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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 !!!
\(2\frac{1}{6}+x:\frac{6}{7}=-75\%\)
\(\frac{13}{6}+x:\frac{6}{7}=-\frac{3}{4}\)
\(x:\frac{6}{7}=-\frac{3}{4}-\frac{13}{6}=-\frac{35}{12}\)
\(x=-\frac{35}{12}x\frac{6}{7}=\frac{5}{2}\)
vậy \(x=\frac{5}{2}\)
\(2\frac{1}{6}+x\div\frac{6}{7}\)\(=-75\%\)
\(\frac{13}{6}+x\div\frac{6}{7}\)\(=-0,75\)
\(\frac{13}{6}+x\div\frac{6}{7}\) \(=-0,75\)
\(\frac{13}{6}+x\) \(=-0,75\times\frac{6}{7}\)
\(\frac{13}{6}+x\) \(=\frac{-9}{14}\)
\(x\) \(=\frac{-9}{14}-\frac{13}{6}\)
\(x\) \(=\frac{-59}{21}\)
Tk cho mk nhé!
đặt \(A=\frac{1}{1.2.3}+\frac{1}{2.3.4}+\frac{1}{3.4.5}+...+\frac{1}{18.19.20}\)
\(=\frac{1}{2}\left(\frac{1}{1.2}-\frac{1}{2.3}\right)+\frac{1}{2}\left(\frac{1}{2.3}-\frac{1}{3.4}\right)+...+\frac{1}{2}\left(\frac{1}{18.19}-\frac{1}{19.20}\right)\)
\(=\frac{1}{2}\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+...+\frac{1}{18.19}-\frac{1}{19.20}\right)\)
\(=\frac{1}{2}\left(\frac{1}{2}-\frac{1}{380}\right)=\frac{189}{760}\)
Đặt \(B=\frac{3}{1.2}+\frac{3}{2.3}+...+\frac{3}{19.20}=\frac{3}{1}-\frac{3}{2}+\frac{3}{2}-\frac{3}{3}+...+\frac{3}{19}-\frac{3}{20}\)
\(=3-\frac{3}{20}=\frac{57}{20}\)
\(D=A-B=\frac{189}{760}-\frac{57}{20}=-\frac{1977}{760}\)
Gọi \(\frac{1}{1.2.3}+\frac{1}{2.3.4}+...+\frac{1}{18.19.20}\)là A
\(\frac{3}{1.2}-\frac{3}{2.3}-...-\frac{3}{19.20}\)là B
\(A=\left[\frac{1}{2}.\left(\frac{1}{1.2}-\frac{1}{2.3}\right)+\frac{1}{2}.\left(\frac{1}{2.3}-\frac{1}{3.4}\right)+...+\frac{1}{2}.\left(\frac{1}{18.19}-\frac{1}{19.20}\right)\right]\)
\(A=\left[\frac{1}{2}.\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{18.19}-\frac{1}{19.20}\right)\right]\)
\(A=\left[\frac{1}{2}.\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{19}-\frac{1}{20}\right)\right]\)
\(A=\left[\frac{1}{2}.\left(1-\frac{1}{20}\right)\right]\)
\(A=\frac{1}{2}.\frac{19}{20}\)
\(A=\frac{19}{40}\)
\(B=\frac{3}{1.2}-\frac{3}{2.3}-...-\frac{3}{19.20}\)
\(B=\left(\frac{3}{1.2}+\frac{3}{2.3}+...+\frac{3}{19.20}\right)\)
\(B=\left[3.\left(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{19.20}\right)\right]\)
\(B=\left[3.\left(1-\frac{1}{2}+\frac{1}{2}-\frac{2}{3}+...+\frac{1}{19}-\frac{1}{20}\right)\right]\)
\(B=\left[3.\left(\frac{19}{20}\right)\right]\)
\(B=\frac{57}{20}\)
Vậy A - B = \(\frac{19}{40}-\frac{57}{20}\)
\(=-\frac{95}{40}=-\frac{19}{8}\)
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