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a) \(\frac{2^{11}.9^2}{3^5.16^2}=\frac{2^{11}.3^4}{3^5.2^8}=\frac{2^3}{3}=\frac{8}{3}\)
a,
\(\dfrac{\left(3^3\right)^{15}.5^3.\left(2^3\right)^4}{\left(5^2\right)^2.\left(3^4\right)^{11}.2^{11}}=\dfrac{3^{45}.5^3.2^{12}}{5^4.3^{44}.2^{11}}=\dfrac{6}{5}\)
b, \(\left(-\dfrac{14}{25}\right)^2.\dfrac{125}{49}+\left(-3\dfrac{11}{36}\right).2\dfrac{2}{17}=\dfrac{4}{5}.\left(-7\right)=-\dfrac{28}{5}\)
c, \(\dfrac{1}{3}-2.1=-\dfrac{5}{3}\)
Đặt A=1+2+22+..............+22017
\(\Rightarrow\)2A =2+22+23+.............+22018
\(\Rightarrow\)2A -A = (2+22+23+............+22018) -(1+2+22 +...............+22017)
\(\Rightarrow\)A= 22018 -1
Lại có :A = ( 23 )672 .22 -1 =(7+1)672 .22 -1= ( B(7) +1).22 -1 =22 .B(7) +22-1=22 .B(7)+3
Vây A chia 7 dư 3
a) \(A=\frac{5^4.20^4}{25^5.4^5}=\frac{5^4.\left(2^2.5\right)^4}{5^{2^5}.\left(2^2\right)^5}=\frac{5^8.2^8}{5^{10}.2^{10}}=\frac{1}{\left(5^{10}:5^8\right).\left(2^{10}:2^8\right)}=\frac{1}{5^2.2^2}=\frac{1}{25.4}=\frac{1}{100}\)
b) \(B=\frac{2^{30}.5^7+2^{13}.5^{27}}{2^{27}.5^7+2^{10}.5^{27}}\)\(=\frac{2^3+2^3}{1}=\frac{8+8}{1}=16\)
c) \(C=\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...........+\frac{1}{2^{100}}\)
\(\Rightarrow2C=1+\frac{1}{2}+\frac{1}{2^2}+..........+\frac{1}{2^{99}}\)
\(\Rightarrow2C-C=\left(1+\frac{1}{2}+\frac{1}{2^2}+.........+\frac{1}{2^{99}}\right)-\left(\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...........+\frac{1}{2^{100}}\right)\)
\(\Rightarrow C=1-\frac{1}{2^{100}}\)
d) \(D=1+\frac{1}{5}+\frac{1}{5^2}+\frac{1}{5^3}+.........+\frac{1}{5^{100}}\)
\(\Rightarrow5D=5+1+\frac{1}{5^2}+\frac{1}{5^3}+...........+\frac{1}{5^{101}}\)
\(\Rightarrow5D-D=\left(5+1+\frac{1}{5^2}+\frac{1}{5^3}+.........+\frac{1}{5^{101}}\right)-\left(1+\frac{1}{5}+\frac{1}{5^2}+\frac{1}{5^3}+..........+\frac{1}{5^{100}}\right)\)
\(\Rightarrow4D=5-\frac{1}{5^{101}}\)
\(\Rightarrow D=\frac{5-\frac{1}{5^{101}}}{4}\)
a) \(A=\frac{5^4x20^4}{25^5x4^5}=\frac{5^4x\left(2^2x5\right)^4}{\left(5^2\right)^5x\left(2^2\right)^5}=\frac{5^8.2^8}{5^{10}.2^{10}}=\frac{1}{5^2x2^2}=\frac{1}{25.4}=\frac{1}{100}\)
b) \(B=\frac{2^{30}x5^7+2^{13}x5^{27}}{2^{27}x5^7+2^{10}x5^{27}}=\frac{2^{13}.5^7.\left(2^{17}+5^{20}\right)}{2^{10}.5^7.\left(2^{17}+5^{20}\right)}=2^3=8\)
c) \(C=\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{100}}\)
\(\Rightarrow2C=1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{99}}\)
\(\Rightarrow2C-C=1-\frac{1}{2^{100}}\)
\(C=1-\frac{1}{2^{100}}\)
phần d bn lm tương tự như phần c nha!
a) 75 - 3.(2/13+1/17-1/19) 3.[25-(2/13+1/17-1/19)]
-------------------------------------- = ---------------------------------------------- = 3/11
275-11.(2/13+1/17-1/19) 11.[25-(2/13+1/17-1/19)]
a) \(\frac{75-\frac{6}{13}+\frac{3}{17}-\frac{3}{19}}{275-\frac{22}{13}+\frac{11}{17}-\frac{11}{19}}=\frac{75-3.\left(\frac{2}{13}+\frac{1}{17}-\frac{1}{19}\right)}{275-11.\left(\frac{2}{13}+\frac{1}{17}-\frac{1}{19}\right)}\)
\(=\frac{75-3}{275-11}\)
\(=\frac{72}{264}=\frac{3}{11}\)
b) \(\frac{2}{3.5}+\frac{7}{5.12}+\frac{9}{4.39}=\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{12}+\frac{3}{52}\)
\(=\frac{1}{3}-\frac{1}{12}+\frac{3}{52}\)
\(=\frac{1}{4}+\frac{3}{52}=\frac{4}{13}\)
\(\text{A = }\frac{\text{-1}}{\text{2011}}-\frac{\text{3}}{\text{11}^2}-\frac{\text{5}}{\text{11}^2.\text{11}}-\frac{\text{7}}{\text{11}^2.\text{11}^2}=\text{ }\frac{\text{-1}}{\text{2011}}-\frac{\text{1}}{\text{11}^2}.\left(3-\frac{\text{5}}{\text{11}}-\frac{\text{7}}{\text{11}^2}\right)\)
\(\text{B = }\frac{\text{-1}}{\text{2011}}-\frac{7}{\text{11}^2}-\frac{5}{\text{11}^2.\text{11}}-\frac{3}{\text{11}^2.\text{11}^2}=\frac{\text{-1}}{\text{2011}}-\frac{\text{1}}{\text{11}^2}.\left(7-\frac{5}{\text{11}}-\frac{3}{\text{11}^2}\right)\)
\(\text{Vì }3-\frac{\text{5}}{\text{11}}-\frac{\text{7}}{\text{11}^2}< 7-\frac{5}{\text{11}}-\frac{3}{\text{11}^2}\)
\(\Rightarrow\frac{\text{-1}}{\text{2011}}-\frac{\text{1}}{\text{11}^2}.\left(3-\frac{\text{5}}{\text{11}}-\frac{\text{7}}{\text{11}^2}\right)>\frac{\text{-1}}{\text{2011}}-\frac{\text{1}}{\text{11}^2}.\left(7-\frac{5}{\text{11}}-\frac{3}{\text{11}^2}\right)\)
=> A > B
Vậy A > B