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\(B=\frac{1}{2^2}.\left(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{100^2}\right)\)
\(\rightarrow\frac{A}{B}=\frac{\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{100^2}^2}{\frac{1}{4}\left(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{100^2}\right)}=\frac{1}{\frac{1}{4}}=4\)
\(B=\frac{1}{4^2}+\frac{1}{6^2}+\frac{1}{8^2}+..+\frac{1}{100^2}=\frac{1}{\left(2.2\right)^2}+\frac{1}{\left(2.3\right)^2}+\frac{1}{\left(2.4\right)^2}+..+\frac{1}{\left(2.100\right)^2}\)
\(B=\frac{1}{2^2.2^2}+\frac{1}{2^2.3^2}+\frac{1}{2^2.4^2}+..+\frac{1}{2^2.100^2}=\frac{1}{2^2}.\frac{1}{2^2}+\frac{1}{2^2}.\frac{1}{3^2}+\frac{1}{2^2}.\frac{1}{4^2}+..+\frac{1}{2^2}.\frac{1}{100^2}\)
\(=\frac{1}{2^2}.\left(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{100^2}\right)=\frac{1}{4}.A\)
\(=>\frac{A}{B}=\frac{A}{\frac{A}{4}}=4\)
xin loi may anh tai tui moi lop 6 thui ha
Ta có :
\(\frac{A}{B}=\frac{\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+......+\frac{1}{100^2}}{\frac{1}{4^2}+\frac{1}{6^2}+\frac{1}{8^2}+......\frac{1}{200^2}}=\frac{4\left(\frac{1}{4^2}+\frac{1}{6^2}+\frac{1}{8^2}+...+\frac{1}{200^2}\right)}{\frac{1}{4^2}+\frac{1}{6^2}+\frac{1}{8^2}+...+\frac{1}{200^2}}=4\)
Vậy \(\frac{A}{B}=4\)
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!
Mai Thanh Hoàng: You are right!