=1/5 . 1/3 .( 1/2+ 1/2.3 + 1/3.4 +... + 1/670.671)
=1/15. ( 1-1/2 + 1/2 - 1/3+...+ 1/670-1/671)
= 1/15 .( 1 - 1/671 )
= 1/15 .670/671
=134/2013
S = 1/5.6 + 1/10.9+....+ 1/3350.2013
=1/5 . 1/3 .( 1/2+ 1/2.3 + 1/3.4 +... + 1/670.671)
=1/15. ( 1-1/2 + 1/2 - 1/3+...+ 1/670-1/671)
= 1/15 .( 1 - 1/671 )
= 1/15 .670/671
=134/2013
Đặt dãy trên là A
Khi đó \(A=\frac{2}{6\times10}+\frac{2}{7\times9}+\frac{1}{64}\)
\(A=\frac{1}{30}+\frac{2}{63}+\frac{1}{64}\)
\(A=\frac{672}{20160}+\frac{640}{20160}+\frac{315}{20160}=\frac{1627}{20160}\)
\(B=\frac{18^6\cdot2^{12}\cdot4^3\cdot9^3}{16^3\cdot6^9\cdot27^3}\)
\(=>B=\frac{\left(3^2\cdot2\right)^6\cdot2^{12}\cdot\left(2^2\right)^3\cdot\left(3^2\right)^3}{\left(2^4\right)^3\cdot\left(2\cdot3\right)^9\cdot\left(3^3\right)^3}\)
\(=>B=\frac{3^{12}\cdot2^6\cdot2^{12}\cdot2^6\cdot3^6}{2^{12}\cdot2^9\cdot3^9\cdot3^9}\)
\(=>B=\frac{\left(3^{12}\cdot3^6\right)\cdot\left(2^6\cdot2^{12}\cdot2^6\right)}{\left(2^{12}\cdot2^9\right)\cdot\left(3^9\cdot3^9\right)}\)
\(=>B=\frac{3^{18}\cdot2^{24}}{2^{21}\cdot3^{18}}\)
\(=>B=\frac{2^{24}}{2^{21}}\)
\(=>B=2^{24-21}\)
\(=>B=2^3\)
\(=>B=8\)
\(\frac{2.4.10+4.6.8+14.16.20}{3.6.15+6.9.12+21.24.30}=\frac{2\left(1.2.5\right)+2\left(2.3.4\right)+2\left(7.8.9\right)}{3\left(1.2.5\right)+3\left(2.3.4\right)+3\left(7.8.9\right)}=\frac{2\left(1.2.5+2.3.4+7.8.9\right)}{3\left(1.2.5+2.3.4+7.8.9\right)}=\frac{2}{3}.\)
Hacker 2k6
Trả lời:
\(\frac{2.4.10+4.6.8+14.16.20}{3.6.15+6.9.12+21.24.30}\)
\(=\frac{2\left(1.2.5\right)+2\left(2.3.4\right)+2\left(7.8.10\right)}{3\left(1.2.5\right)+3\left(2.3.4\right)+3\left(7.8.10\right)}\)
\(=\frac{2\left(1.2.5+2.3.4+7.8.10\right)}{3\left(1.2.5+2.3.4+7.8.10\right)}\)
\(=\frac{2}{3}\)(Vì\(1.2.5+2.3.4+7.8.10\ne0\))
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Good girl
a) \(A=\frac{1}{1\cdot3\cdot5}+\frac{1}{3\cdot5\cdot7}+...+\frac{1}{25\cdot27\cdot29}\)
\(\Rightarrow4A=\frac{4}{1\cdot3\cdot5}+\frac{4}{3\cdot5\cdot7}+...+\frac{4}{25\cdot27\cdot29}\)
\(\Rightarrow4A=\frac{1}{1\cdot3}-\frac{1}{3\cdot5}+\frac{1}{3\cdot5}-\frac{1}{5\cdot7}+...+\frac{1}{25\cdot27}-\frac{1}{27\cdot29}\)
\(\Rightarrow4A=\frac{1}{1\cdot3}-\frac{1}{27\cdot29}=\frac{1}{3}-\frac{1}{783}=\frac{261}{783}-\frac{1}{783}=\frac{260}{783}\)
\(\Rightarrow A=\frac{\frac{260}{783}}{4}=\frac{65}{783}\)
b) \(\left(\frac{1}{1\cdot101}+\frac{1}{2\cdot102}+...+\frac{1}{10\cdot110}\right)x=\frac{1}{1\cdot11}+\frac{1}{2\cdot12}+...+\frac{1}{100\cdot110}\)
\(\Rightarrow100\cdot\left(\frac{1}{1\cdot101}+\frac{1}{2\cdot102}+...+\frac{1}{10\cdot110}\right)x=100\cdot\left(\frac{1}{1\cdot11}+\frac{1}{2\cdot12}+...+\frac{1}{100\cdot110}\right)\)
\(\Rightarrow\left(\frac{100}{1\cdot101}+\frac{100}{2\cdot102}+...+\frac{100}{10\cdot110}\right)x=10\cdot\left(\frac{10}{1\cdot11}+\frac{10}{2\cdot12}+...+\frac{10}{100\cdot110}\right)\)
\(\Rightarrow\left(1-\frac{1}{101}+\frac{1}{2}-\frac{1}{102}+...+\frac{1}{10}-\frac{1}{110}\right)x=10\cdot\left(1-\frac{1}{10}+\frac{1}{2}-\frac{1}{12}+...+\frac{1}{100}-\frac{1}{110}\right)\)
\(\Rightarrow\left(1-\frac{1}{101}+\frac{1}{2}-\frac{1}{102}+...+\frac{1}{10}-\frac{1}{110}\right)x=10\cdot\left(1-\frac{1}{101}+\frac{1}{2}-\frac{1}{102}+...+\frac{1}{10}-\frac{1}{110}\right)\)
\(\Rightarrow x=10\cdot\)
\(A=\frac{7}{10.11}+\frac{7}{11.12}+\frac{7}{12.13}+...+\frac{7}{69.70}\)
\(=7.\frac{1}{10.11}+7.\frac{1}{11.12}+7.\frac{1}{12.13}+...+7.\frac{1}{69.70}\)
\(=7.\left(\frac{1}{10.11}+\frac{1}{11.12}+\frac{1}{12.13}+...+\frac{1}{69.70}\right)\)
\(=7.\left(\frac{1}{10}-\frac{1}{11}+\frac{1}{11}-\frac{1}{12}+\frac{1}{12}-\frac{1}{13}+...+\frac{1}{69}-\frac{1}{70}\right)\)
\(=7.\left(\frac{1}{10}-\frac{1}{70}\right)=7.\frac{3}{35}=\frac{3}{5}\)
\(A=7.\left(\frac{1}{10}-\frac{1}{70}\right)\)
\(=\frac{6}{70}\)
\(=\frac{3}{35}\)
= \(\frac{1}{3.6}-\frac{1}{6.9}+\frac{1}{6.9}-\frac{1}{9.12}+....+\frac{1}{23.26}-\frac{1}{26.29}\)
= \(\frac{1}{3.6}-\frac{1}{26.29}\)
= \(\frac{23}{26}\).