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bài 1 :
\(\frac{2}{3}\)+\(\frac{1}{3}\)=\(\frac{3}{3}\)=1
\(\frac{3}{4}\)+\(\frac{2}{4}\)+\(\frac{1}{4}\)=\(\frac{4}{4}\)=1
\(\frac{4}{5}\)+\(\frac{3}{5}\)+\(\frac{2}{5}\)+\(\frac{1}{5}\)=\(\frac{10}{5}\)= 2
chúc bạn học tốt !!!
\(\left(30:\frac{7}{2}+0,5\times3-1,5\right)\times\left(\frac{9}{2}-\frac{9}{2}\right):\left(4,5\times100\right)\)
\(=\left(30:\frac{7}{2}+0,5\times3-1,5\right)\times0:\left(4,5\times100\right)\)
\(=0\)
2/20x22 + 2/22x24 + 2/24x26 +...+2/78x80
=2 x(1/20x22 + 1/22x24 + 1/24x26 +...+1/78x80)
=2 x(1/20 - 1/22 + 1/22 -....+1/78 - 1/80)
=2 x (1/20 - 1/80)
=2 x 3/80
=3/40
Vì 3/40 < 1/9 nên 2/20x22 + 2/22x24 + 2/24x26 +...+2/78x80 < 1/9
Ta có : \(N=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{1000.1001}\)
\(=\frac{2-1}{1.2}+\frac{3-2}{2.3}+\frac{4-3}{3.4}+...+\frac{1001-1000}{1000.1001}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{1000}-\frac{1}{1001}\)
\(=1-\frac{1}{1001}=\frac{1000}{1001}\)
Ta thấy : \(1001< 2020\Rightarrow\frac{1}{1001}>\frac{1}{2020}\)
\(\Rightarrow-\frac{1}{1001}< -\frac{1}{2020}\)
\(\Rightarrow1-\frac{1}{1001}< 1-\frac{1}{2020}\Rightarrow\frac{1000}{1001}< \frac{2019}{2020}\)
Hay : \(N< M\)
\(A=\frac{1}{2}.\frac{3}{4}.\frac{5}{6}...\frac{99}{100}\)
\(\Rightarrow A>\frac{1}{2}.\frac{2}{3}.\frac{4}{5}...\frac{98}{99}\)
\(\Rightarrow A^2>\frac{1}{2}.\frac{2}{3}.\frac{3}{4}.\frac{4}{5}...\frac{98}{99}.\frac{99}{100}\)
\(\Rightarrow A^2>\frac{1}{100}=\frac{1}{10^2}\)
Vậy \(A>\frac{1}{10}\)
\(A=\frac{1}{2}.\frac{3}{4}.\frac{5}{6}...\frac{9999}{10000}\)
\(\Rightarrow A>\frac{1}{2}.\frac{2}{3}.\frac{4}{5}...\frac{9998}{9999}\)
\(\Rightarrow A^2>\frac{1}{2}.\frac{2}{3}.\frac{3}{4}.\frac{4}{5}...\frac{9998}{9999}.\frac{9999}{10000}\)
\(\Rightarrow A^2>\frac{1}{10000}=\frac{1}{100^2}\)
\(VayA>\frac{1}{100}=B\)
Đặt \(B=\frac{2}{3}\cdot\frac{4}{5}\cdot\frac{6}{7}\cdot....\cdot\frac{10000}{10001}\)
\(\Rightarrow A< B\)
\(\Rightarrow A^2< AB\)
\(\Rightarrow A^2< \left(\frac{1}{2}\cdot\frac{3}{4}\cdot\frac{5}{6}\cdot....\cdot\frac{9999}{10000}\right)\left(\frac{2}{3}\cdot\frac{4}{5}\cdot\frac{6}{7}\cdot....\cdot\frac{10000}{10001}\right)\)
\(=\frac{1}{10001}< \frac{1}{10000}=0.0001\)
\(\Rightarrow A^2< 0.0001\)
\(\Rightarrow A< 0.1\)