\(\frac{3}{5\cdot7}+\frac{3}{7\cdot9}+........+\frac{3}{59\cdot61}\)
Tính tổng
\(\frac{5}{5\cdot7}+\frac{5}{7\cdot9}+...+\frac{5}{59\cdot61}\)
\(\frac{5}{5.7}+\frac{5}{7.9}+...+\frac{5}{59.61}\)
\(=\frac{5}{2}\left(\frac{2}{5.7}+\frac{2}{7.9}+...+\frac{2}{59.61}\right)\)
\(=\frac{5}{2}\left(\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{59}-\frac{1}{61}\right)\)
\(=\frac{5}{2}\left(\frac{1}{5}-\frac{1}{61}\right)=\frac{5}{2}.\frac{56}{305}=\frac{28}{61}\)
\(\frac{3}{5\cdot7}\)+ \(\frac{3}{7\cdot9}\) + ......+\(\frac{3}{59\cdot61}\)
S=\(\frac{1}{2}\) + \(\frac{1}{2^2}\) +......+\(\frac{1}{2^{20}}\)
\(\frac{3}{5.7}+\frac{3}{7.9}+\frac{3}{9.11}+...+\frac{3}{59.61}\)
\(=\)\(\frac{3}{2}\left(\frac{2}{5.7}+\frac{2}{7.9}+\frac{2}{9.11}+...+\frac{2}{59.61}\right)\)
\(=\)\(\frac{3}{2}\left(\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11}+...+\frac{1}{59}-\frac{1}{61}\right)\)
\(=\)\(\frac{3}{2}\left(\frac{1}{5}-\frac{1}{61}\right)\)
\(=\)\(\frac{3}{2}.\frac{56}{305}\)
\(=\)\(\frac{84}{305}\)
Chúc bạn học tốt ~
\(\frac{3}{5.7}+\frac{3}{7.9}+...+\frac{3}{59.61}\)
\(=\frac{3}{2}.\left(\frac{2}{5.7}+\frac{2}{7.9}+...+\frac{2}{59.61}\right)\)
\(=\frac{3}{2}.\left(\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{59}-\frac{1}{61}\right)\)
\(=\frac{3}{2}.\left(\frac{1}{5}-\frac{1}{61}\right)\)
\(=\frac{3}{2}.\left(\frac{61}{305}-\frac{5}{305}\right)\)
\(=\frac{3}{2}.\frac{56}{305}\)
\(=\frac{84}{305}\)
\(S=\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{20}}\)
\(\Rightarrow2S=1+\frac{1}{2}+...+\frac{1}{2^{19}}\)
\(\Rightarrow2S-S=1-\frac{1}{2^{20}}\)
\(\Rightarrow S=1-\frac{1}{2^{20}}\)
Chúc bạn học tốt !!!
giải chi tiết giúp mình nha mình cần gấp lắm mình cảm ơn nhìu
\(\frac{4}{5\cdot7}+\frac{4}{7\cdot9}+\frac{4}{9\cdot11}+...+\frac{4}{59\cdot61}\)
\(=\frac{4}{5.7}+\frac{4}{7.9}+\frac{4}{9.11}+...+\frac{4}{59.61}\)
\(=2.\left(\frac{2}{5.7}+\frac{2}{7.9}+...+\frac{2}{59.61}\right)\)
\(=2.\left(\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11}+...+\frac{1}{59}-\frac{1}{61}\right)\)
=\(2.\left(\frac{1}{5}-\frac{1}{61}\right)\)
\(=2.\left(\frac{36}{505}\right)\)
\(=\frac{72}{505}\)
TK nha !!
Ta có : \(\frac{4}{5.7}+\frac{4}{7.9}+\frac{4}{9.11}+....+\frac{4}{59.61}\)
\(=2\left(\frac{2}{5.7}+\frac{2}{7.9}+\frac{2}{9.11}+.....+\frac{2}{59.61}\right)\)
\(=2\left(\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+.....+\frac{1}{59}-\frac{1}{61}\right)\)
\(=2\left(\frac{1}{5}-\frac{1}{61}\right)\)
\(=2.\frac{56}{305}=\frac{112}{305}\)
Mk nhân lộn ở cái phần \(2.\left(\frac{1}{5}-\frac{1}{61}\right)\)
\(=2.\left(\frac{56}{305}\right)\)
\(=\frac{112}{305}\)
Tk nha cảm ơn !!
