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a) \(\frac{-1}{21}+\frac{-1}{28}-\frac{-1}{21}-\frac{3}{14}\)
\(=\frac{-1}{21}+\frac{-1}{28}+\frac{1}{21}+\frac{-3}{14}\)
\(=\left(\frac{-1}{21}+\frac{1}{21}\right)+\left(\frac{-1}{28}+\frac{-3}{14}\right)\)
\(=0+\left(\frac{-1}{28}+\frac{-6}{28}\right)\)
\(=0+\frac{-1}{4}=\frac{-1}{4}\)
b) \(\left(\frac{-1}{5}+\frac{3}{12}\right)+\frac{-3}{4}\)
\(=\frac{-1}{5}+\frac{1}{4}+\frac{-3}{4}\)
\(=\left(\frac{1}{4}+\frac{-3}{4}\right)+\frac{-1}{5}\)
\(=\frac{-1}{2}+\frac{-1}{5}\)
\(=\frac{-5}{10}+\frac{-2}{10}=\frac{-7}{10}\)
c) \(\frac{-4}{11}\cdot\frac{2}{5}+\frac{6}{11}\cdot\frac{-3}{10}\)
\(=\frac{-8}{55}+\frac{-9}{55}=\frac{-17}{55}\)
a. \(x\in B\left(12\right)=\left\{0;12;24;36;48;60;...\right\}\)
Mà 20 < x < 50
=> \(x\in\left\{24;36;48\right\}\)
b. \(\Rightarrow x\in B\left(15\right)=\left\{0;15;30;45;...\right\}\)
Mà 0 < x < 40
=> x \(\in\left\{15;30\right\}\)
c. \(x\inƯ\left(20\right)=\left\{1;2;4;5;10;20\right\}\)
Mà x > 8
=> x \(\in\left\{10;20\right\}\)
d. \(\Rightarrow x\inƯ\left(16\right)=\left\{1;2;4;8;16\right\}\)
C= (1.3.5.....199)/(2.4.6.....200)
=> C^2= (1^2. 3^2. 5^2......199^2)/(2^2. 4^2. 6^2......200^2)
Ta có k^2 > k^-1 = (k-1)(k+1) nên 2^2 > 1.3
4^2 > 3.5
....
200^2 > 199.201
=> C^2 < (1^2.3^2.5^2.....199^2) / (1.3)(3.5)(5.7).....(199.201)
ta có: (1^2.3^2.5^2.....199^2) / (1.3)(3.5)(5.7).....(199.201)
=1/201
Do đó C^2 <1/201
Vậy C^2 < 1/201
Ta có : \(C=\frac{1}{2}\times\frac{3}{4}\times.....\times\frac{199}{200}\)
\(\Rightarrow C< \frac{2}{3}\times\frac{4}{5}\times.......\times\frac{200}{201}\)
\(\Rightarrow C^2< \frac{2}{3}\times\frac{4}{5}\times......\times\frac{200}{201}\times\frac{1}{2}\times\frac{3}{4}\times.....\times\frac{199}{200}\)
\(\Rightarrow C^2< \frac{1}{2}\times\frac{2}{3}\times\frac{3}{4}\times........\times\frac{199}{200}\times\frac{200}{201}\)
\(\Rightarrow C^2< \frac{1}{201}\left(đ.p.c.m\right)\)
Mình cũng có bài đấy k bt lm