Tính giá trị biểu thức:
\(B=\frac{2^2}{1.3}.\frac{3^2}{2.4}.\frac{4^2}{3.5}...\frac{10^2}{9.11}\)
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C
Tính giá trị biểu thức \(B=\frac{2^2}{1.3}.\frac{3^2}{2.4}.\frac{4^2}{3.5}...\frac{9.9^2}{9.8.100}\)
0
CN
1
11 tháng 6 2017
\(\frac{1.3}{2^2}.\frac{2.4}{3^2}.\frac{3.5}{4^2}...\frac{98.100}{99^2}\)
\(=\frac{1.3}{2.2}.\frac{2.4}{3.3}.\frac{3.5}{4.4}...\frac{98.100}{99.99}\)
\(=\frac{1.2.3...98}{2.3.4...99}.\frac{3.4.5...100}{2.3.4...99}\)
\(=\frac{1}{99}.\frac{100}{2}\)
\(=\frac{1}{99}.50=\frac{50}{99}\)
24 tháng 3 2019
\(\Leftrightarrow N=\frac{\left(2.3.4....50\right)\left(2.3.4...........50\right)}{\left(1.2.3.........49\right)\left(3.4.5...........51\right)}=\frac{50.2}{51}=\frac{100}{51}\)
25 tháng 3 2019
\(\frac{2^2}{1.3}+\frac{3^2}{2.4}+\frac{4^2}{3.5}+....+\frac{50^2}{49.51}\)
\(=\frac{2^2-1}{1.3}+\frac{3^2-1}{2.4}+....+\frac{50^2-1}{49.51}+\frac{1}{1.3}+\frac{1}{2.4}+....+\frac{1}{49.51}\)
\(=\frac{1}{2}.\left(1+1+...+1\right)+\frac{1}{1}-\frac{1}{3}+\frac{1}{2}-\frac{1}{4}+\frac{1}{3}-\frac{1}{5}+....+\frac{1}{49}-\frac{1}{51}\)
Tự làm tiếp :))
25 tháng 3 2019
tớ nhầm đoạn này tí :((
\(=\left(1+1+....+1\right)+\frac{1}{2}.\left(\frac{1}{1}-\frac{1}{3}+\frac{1}{2}-\frac{1}{4}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{49}-\frac{1}{51}\right)\)(49 chữ số 1)
\(=49+\frac{1}{2}.\left[\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{49}\right)-\left(\frac{1}{3}+\frac{1}{4}+...+\frac{1}{51}\right)\right]\)
\(=49+\left(\frac{3}{2}-\frac{1}{50}-\frac{1}{51}\right):2\)Tự tính
9 tháng 4 2017
2A=\(\left(1+\frac{1}{3}\right)\)\(\left(1+\frac{1}{8}\right)\)\(\left(1+\frac{1}{15}\right)\)\(.......\)\(\left(1+\frac{1}{4064255}\right)\)
2A = \(\frac{4}{3}\)\(.\)\(\frac{9}{8}\)\(.\)\(\frac{16}{15}\)\(......\)\(\frac{4064256}{4064255}\)
2A = \(\frac{2.2}{1.3}\)\(.\)\(\frac{3.3}{2.4}\)\(.\)\(\frac{4.4}{3.5}\)\(......\)\(\frac{2016.2016}{2015.2017}\)
2A = \(\frac{2.3.4....2016}{1.2.3.....2015}\)\(.\)\(\frac{2.3.4....2016}{3.4.5....2017}\)
2A = \(\frac{2016}{1}\)\(.\)\(\frac{2}{2017}\)
2A = \(\frac{4032}{2017}\)
A = \(\frac{4032}{2017}\)\(:2\)
A = \(\frac{2016}{2017}\)
8 tháng 11 2017
\(A=\frac{1^2}{1.3}+\frac{2^2}{3.5}+...+\frac{1006^2}{2011.2013}\)
\(\Leftrightarrow4A=\frac{2^2.1^2}{2^2-1}+\frac{2^2.2^2}{4^2-1}+...+\frac{2^2.1006^2}{2012^2-1}\)
\(=1006+\left(\frac{1}{1.3}+\frac{1}{3.5}+...+\frac{1}{2011.2013}\right)\)
\(=1006+\frac{1}{2}\left(\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{2011}-\frac{1}{2013}\right)\)
\(=1006+\frac{1}{2}\left(1-\frac{1}{2013}\right)=\frac{2026084}{2013}\)
\(\Rightarrow A=\frac{506521}{2013}\)
TH
3
13 tháng 5 2018
\(A=\frac{2^2}{1.3}\cdot\frac{2^2}{2.4}\cdot\frac{2^2}{3.5}\cdot\frac{2^2}{4.6}\)
\(A=\frac{4}{3}\cdot\frac{1}{2}\cdot\frac{4}{15}\cdot\frac{1}{6}\)
\(A=\frac{4.1.4.1}{3.2.15.6}\)
\(A=\frac{4}{135}\)
13 tháng 5 2018
\(\frac{2^2}{1.3}.\frac{3^2}{2.4}.\frac{4^2}{3.5}.\frac{5^2}{4.6}\)
\(=\frac{2.2}{1.3}.\frac{3.3}{2.4}.\frac{4.4}{3.5}.\frac{5.5}{4.6}\)
\(=\frac{2.3.4.5}{1.2.3.4}.\frac{2.3.4.5}{3.4.5.6}\)
\(=\frac{5}{1}.\frac{2}{6}\)
\(=\frac{5}{1}.\frac{1}{3}\)
\(=\frac{5}{3}\)
\(B=\frac{2^2}{1.3}.\frac{3^2}{2.4}.\frac{4^2}{3.5}......\frac{10^2}{9.11}=\frac{\left(1.2.3.....10\right)^2}{\left(1.2.3.....9\right).\left(3.4.5....9.10.11\right)}=\frac{\left(1.2.3....10\right)^2}{\left(1.2\right)\left(3.4.5.....9\right)^2\left(10.11\right)}=\frac{\left(1.2.10\right)^2}{\left(1.2\right).\left(10.11\right)}=\frac{1.2.10}{11}=\frac{20}{11}\)