\(\frac{20^5\cdot5^{10}}{100^5}\)
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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) \(\left(\frac{1}{1\cdot101}+\frac{1}{2\cdot102}+...+\frac{1}{10\cdot110}\right)x=\frac{1}{1\cdot11}+\frac{1}{2\cdot12}+...+\frac{1}{100\cdot110}\)
\(\Rightarrow100\cdot\left(\frac{1}{1\cdot101}+\frac{1}{2\cdot102}+...+\frac{1}{10\cdot110}\right)x=100\cdot\left(\frac{1}{1\cdot11}+\frac{1}{2\cdot12}+...+\frac{1}{100\cdot110}\right)\)
\(\Rightarrow\left(\frac{100}{1\cdot101}+\frac{100}{2\cdot102}+...+\frac{100}{10\cdot110}\right)x=10\cdot\left(\frac{10}{1\cdot11}+\frac{10}{2\cdot12}+...+\frac{10}{100\cdot110}\right)\)
\(\Rightarrow\left(1-\frac{1}{101}+\frac{1}{2}-\frac{1}{102}+...+\frac{1}{10}-\frac{1}{110}\right)x=10\cdot\left(1-\frac{1}{10}+\frac{1}{2}-\frac{1}{12}+...+\frac{1}{100}-\frac{1}{110}\right)\)
\(\Rightarrow\left(1-\frac{1}{101}+\frac{1}{2}-\frac{1}{102}+...+\frac{1}{10}-\frac{1}{110}\right)x=10\cdot\left(1-\frac{1}{101}+\frac{1}{2}-\frac{1}{102}+...+\frac{1}{10}-\frac{1}{110}\right)\)
\(\Rightarrow x=10\cdot\)
VT = 1/2.( 1-1/3+1/3-1/5+...+ 2/49-1/51)
= 1/2. 50/51
=> 6x-5/10+10 = 25/51
............. Tụ làm phàn còn lại nhé
Nhân cả 2 vê với 2 ta được:
\(\frac{2.\left(6x-5\right)}{20}\)=\(\frac{2}{1.3}\)+\(\frac{2}{3.5}\)+...+\(\frac{2}{49.51}\)
<=>\(\frac{6x-5}{10}\)=\(1-\frac{1}{3}+\)\(\frac{1}{3}-\frac{1}{5}+...+\frac{1}{49}-\frac{1}{51}\)
<=>\(\frac{6x-5}{10}=1-\frac{1}{51}\)
<=>\(6x-5=\frac{50}{51}.10\)
<=>\(x=\frac{755}{306}\)
\(\frac{10^3+2.5^3+5^3}{65}=\frac{10^3+5^3.\left(2+1\right)}{65}=\frac{10^3+5^3.3}{65}\)
= \(\frac{10^3+375}{65}=\frac{1375}{65}\)
\(\frac{10^3+2.5^3+5^3}{65}=\frac{1000+2.125+125}{65}=\frac{8.125+2.125+125.1}{65}=\frac{125\left(8+2+1\right)}{65}=\frac{125.11}{65}=\frac{1375}{65}=\frac{275}{13}\)
\(\frac{10^3+2.5^3+5^3}{65}=\frac{1000+5^3.3}{65}=\frac{1000+375}{65}\)
= \(\frac{1375}{65}=\frac{275}{13}\)
\(\frac{10^3+2.5^5+5^3}{65}\)
= \(\frac{\left(2.5\right)^3+2.5^5+5^3}{5.13}\)
= \(\frac{2^3.5^3+2.5^5+5^3}{5.13}\)
= \(\frac{5^3\left(2^3+2+1\right)}{5.13}\)
= \(\frac{5^2.11}{13}\)
= \(\frac{275}{13}\)
\(\frac{2^{15}.5^8+2^{14}.5^9}{2^{16}.5^7+2^{16}.5^8}=\frac{2^{14}.5^8\left(2+5\right)}{2^{16}.5^7\left(1+5\right)}=\frac{2^{14}.5^8.7}{2^{16}.5^7.6}=\frac{5.7}{4.6}=\frac{35}{24}\)
\(\frac{20^5.5^{10}}{100^5}=\frac{5^5.4^5.25^5}{100^5}=\frac{5^5.100^5}{100^5}=5^5=3125\)
\(\frac{20^5.5^{10}}{100^5}\)= \(\frac{5^5.4^5.5^5.5^5}{5^5.4^5.5^5}\)= \(5^5\)= \(3125\)