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\(A=\frac{1}{15}+\frac{1}{35}+...+\frac{1}{2499}\)
\(A=\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{49.51}\)
\(A=\frac{1}{2}\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{49}-\frac{1}{51}\right)\)
\(A=\frac{1}{2}\left(\frac{1}{3}-\frac{1}{51}\right)\)
\(A=\frac{1}{2}.\frac{16}{51}\)
\(A=\frac{8}{51}\)
\(A=\frac{1}{15}+\frac{1}{35}+...+\frac{1}{2499}\)
\(A=\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{49.51}\)
\(2A=\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{49.51}\)
\(2A=\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{49}-\frac{1}{50}\)
\(2A=\frac{1}{3}-\frac{1}{50}\)
\(A=\frac{1}{2}.\left(\frac{1}{3}-\frac{1}{50}\right)\)
\(A=\frac{1}{2}.\frac{1}{3}-\frac{1}{2}.\frac{1}{50}\)
\(A=\frac{1}{6}-\frac{1}{100}=\frac{50}{300}-\frac{3}{300}=\frac{47}{300}\)
=1/2(2/3.5 + 2/5.7 +.....+2/49.51
=1/2(1/3 - 1/5+1/5-1/7+....+1/49-1/51)
=1/2(1/3-1/51)
=1/2.16/51
=8/51
HỌC TỐT NHÉ BẠN!
\(=\dfrac{2}{1\cdot3}+\dfrac{2}{3\cdot5}+...+\dfrac{2}{13\cdot15}\)
\(=1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+...+\dfrac{1}{13}-\dfrac{1}{15}=\dfrac{14}{15}\)
Ta có : M=1/3+1/15+1/35+...+1/2499
=> M=1/1.3+1/3.5+1/5.7+...+1/49.51
=>2M=2/1.3+2/3.5+2/5.7+...+2/49.51
=>2M=1-1/3+1/3-1/5+1/5-1/7+...+1/49-1/51
=>2M=1-1/51
=>2M=50/51
=> M=50/51:2
=> M=25/51
Vậy M=25/51
Mọi người tk giúp mik nha mik đang âm điểm nek !!!
\(E=\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{49.51}\)
\(=\frac{1}{2}\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{49}-\frac{1}{51}\right)\)
\(=\frac{1}{2}\left(\frac{1}{3}-\frac{1}{51}\right)=\frac{1}{2}.\frac{16}{51}=\frac{8}{51}\)
Tham khảo
=1/2(2/3.5 + 2/5.7 +.....+2/49.51
=1/2(1/3 - 1/5+1/5-1/7+....+1/49-1/51)
=1/2(1/3-1/51)
=1/2.16/51
=8/51
Tham khảo
=1/2(2/3.5 + 2/5.7 +.....+2/49.51
=1/2(1/3 - 1/5+1/5-1/7+....+1/49-1/51)
=1/2(1/3-1/51)
=1/2.16/51
=8/51
C=1/15+1/35+1/63+..+1/2499
=1/3.5+1/5.7+1/7.9+...+1/49.51
=1/2(2/3.5+2/5.7+2/7.9+...+2/49.51)
=1/2(1/3-1/5+1/5-1/7+1/7-1/9+...+1/49-1/51)
= 1/2.(1/3-1/51)
=1/2.16/51
=8/51
\(\frac{1}{15}+\frac{1}{35}+...+\frac{1}{2499}\)
= \(\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{49.51}\)
= \(\frac{1}{2}\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{49}-\frac{1}{51}\right)\)
= \(\frac{1}{2}\left(\frac{1}{3}-\frac{1}{51}\right)\)
= \(\frac{1}{2}.\frac{16}{51}=\frac{8}{51}\)
Bài giải:
\(\frac{1}{15}+\frac{1}{35}+...+\frac{1}{2499}\)
\(=\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{49.51}\)
\(=\frac{1}{2}\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{49}-\frac{1}{51}\right)\)
\(=\frac{1}{2}\left(\frac{1}{3}-\frac{1}{51}\right)\)
\(=\frac{8}{51}\)
\(M=\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{49.51}\)
\(M=\frac{1}{2}\left(1-\frac{1}{51}\right)\)
M=\(\frac{1}{2}.\frac{50}{51}=\frac{25}{51}\)
\(M=\frac{1}{3}+\frac{1}{15}+\frac{1}{35}+...+\frac{1}{2499}\)
\(M=\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{49.51}\)
\(M=\frac{1}{2}.\left(\frac{1}{1}-\frac{1}{3}+...+\frac{1}{49}-\frac{1}{51}\right)\)
\(M=\frac{1}{2}.\left(\frac{1}{1}-\frac{1}{51}\right)\)
\(M=\frac{1}{2}.\frac{50}{51}\)
\(M=\frac{25}{51}\)