\(\frac{3}{5\cdot7}+\frac{3}{7\cdot9}+...+\frac{5}{53\cdot55}\)
\(\frac{3}{5\cdot7}+\frac{3}{7\cdot9}+...+\frac{5}{53\cdot55}\)
\(=\frac{3}{2}\cdot\left(\frac{2}{5\cdot7}+\frac{2}{7\cdot9}+...+\frac{2}{53\cdot55}\right)\)
\(=\frac{3}{2}\cdot\left(\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{53}-\frac{1}{55}\right)\)
\(=\frac{3}{2}\cdot\left(\frac{1}{5}-\frac{1}{55}\right)\)
\(=\frac{3}{2}\cdot\left(\frac{11}{55}-\frac{1}{55}\right)\)
\(=\frac{3}{2}\cdot\frac{10}{55}\)
\(=\frac{3}{2}\cdot\frac{2}{11}\)
\(=\frac{3}{11}\)
\(\frac{4}{1\cdot3\cdot5}+\frac{4}{3\cdot5\cdot7}+\frac{4}{5\cdot7\cdot9}+\frac{4}{7\cdot9\cdot11}+\frac{4}{9\cdot11\cdot13}\)
giúp mk nha các bn
\(\frac{4}{1\cdot3\cdot5}+\frac{4}{3\cdot5\cdot7}+\frac{4}{5\cdot7\cdot9}+\frac{4}{7\cdot9\cdot11}+\frac{4}{9\cdot11\cdot13}\)
\(=\frac{1}{1.3}-\frac{1}{3.5}+\frac{1}{3.5}-\frac{1}{5.7}+...+\frac{1}{9.11}-\frac{1}{11.13}\)
\(=\frac{1}{1.3}-\frac{1}{11.13}\)
\(=\frac{1}{3}-\frac{1}{143}\)
\(=\frac{140}{429}\)
\(\frac{1}{3\cdot5}+\frac{1}{5\cdot7}+\frac{1}{7\cdot9}+......+\frac{1}{97\cdot99}\)
đặt \(\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+...+\frac{1}{97.99}\) là A
A = \(\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+...+\frac{1}{97.99}\)
2A = 2 . (\(\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+...+\frac{1}{97.99}\))
2A = \(\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+...+\frac{2}{97.99}\)
2A = \(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{97}-\frac{1}{99}\)
2A = \(\frac{1}{3}-\frac{1}{99}\)
2A = \(\frac{33}{99}-\frac{1}{99}\)
2A = \(\frac{32}{99}\)
A = \(\frac{32}{99}:2\)
A = \(\frac{32}{99}.\frac{1}{2}\)
A = \(\frac{32}{198}\)
A = \(\frac{16}{99}\)
HỌC TỐT
Tính (theo mẫu)
Mẫu:\(\frac{5\cdot6\cdot7\cdot9}{12\cdot7\cdot27}\)=\(\frac{5\cdot6\cdot7\cdot9}{6\cdot2\cdot7\cdot9\cdot3}\)=\(\frac{5}{6}\)
a.\(\frac{3\cdot4\cdot7}{12\cdot8\cdot9}\)=.........................................................................
b.\(\frac{4\cdot5\cdot6}{12\cdot10\cdot8}\)=.......................................................................
c\(\frac{5\cdot6\cdot7}{12\cdot14\cdot15}\)=......................................................................
a.\(\frac{3\cdot4\cdot7}{12\cdot8\cdot9}\)= \(\frac{3\cdot4\cdot7}{3\cdot4\cdot8\cdot9}\)= \(\frac{7}{72}\)
b. \(\frac{4\cdot5\cdot6}{12\cdot10\cdot8}\)= \(\frac{4\cdot5\cdot2\cdot3}{3\cdot4\cdot5\cdot2\cdot8}\)= \(\frac{1}{8}\)
c.\(\frac{5\cdot6\cdot7}{12\cdot14\cdot15}\)= \(\frac{5\cdot6\cdot7}{2\cdot6\cdot2\cdot7\cdot3\cdot5}\)= \(\frac{1}{12}\)
Tính nhanh
1, A = \(\frac{4}{5\cdot7}\)+ \(\frac{4}{7\cdot9}\) + ..... + \(\frac{4}{59\cdot61}\)
2, B = \(\frac{3^2}{2\cdot5}\) + \(\frac{3^2}{58}\) + \(\frac{3^2}{8\cdot11}\) + ..... + \(\frac{3^2}{38\cdot41}\)
3, C = ( \(\frac{1}{2}\) + 1 ) . ( \(\frac{1}{3}\)+ 1 ) . ( \(\frac{1}{4}\)+ 1 ) . .... . ( \(\frac{1}{99}\)+ 1 )
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
Cho \(B=\frac{2^3}{3\cdot5}+\frac{2^3}{5\cdot7}+\frac{2^3}{7\cdot9}+...+\frac{2^3}{101\cdot103}\)
Tính B
B=22(\(\frac{2}{3.5}\)+\(\frac{2}{5.7}\)+\(\frac{2}{7.9}\)+...+\(\frac{2}{101.103}\))
B=4[1/3-1/5+1/5-1/7+1/7-1/9 +...+1/101-1/103]
B=4[1/3-1/103]
B=4.(100/309)
B=400/309