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\(\frac{1}{5.7}+\frac{1}{7.9}+\frac{1}{9.11}+\frac{1}{49.51}\)
\(=\frac{1}{2}.\left(\frac{2}{5}.7+\frac{2}{7}.9+\frac{2}{9}.11+...+\frac{2}{49}.51\right)\)
\(=\frac{1}{2}.\left(\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-...+\frac{1}{49}-\frac{1}{51}\right)\)
\(=\frac{1}{2}.\left(\frac{1}{5}-\frac{1}{51}\right)\)
\(=\frac{1}{2}.\frac{46}{255}\)
\(=\frac{23}{255}\)
\(\frac{1}{5.7}+\frac{1}{7.9}+...+\frac{1}{49.51}\)
\(\Rightarrow2 \left(\frac{1}{5.7}+\frac{1}{7.9}+\frac{1}{9.11}+...+\frac{1}{49.51}\right)\)
\(\Rightarrow\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-...+\frac{1}{49}-\frac{1}{51}\)
\(\Rightarrow\frac{1}{5}-\frac{1}{51}=\frac{46}{255}\)
Vì biểu thức đã được nhân 2 nên giá trị của biểu thức là:
\(\frac{46}{255}:2=\frac{23}{255}\)
NHẦM GIẢI LẠI :
\(A=\frac{3}{2}.\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{49}-\frac{1}{51}\right)=\frac{3}{2}.\left(\frac{1}{3}-\frac{1}{51}\right)=\frac{3}{2}.\frac{16}{51}=\frac{8}{17}\)
a, Ta có \(A=\frac{3}{3.5}+\frac{3}{5.7}+....+\frac{3}{49.51}\)
\(=\frac{3}{2}.\left(\frac{2}{3.5}+\frac{2}{5.7}+.....+\frac{2}{49.51}\right)\)
\(=\frac{3}{2}.\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+....+\frac{1}{49}-\frac{1}{51}\right)\)
\(=\frac{3}{2}.\left(\frac{1}{3}-\frac{1}{51}\right)\)
\(=\frac{1}{2}-\frac{3}{102}=\frac{48}{102}=\frac{24}{51}\)
b,Ta có \(\frac{1}{2}+\frac{2}{2.4}+\frac{3}{4.7}+\frac{4}{7.11}+\frac{5}{11.16}\)
\(=\frac{2-1}{2}+\frac{4-2}{2.4}+\frac{7-4}{4.7}+\frac{11-7}{7.11}+\frac{16-11}{11.16}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{11}+\frac{1}{11}-\frac{1}{16}\)
\(=\frac{15}{16}\)
!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!
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=1/2.(2/1.3 + 2/3.5 + 2/5.7 +.....+ 2/49.51)
=1/2.(1-1/3+1/3-1/5+1/5-1/7+.....+1/49-1/51)
=1/2.(1-1/51)
=1/2.50/51
=25/51
=1/2.(2/1.3 + 2/3.5 + 2/5.7 +.....+ 2/49.51)
=1/2.(1-1/3+1/3-1/5+1/5-1/7+.....+1/49-1/51)
=1/2.(1-1/51)
=1/2.50/51
=25/51
\(\frac{1}{3\cdot5}+\frac{1}{5\cdot7}+\frac{1}{7\cdot9}+...+\frac{1}{13\cdot15}\)
\(=\frac{1}{2}\left(\frac{2}{3\cdot5}+\frac{2}{5\cdot7}+\frac{2}{7\cdot9}+...+\frac{2}{13\cdot15}\right)\)
\(=\frac{1}{2}\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{13}-\frac{1}{15}\right)\)
\(=\frac{1}{2}\left(\frac{1}{3}-\frac{1}{15}\right)\)
\(=\frac{1}{2}\cdot\frac{4}{15}\)
\(=\frac{2}{15}\)
\(\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+...+\frac{1}{13.15}\)
\(=\frac{1}{2}.\left(\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+...+\frac{1}{13.15}\right)\)
\(=\frac{1}{2}.\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{13}-\frac{1}{15}\right)\)
\(=\frac{1}{2}.\left(\frac{1}{3}-\frac{1}{15}\right)\)
\(=\frac{1}{2}.\)4/15
=2/15
1/5.7 + 1/7.9 + 1/9.11 + ... + 1/49.51
= 1/2 . (2/5.7 + 2/7.9 + 2/9.11 + ... + 2/49.51)
= 1/2 . (1/5 - 1/7 + 1/7 - 1/9 + 1/9 - 1/11 + ... + 1/49 - 1/51)
= 1/2 . (1/5 - 1/51)
= 1/2 . 46/255
= 23/255
S = \(\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11}+...\frac{1}{49}-\frac{1}{51}\)
S = \(\frac{1}{5}-\frac{1}{51}=\frac{46}{255}\)
\(M=\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+\frac{1}{9.11}+\frac{1}{11.13}\)
\(M=\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11}+\frac{1}{11}-\frac{1}{13}\)
\(M=\frac{1}{3}-\frac{1}{13}\)
\(M=\frac{10}{39}\)
\(M=\frac{1}{2}\left(\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+\frac{2}{9.11}+\frac{2}{11.13}\right)\)
\(M=\frac{1}{2}\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{11}-\frac{1}{13}\right)\)
\(M=\frac{1}{2}\left(\frac{1}{3}-\frac{1}{13}\right)\)
\(M=\frac{1}{2}.\frac{10}{39}\)
\(M=\frac{5}{39}\)
tk mk nha bn
\(\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+...+\frac{1}{87.89}\)
= \(\frac{1}{2}.\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{87}-\frac{1}{89}\right)\)
= \(\frac{1}{2}.\left(\frac{1}{3}-\frac{1}{89}\right)\)
= \(\frac{1}{2}.\frac{86}{267}=\frac{43}{267}\)
~~~
Đáp số to quá, tớ không chắc là mình đúng đâu.
#Sunrise
\(\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{49.51}=\frac{1}{2}.\left(\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{49.51}\right)\)
\(=\frac{1}{2}.\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{49}-\frac{1}{51}\right)=\frac{1}{2}.\left(\frac{1}{3}-\frac{1}{51}\right)\)
\(=\frac{1}{2}.\frac{16}{51}=\frac{8}{51}\)
\(=\frac{1}{2}.\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{49}-\frac{1}{51}\right)\)
\(=\frac{1}{2}.\left(\frac{1}{3}-\frac{1}{51}\right)=\frac{1}{2}.\frac{16}{51}=\frac{8}{51}\)