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\(=\frac{4}{5.7}+\frac{4}{7.9}+\frac{4}{9.11}+...+\frac{4}{59.61}\)
\(=2.\left(\frac{2}{5.7}+\frac{2}{7.9}+...+\frac{2}{59.61}\right)\)
\(=2.\left(\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11}+...+\frac{1}{59}-\frac{1}{61}\right)\)
=\(2.\left(\frac{1}{5}-\frac{1}{61}\right)\)
\(=2.\left(\frac{36}{505}\right)\)
\(=\frac{72}{505}\)
TK nha !!
Ta có : \(\frac{4}{5.7}+\frac{4}{7.9}+\frac{4}{9.11}+....+\frac{4}{59.61}\)
\(=2\left(\frac{2}{5.7}+\frac{2}{7.9}+\frac{2}{9.11}+.....+\frac{2}{59.61}\right)\)
\(=2\left(\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+.....+\frac{1}{59}-\frac{1}{61}\right)\)
\(=2\left(\frac{1}{5}-\frac{1}{61}\right)\)
\(=2.\frac{56}{305}=\frac{112}{305}\)
\(\frac{4}{1\cdot3\cdot5}+\frac{4}{3\cdot5\cdot7}+\frac{4}{5\cdot7\cdot9}+\frac{4}{7\cdot9\cdot11}+\frac{4}{9\cdot11\cdot13}\)
\(=\frac{1}{1.3}-\frac{1}{3.5}+\frac{1}{3.5}-\frac{1}{5.7}+...+\frac{1}{9.11}-\frac{1}{11.13}\)
\(=\frac{1}{1.3}-\frac{1}{11.13}\)
\(=\frac{1}{3}-\frac{1}{143}\)
\(=\frac{140}{429}\)
\(\frac{5}{5.7}+\frac{5}{7.9}+...+\frac{5}{59.61}\)
\(=\frac{5}{2}\left(\frac{2}{5.7}+\frac{2}{7.9}+...+\frac{2}{59.61}\right)\)
\(=\frac{5}{2}\left(\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{59}-\frac{1}{61}\right)\)
\(=\frac{5}{2}\left(\frac{1}{5}-\frac{1}{61}\right)=\frac{5}{2}.\frac{56}{305}=\frac{28}{61}\)
\(=\frac{3}{2}\left(\frac{2}{5.7}+\frac{2}{7.9}+.....+\frac{2}{59.61}\right)\)
\(=\frac{3}{2}\left(\frac{1}{5}-\frac{1}{61}\right)\)
\(=\frac{3}{2}.\frac{56}{305}\)
\(=\frac{84}{305}\)
tk cho minh nhe >.<
Ta có: \(\frac{2}{3}\times\left(\frac{3}{5.7}+\frac{3}{7.9}+.....+\frac{3}{59.61}\right):\frac{2}{3}\)
\(\Rightarrow\left(\frac{2}{5.7}+\frac{2}{7.9}+.....+\frac{2}{59.61}\right):\frac{2}{3}\)
\(=\left(\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+....+\frac{1}{59}-\frac{1}{61}\right):\frac{2}{3}\)
\(\Rightarrow\left(\frac{1}{5}-\frac{1}{61}\right):\frac{2}{3}=\frac{56}{305}:\frac{2}{3}=\frac{84}{305}\)
1) \(A=\frac{4}{5.7}+\frac{4}{7.9}+...+\frac{4}{59.61}\)
\(A=2.\left(\frac{2}{5.7}+\frac{2}{7.9}+...+\frac{2}{59.61}\right)\)
\(A=2.\left(\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+..+\frac{1}{59}-\frac{1}{61}\right)\)
tiếp theo bạn tính kết quả trong ngoặc rồi nhân với 2 là ra kết quả của phần 1
phần 2 tách 3^2 = 3.3 sau đó lấy thừa số chung là 3,tiếp theo làm như phần 1 là ra kết quả
\(\frac{3}{5.7}+\frac{3}{7.9}+\frac{3}{9.11}+...+\frac{3}{59.61}\)
\(=\)\(\frac{3}{2}\left(\frac{2}{5.7}+\frac{2}{7.9}+\frac{2}{9.11}+...+\frac{2}{59.61}\right)\)
\(=\)\(\frac{3}{2}\left(\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11}+...+\frac{1}{59}-\frac{1}{61}\right)\)
\(=\)\(\frac{3}{2}\left(\frac{1}{5}-\frac{1}{61}\right)\)
\(=\)\(\frac{3}{2}.\frac{56}{305}\)
\(=\)\(\frac{84}{305}\)
Chúc bạn học tốt ~
\(\frac{19}{37}+\left(1-\frac{19}{37}\right)\)
\(=\frac{19}{37}+1-\frac{19}{37}\)
\(=\left(\frac{19}{37}-\frac{19}{37}\right)+1\)
\(=0+1=1\)
tớ ko chép lại đề đâu
\(\frac{1}{2}M=\frac{2}{5.7}+\frac{2}{7.9}+\frac{2}{9.11}+...+\frac{2}{101.103}\)
\(=\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11}+...+\frac{1}{101}-\frac{1}{103}\)
\(=\frac{1}{5}-\frac{1}{103}\)
=\(\frac{98}{515}\)
=> \(M=\frac{98}{515}:\frac{1}{2}=\frac{196}{515}\)
Vậy \(M=\frac{196}{515}\)
1/2A=2/5.7+2/7.9+...+2/2017.2019
1/2A=1/5-1/7+1/7-1/9+...+1/2017-1/2019
1/2A=1/5-1/2019
1/2A=2014/10095
A=4028/10095
M = \(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+\frac{2}{9.11}\)
M = \(\frac{2}{1}-\frac{2}{3}+\frac{2}{3}-\frac{2}{5}+\frac{2}{5}-\frac{2}{7}+\frac{2}{7}-\frac{2}{9}+\frac{2}{9}-\frac{2}{11}\)
M = \(\frac{2}{1}-\frac{2}{11}\)
M = \(\frac{20}{11}\)
112 phần 305
= \(2.\left(\frac{2}{5.7}+\frac{2}{7.9}+\frac{2}{9.11}+\cdots+\frac{2}{59.61}\right)\)
= \(2\left(\frac15-\frac17+\frac17-\frac19+\frac19-\frac{1}{11}+.\ldots+\frac{1}{59}-\frac{1}{61}\right)\)
\(=2\left(\frac15-\frac{1}{61}\right)\)
= \(2\cdot\frac{56}{305}=\frac{112}{305}\)
Khó thế 👀