Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
2:
\(B=\left(\dfrac{1}{2^2}-1\right)\left(\dfrac{1}{3^2}-1\right)\cdot...\cdot\left(\dfrac{1}{100^2}-1\right)\)
\(=\left(\dfrac{1}{2}-1\right)\left(\dfrac{1}{2}+1\right)\left(\dfrac{1}{3}-1\right)\left(\dfrac{1}{3}+1\right)\cdot...\cdot\left(\dfrac{1}{100}-1\right)\left(\dfrac{1}{100}+1\right)\)
\(=\left(\dfrac{1}{2}-1\right)\left(\dfrac{1}{3}-1\right)\cdot...\cdot\left(\dfrac{1}{100}-1\right)\left(\dfrac{1}{2}+1\right)\left(\dfrac{1}{3}+1\right)\cdot...\cdot\left(\dfrac{1}{100}+1\right)\)
\(=\dfrac{-1}{2}\cdot\dfrac{-2}{3}\cdot...\cdot\dfrac{-99}{100}\cdot\dfrac{3}{2}\cdot\dfrac{4}{3}\cdot...\cdot\dfrac{101}{100}\)
\(=-\dfrac{1}{100}\cdot\dfrac{101}{2}=\dfrac{-101}{200}< -\dfrac{100}{200}=-\dfrac{1}{2}\)
Lời giải:
\(A=\frac{1}{3}-\frac{2}{3^2}+\frac{3}{3^3}-\frac{4}{3^4}+...+\frac{99}{3^{99}}-\frac{100}{3^{100}}\)
\(3A=1-\frac{2}{3}+\frac{3}{3^2}-\frac{4}{3^3}+...+\frac{99}{3^{98}}-\frac{100}{3^{99}}\)
\(4A=A+3A=1-\frac{1}{3}+\frac{1}{3^2}-\frac{1}{3^3}+....-\frac{1}{3^{99}}-\frac{100}{3^{100}}\)
\(12A=3-1+\frac{1}{3}-\frac{1}{3^2}+....-\frac{1}{3^{98}}-\frac{100}{3^{99}}\)
\(\Rightarrow 16A=12A+4A=3-\frac{101}{3^{99}}-\frac{100}{3^{100}}<3\)
\(\Rightarrow A< \frac{3}{16}\)
a: \(\dfrac{4^5\cdot9^4-2\cdot6^9}{2^{10}\cdot3^8+6^8\cdot20}\)
\(=\dfrac{2^{10}\cdot3^8-2\cdot2^9\cdot3^9}{2^{10}\cdot3^8+2^8\cdot3^8\cdot2^2\cdot5}\)
\(=\dfrac{2^{10}\cdot3^8-2^{10}\cdot3^9}{2^{10}\cdot3^8+2^{10}\cdot3^8\cdot5}\)
\(=\dfrac{2^{10}\cdot3^8\left(1-3\right)}{2^{10}\cdot3^8\left(1+5\right)}=\dfrac{-2}{6}=-\dfrac{1}{3}\)
\(3C=1+\dfrac{2}{3}+\dfrac{2}{3^2}+...+\dfrac{100}{3^{99}}\\ \Rightarrow2C=3C-C=1+\dfrac{1}{3}+\dfrac{1}{3^2}+...+\dfrac{1}{3^{99}}-\dfrac{100}{3^{100}}\\ D=\dfrac{1}{3}+\dfrac{1}{3^2}+...+\dfrac{1}{3^{99}}\\ 2D=3D-D=1+\dfrac{1}{3}+\dfrac{1}{3^2}+...+\dfrac{1}{3^{98}}-\left(\dfrac{1}{3}+\dfrac{1}{3^2}+...+\dfrac{1}{3^{99}}\right)=1-\dfrac{1}{3^{99}}< 1\\ \Rightarrow D< \dfrac{1}{2}\\ \Rightarrow2C< 1+\dfrac{1}{2}\\ \RightarrowĐPCM\)
2C < \(1+\dfrac{1}{2}\)
\(\Rightarrow\)C < \(\dfrac{3}{4}\)