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\(A=\frac{2^2.3^2.4^2............99^2}{1.3.2.4.3.5................998.1000}\)
\(A=\frac{1.2.3.4.5................999.1.2.3.4................999}{1.2.3.4.5.6.7..........1000.1.2.3.4..............998}\)
\(A=\frac{999.999}{1000.998}\)
\(Ko\) \(\text{chắc lắm}\)
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
\(A=\frac{2^2}{1.3}\cdot\frac{2^2}{2.4}\cdot\frac{2^2}{3.5}\cdot\frac{2^2}{4.6}\)
\(A=\frac{4}{3}\cdot\frac{1}{2}\cdot\frac{4}{15}\cdot\frac{1}{6}\)
\(A=\frac{4.1.4.1}{3.2.15.6}\)
\(A=\frac{4}{135}\)
\(\frac{2^2}{1.3}.\frac{3^2}{2.4}.\frac{4^2}{3.5}.\frac{5^2}{4.6}\)
\(=\frac{2.2}{1.3}.\frac{3.3}{2.4}.\frac{4.4}{3.5}.\frac{5.5}{4.6}\)
\(=\frac{2.3.4.5}{1.2.3.4}.\frac{2.3.4.5}{3.4.5.6}\)
\(=\frac{5}{1}.\frac{2}{6}\)
\(=\frac{5}{1}.\frac{1}{3}\)
\(=\frac{5}{3}\)
\(A=\frac{2^2}{1.3}\cdot\frac{3^2}{2.4}....\frac{999^2}{998.1000}\)
\(A=\frac{2^2.3^2....999^2}{1.3.2.4.998.100}=\frac{\left(2.3.....999\right)\left(2.3....999\right)}{\left(1.2....998\right)\left(3.4....1000\right)}\)
\(A=999\cdot\frac{1}{500}=\frac{999}{500}\)( khúc này mk làm tắt, bn bỏ dấu ở trên rồi bỏ từng tử)
=?????????????????????????????????????????????????????????????????????????????????????????????????????????????????
A = \(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{49.50}=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{49}-\frac{1}{50}\)
\(=1-\frac{1}{50}=\frac{49}{50}\)
B = \(\frac{2^2}{1.3}.\frac{3^2}{2.4}.\frac{4^2}{3.5}.\frac{5^2}{4.6}=\frac{\left(2.3.4.5\right).\left(2.3.4.5\right)}{\left(1.2.3.4\right).\left(3.4.5.6\right)}=\frac{5.2}{1.6}=\frac{5}{3}\)
C = \(\frac{3}{5.7}+\frac{3}{7.9}+...+\frac{3}{59.61}=\frac{3}{2}.\left(\frac{2}{5.7}+\frac{2}{7.9}+...+\frac{2}{59.61}\right)\)
\(=\frac{3}{2}.\left(\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{59}-\frac{1}{61}\right)=\frac{3}{2}.\left(\frac{1}{5}-\frac{1}{61}\right)=\frac{3}{2}.\frac{56}{305}=\frac{74}{305}\)
Bài làm:
1) \(A=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{49.50}\)
\(A=\frac{2-1}{1.2}+\frac{3-2}{2.3}+\frac{4-3}{3.4}+...+\frac{50-49}{49.50}\)
\(A=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{49}-\frac{1}{50}\)
\(A=1-\frac{1}{50}=\frac{49}{50}\)
2) \(B=\frac{2^2.3^2.4^2.5^2}{1.2.3^2.4^2.5.6}=\frac{2.5}{6}=\frac{5}{3}\)
3) \(C=\frac{3}{5.7}+\frac{3}{7.9}+...+\frac{3}{59.61}\)
\(C=\frac{3}{2}\left(\frac{2}{5.7}+\frac{2}{7.9}+...+\frac{2}{59.61}\right)\)
\(C=\frac{3}{2}\left(\frac{7-5}{5.7}+\frac{9-7}{7.9}+...+\frac{61-59}{59.61}\right)\)
\(C=\frac{3}{2}\left(\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{59}-\frac{1}{61}\right)\)
\(C=\frac{3}{2}\left(\frac{1}{5}-\frac{1}{61}\right)\)
\(C=\frac{3}{2}.\frac{56}{305}=\frac{84}{305}\)
\(\frac{2^2}{1.3}.\frac{3^2}{2.4}.\frac{4^2}{3.5}.....\frac{50^2}{49.51}=\frac{2.2}{1.3}.\frac{3.3}{2.4}.\frac{4.4}{3.5}.....\frac{50.50}{49.51}\)
\(=\frac{2.2.3.3.4.4......50.50}{1.3.2.4.3.5....49.51}=\frac{\left(2.3.4.....50\right).\left(2.3.4......50\right)}{\left(1.2.4.....49\right).\left(3.4.5.....51\right)}\)
\(=\frac{50.2}{1.51}=\frac{100}{51}\)
Cách làm:
tách tử thành 2.2;3.3;4.4;...;50.50
Sau đó ta nhân tử với tử,mẫu với mẫu theo thứ tự chữ số 1 trước như sau:
Tử: 2.3.4...50/1.2.3....49 . 2.3.4...50/3.4.5...51
=50.2/51=100/51
*Cho tôi biết cách viết dấu gạch ngang phân số nhé!
\(I=\frac{2^2}{1.3}.\frac{3^2}{2.4}.\frac{4^2}{3.5}...\frac{999^2}{998.1000}\)
\(I=\frac{2^2.3^2.4^2...999^2}{2.3^2.4^2...998^2.1000}\)
\(I=\frac{2}{1000}=\frac{1}{500}\)
\(=\frac{2\cdot2}{1\cdot3}\cdot\frac{3\cdot3}{2\cdot4}\cdot\frac{4\cdot4}{3\cdot5}\cdot...\cdot\frac{999\cdot999}{998\cdot1000}\)
\(=\frac{2\cdot2\cdot3\cdot3\cdot4\cdot4\cdot...\cdot999\cdot999}{1\cdot3\cdot2\cdot4\cdot3\cdot5\cdot...\cdot998\cdot1000}\)
\(=\frac{2\cdot3\cdot4\cdot...\cdot999}{1\cdot2\cdot3\cdot...\cdot998}\cdot\frac{2\cdot3\cdot4\cdot...\cdot999}{3\cdot4\cdot5\cdot...\cdot1000}\)
\(=\frac{999}{1}\cdot\frac{2}{1000}\)
\(=\frac{999}{500}\)