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a) 6B = 2.4.6 + 4.6.(8-2) + 6.8.(10-4) + ... + 18.20.(22-16)
6B = 2.4.6 + 4.6.8 - 2.4.6 + 6.8.10 - 4.6.8 +...+ 18.20.22 - 16.18.20
6B = 18.20.
B = (18.20.22) : 6
B = 1320
Mấy bài kia tương tự, cần giải luôn không bạn? Nhưng hơi mất thời gian
a) \(A=2.4+4.6+6.8+...+18.20\)
\(6A=2.4.6+4.6.\left(8-2\right)+6.8.\left(10-4\right)+...+18.20.\left(22-16\right)\)
\(6A=2.4.6+4.6.8-2.4.6+6.8.10-4.6.8+...+18.20.22-16.18.20\)
\(6A=18.20.22\)
\(A=\frac{18.20.22}{6}=\frac{7920}{6}=1320\)
d/ Đặt : A = 1.2 + 2.3 + 3.4 + ......... + 99.100
=> 3A = 1.2.(3 - 0) + 2.3.(4 - 1) + ..... + 99.100.(101 - 98)
=> 3A = 1.2.3 - 1.2.3 + 2.3.4 - 2.3.4 + ..... + 99.100.101
=> 3A = 99.100.101
=> A = 99.100.101 / 3
=> A = 333300
a) =1-1/3+1/3-1/5+1/5-1/7+...+1/99-1/101
=1-1/101
=100/101
b) =(2/1.3+2/3.5+2/5.7+...+2/99.101).2,5
=(1-1/3+1/3-1/5+1/5-1/7+...+1/99-1/101).2,5
=(1-1/101).2,5
=100/101.2,5
=250/101
c) =(2/2.4+2/4.6+2/6.8+...+2/2008-2/2010).2
=(1/2-1/4+1/4-1/6+1/6-1/8+...+1/2008-1/2010).2
=(1/2-1/2010).2
=1004/1005
\(S=\frac{1}{1.3}-\frac{1}{2.4}+\frac{1}{3.5}-\frac{1}{4.6}+\frac{1}{5.7}-\frac{1}{6.8}+\frac{1}{7.9}-\frac{1}{8.10}\)
\(S=\frac{1}{2}\left(1-\frac{1}{3}+\frac{1}{2}-\frac{1}{4}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{7}-\frac{1}{9}=\frac{1}{8}-\frac{1}{10}\right)\)
\(S=\frac{1}{2}\left(1+\frac{1}{2}-\frac{1}{9}-\frac{1}{10}\right)\)
\(S=\frac{1}{2}.\left(\frac{58}{45}\right)\)
\(S=\frac{29}{45}\)
\(S=\frac{1}{1.3}-\frac{1}{2.4}+\frac{1}{3.5}-\frac{1}{4.6}+\frac{1}{5.7}-\frac{1}{6.8}+\frac{1}{7.9}-\frac{1}{8.10}\)
\(=\left(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}\right)-\left(\frac{1}{2.4}-\frac{1}{4.6}-\frac{1}{6.8}-\frac{1}{8.10}\right)\)
\(=\frac{1}{2}\left(1-\frac{1}{3}+\frac{1}{3}-...+\frac{1}{7}-\frac{1}{9}\right)-\frac{1}{2}\left(\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-...+\frac{1}{8}-\frac{1}{10}\right)\)
\(=\frac{1}{2}\left(1-\frac{1}{9}\right)-\frac{1}{2}\left(\frac{1}{2}-\frac{1}{10}\right)\)
\(=\frac{1}{2}.\frac{8}{9}-\frac{1}{2}.\frac{2}{5}\)
\(=\frac{4}{9}-\frac{1}{5}\)
\(=\frac{11}{45}\)
\(A=\frac{3}{2\cdot4}+\frac{3}{4\cdot6}+...+\frac{3}{48\cdot50}\)---> Mik nghĩ bn ghi nhầm :]
\(A=\frac{3}{2}\left[\frac{1}{2\cdot4}+\frac{1}{4\cdot6}+...+\frac{1}{48\cdot50}\right]\)
\(A=\frac{3}{2}\left[\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+...+\frac{1}{48}-\frac{1}{50}\right]\)
\(A=\frac{3}{2}\left[\frac{1}{2}-\frac{1}{50}\right]=\frac{3}{2}\cdot\frac{12}{25}=\frac{18}{25}\)
Vậy A = 18/25
\(B=\frac{5}{1\cdot3}+\frac{5}{3\cdot5}+...+\frac{5}{49\cdot51}\)
\(B=\frac{5}{2}\left[\frac{1}{1\cdot3}+\frac{1}{3\cdot5}+...+\frac{1}{49\cdot51}\right]\)
\(B=\frac{5}{2}\left[\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{49}-\frac{1}{51}\right]\)
\(B=\frac{5}{2}\left[1-\frac{1}{51}\right]=\frac{5}{2}\cdot\frac{50}{51}=\frac{125}{51}\)
Ta có:
\(\frac{1}{1.3}+\frac{1}{2.4}+\frac{1}{3.5}+\frac{1}{4.6}+\frac{1}{5.7}+\frac{1}{6.8}+\frac{1}{7.9}+\frac{1}{8.10}\)
\(=\frac{1}{2}.\left(\frac{1}{1}-\frac{1}{3}+\frac{1}{2}-\frac{1}{4}+\frac{1}{3}-\frac{1}{5}+....+\frac{1}{8}-\frac{1}{10}\right)\)
\(=\frac{1}{2}.\left(\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}....+\frac{1}{7}-\frac{1}{9}\right)+\frac{1}{2}.\left(\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+...+\frac{1}{8}-\frac{1}{10}\right)\)
\(=\frac{1}{2}.\frac{8}{9}+\frac{1}{2}.\frac{2}{5}=\frac{1}{2}.\left(\frac{8}{9}+\frac{2}{5}\right)=\frac{1}{2}.\frac{58}{45}=\frac{29}{45}\)