\(\frac{2^2}{1.3}.\frac{3^2}{2.4}.\frac{4^2}{3.5}.....\frac{992^2}{998.1000}\)
Giúp mình với :D Mình vote cho :))
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![](https://rs.olm.vn/images/avt/0.png?1311)
\(A=\frac{2^2}{1.3}\cdot\frac{3^2}{2.4}....\frac{999^2}{998.1000}\)
\(A=\frac{2^2.3^2....999^2}{1.3.2.4.998.100}=\frac{\left(2.3.....999\right)\left(2.3....999\right)}{\left(1.2....998\right)\left(3.4....1000\right)}\)
\(A=999\cdot\frac{1}{500}=\frac{999}{500}\)( khúc này mk làm tắt, bn bỏ dấu ở trên rồi bỏ từng tử)
=?????????????????????????????????????????????????????????????????????????????????????????????????????????????????
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\frac{2^2}{1.3}+\frac{3^2}{2.4}+...+\frac{100^2}{99.101}\\ =\frac{2.2}{1.3}+\frac{3.3}{2.4}+...+\frac{100.100}{99.101}\\ =\frac{2.}{1.}\frac{3.}{2.}\frac{...}{...}\frac{100}{99}+\frac{2.}{3.}\frac{3.}{4.}\frac{...}{...}\frac{100}{101}\\ =\frac{100}{1}+\frac{2}{101}\\ =\frac{10100}{101}+\frac{2}{101}\\ =\frac{10102}{101}\)
\(\frac{2^2}{1.3}+\frac{3^2}{2.4}+\frac{4^2}{3.5}+...+\frac{100^2}{99.101}\)
\(=\frac{2.2}{1.3}+\frac{3.3}{2.4}+\frac{4.4}{3.5}+...+\frac{100.100}{99.101}\)
\(=\frac{2.3.4...100}{1.2.3...99}.\frac{2.3.4...100}{3.4.5...101}\)
\(=100.\frac{2}{101}\)
\(=\frac{200}{101}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(A=\frac{2^2.3^2.4^2............99^2}{1.3.2.4.3.5................998.1000}\)
\(A=\frac{1.2.3.4.5................999.1.2.3.4................999}{1.2.3.4.5.6.7..........1000.1.2.3.4..............998}\)
\(A=\frac{999.999}{1000.998}\)
\(Ko\) \(\text{chắc lắm}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
mk ko ghi đb nhé
\(=\frac{1\cdot3+1}{1\cdot3}+\frac{2\cdot4+1}{2\cdot4}+...+\frac{99\cdot101+1}{99\cdot101}.\)
\(=1+\frac{1}{1\cdot3}+1+\frac{1}{2\cdot4}+...+1+\frac{1}{99\cdot101}\)
\(=99+\frac{1}{2}\left(\frac{2}{1\cdot3}+\frac{2}{2\cdot4}+...+\frac{2}{99\cdot101}\right)\)
\(=99+\frac{1}{2}\left(1-\frac{1}{3}+\frac{1}{2}-\frac{1}{4}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{98}-\frac{1}{100}+\frac{1}{99}-\frac{1}{101}\right)\)
\(=99+\frac{1}{2}\left(1+\frac{1}{2}-\frac{1}{100}-\frac{1}{101}\right)\)
phần còn lại bn tự tính nha
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\frac{2^2}{1.3}.\frac{3^2}{2.4}.\frac{4^2}{3.5}.,,\frac{50^2}{49.51}\)
=\(\frac{2.2}{1.3}.\frac{3.3}{2.4}.\frac{4.4}{3.5}.,,\frac{50.50}{49.51}\)
=\(\frac{\left(2.3.4...50\right).\left(2.3.4...50\right)}{\left(1.2.3....49\right).\left(3.4.5....51\right)}\)
=\(\frac{50.2}{1.51}\)
=\(\frac{100}{51}\)
\(=\frac{2.3.4...50}{1.2.3...49}.\frac{2.3.4...50}{3.4.5...51}=50.\frac{2}{51}=\frac{100}{51}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a, \(A=\frac{2^2}{1.3}.\frac{3^2}{2.4}.\frac{4^2}{3.5}...\frac{999^2}{998.1000}\)
\(=\frac{2.2}{1.3}.\frac{3.3}{2.4}.\frac{4.4}{3.5}...\frac{999.999}{998.1000}\)
\(=\frac{2.3.4...999}{1.2.3...998}.\frac{2.3.4...999}{3.4.5...1000}\)
\(=\frac{999}{1}.\frac{2}{1000}\)
\(=\frac{999.2}{1000.1}=\frac{999.2}{500.2.1}\)
\(=\frac{999}{500}\)
Vậy \(A=\frac{999}{500}\)
chúc bạn học giỏi
![](https://rs.olm.vn/images/avt/0.png?1311)
Xét : \(\frac{x^2}{\left(x-1\right)\left(x+1\right)}=\frac{x^2}{x^2-1}=\frac{x^2-1+1}{x^2-1}=1+\frac{1}{x^2-1}\)
=> \(\left[\frac{x^2}{x^2-1}\right]=1\) vì \(0< \frac{1}{x^2-1}< 1\)
Do đó : \(\left[D\right]=1.98=98\)