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hình như là 32 chứ k f 33
\(B=\frac{1^2}{1\cdot2}\cdot\frac{2^2}{2\cdot3}\cdot\frac{3^2}{3\cdot4}\cdot\frac{4^2}{4\cdot5}\)
\(B=\frac{\left(1\cdot1\right)\left(2\cdot2\right)\left(3\cdot3\right)\left(4\cdot4\right)}{\left(1\cdot2\right)\left(2\cdot3\right)\left(3\cdot4\right)\left(4\cdot5\right)}\)
\(B=\frac{\left(1\cdot2\cdot3\cdot4\right)\left(1\cdot2\cdot3\cdot4\right)}{\left(1\cdot2\cdot3\cdot4\right)\left(2\cdot3\cdot4\cdot5\right)}\)
\(=\frac{1}{5}\)
\(B=\frac{1^2}{1\cdot2}\cdot\frac{2^2}{2\cdot3}\cdot\frac{3^2}{3\cdot4}\cdot\frac{4^2}{4\cdot5}\)
\(B=\frac{1^2\cdot2^2\cdot3^2\cdot4^2}{1\cdot2\cdot2\cdot3\cdot3\cdot4\cdot4\cdot5}\)
\(B=\frac{1^2\cdot2^2\cdot3^2\cdot4^2}{1^2\cdot2^2\cdot3^2\cdot4^2\cdot5}=\frac{1}{5}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
sorry mình nhầm
ta có:
M=\(\frac{1^2}{1.2}\).\(\frac{2^2}{2.3}\).\(\frac{3^2}{3.4}\).\(\frac{4^2}{4.5}\)
=\(\frac{1.1.2.2.3.3.4.4}{1.2.2.3.3.4.4.5}\)
=\(\frac{1}{5}\)
vậy M=\(\frac{1}{5}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
sưả đề \(P=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{2021}-\dfrac{1}{2022}\)
\(=1-\dfrac{1}{2022}=\dfrac{2021}{2022}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
C=1.2+2.3+...+99.100
3C=1.2.3+2.3.3+...+99.100.3
3C=1.2(3-0)+2.3(4-1)+...+99.100(101-98)
C=99.100.101 phần 3
C=333 300
![](https://rs.olm.vn/images/avt/0.png?1311)
\(A=\frac{9}{1.2}+\frac{9}{2.3}+\frac{9}{3.4}+...+\frac{9}{98.99}+\frac{9}{99.100}\)
\(A=9.\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{98.99}+\frac{1}{99.100}\right)\)
\(A=9.\left(\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{98}-\frac{1}{99}+\frac{1}{99}-\frac{1}{100}\right)\)
\(A=9.\left(\frac{1}{1}-\frac{1}{100}\right)\)
\(A=9.\frac{99}{100}\)
\(A=\frac{891}{100}\)
\(B=\frac{15^3.4^4.30}{5^4.16^2.3^2}\)
\(\Rightarrow B=\frac{3.5.3.5.3.5.4^4.3.5.2}{5.5.5.5.4^4.3.3}\)
\(\Rightarrow B=\frac{3.3.2}{1}\)
\(\Rightarrow B=18\)