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Bài 1:
a) Ta có: \(\dfrac{7^4\cdot3-7^3}{7^4\cdot6-7^3\cdot2}\)
\(=\dfrac{7^3\cdot\left(7\cdot3-1\right)}{7^3\cdot2\left(7\cdot3-1\right)}\)
\(=\dfrac{1}{2}\)
c) Ta có: \(E=1+\dfrac{1}{3}+\dfrac{1}{3^2}+...+\dfrac{1}{3^{100}}\)
\(\Leftrightarrow\dfrac{1}{3}\cdot E=\dfrac{1}{3}+\dfrac{1}{3^2}+\dfrac{1}{3^3}+...+\dfrac{1}{3^{101}}\)
\(\Leftrightarrow E-\dfrac{1}{3}\cdot E=1+\dfrac{1}{3}+\dfrac{1}{3^2}+...+\dfrac{1}{3^{100}}-\left(\dfrac{1}{3}+\dfrac{1}{3^2}+\dfrac{1}{3^3}+...+\dfrac{1}{3^{101}}\right)\)
\(\Leftrightarrow E\cdot\dfrac{2}{3}=1-\dfrac{1}{3^{101}}\)
\(\Leftrightarrow E=\dfrac{3-\dfrac{3}{3^{101}}}{2}=\dfrac{1-\dfrac{1}{3^{100}}}{2}\)
\(A=\left(1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}\right)\)
\(B=\frac{1}{2}\left(\frac{1}{2}-\frac{1}{4}+\frac{1}{6}-\frac{1}{8}+....+\frac{1}{198}-\frac{1}{200}\right)\)
= \(\frac{1}{2}.\frac{1}{2}\left(1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+....+\frac{1}{99}-\frac{1}{100}\right)\)
Giờ thì A/B bằng mấy e ?
\(\dfrac{1.2+2.4+3.6+4.8+5.10}{3.4+6.8+9.12+12.16+15.20}\)
\(=\dfrac{1.2+2.4+3.6+4.8+5.10}{3.4+3.4.2.2+3.4.3.3+3.4.2.8+3.4.5.5}\)
\(=\dfrac{1.2.\left(4+3^2+2.8+5^2\right)}{3.4.\left(4+3^2+2.8+5^2\right)}\)
\(=\dfrac{1}{6}\)
\(\dfrac{1.2+2.4+3.6+4.8+5.10}{3.4+6.8+9.12+12.16+15.20}\)
\(=\dfrac{1.2+1.2.2^2+1.2.3^2+1.2.4^2+1.2.5^2}{3.4+3.4.2^2+3.4.3^2+3.4.4^2+3.4.5^2}\)
\(=\dfrac{1.2\left(1+2^2+3^2+4^2+5^2\right)}{3.4\left(1+2^2+3^2+4^2+5^2\right)}\\ =\dfrac{2}{12}=\dfrac{1}{6}\)