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1/1.5+/5.9+1/9.13..........+1/101.103
=1-1/5+1/5-1/7+1/9-1/13.........+1/101-1/103
=1-1/103
=102/103
XIN 5 TÍCH VÌ MẤT 5 PHÚT
OK

\(A=\frac{1}{1.5}+\frac{1}{5.9}+\frac{1}{9.13}+...+\frac{1}{93.97}\)
\(A=\frac{1}{4}.\left(\frac{4}{1.5}+\frac{4}{5.9}+\frac{4}{9.13}+...+\frac{1}{93.97}\right)\)
\(A=\frac{1}{4}.\left(1-\frac{1}{5}+\frac{1}{5}-\frac{1}{9}+\frac{1}{9}-\frac{1}{13}+...+\frac{1}{93}-\frac{1}{97}\right)\)
\(A=\frac{1}{4}.\left(1-\frac{1}{97}\right)\)
\(A=\frac{1}{4}.\frac{96}{97}=\frac{24}{97}\)
\(A=\frac{1}{1.5}+\frac{1}{5.9}+\frac{1}{9.13}+...+\frac{1}{93.97}\)
\(A=\frac{1}{4}.\left(\frac{4}{1.5}+\frac{4}{5.9}+\frac{4}{9.13}+...+\frac{1}{93.97}\right)\)
\(A=\frac{1}{4}.\left(1-\frac{1}{5}+\frac{1}{5}-\frac{1}{9}+\frac{1}{9}-\frac{1}{13}+...+\frac{1}{93}-\frac{1}{97}\right)\)
\(A=\frac{1}{4}.\left(1-\frac{1}{97}\right)\)
\(A=\frac{1}{4}.\frac{96}{97}=\frac{24}{97}\)

\(=\frac{1}{4}.\left(\frac{4}{5.9}+\frac{4}{9.13}+...+\frac{4}{37.41}\right)\)
\(=\frac{1}{4}.\left(\frac{1}{5}-\frac{1}{9}+\frac{1}{9}-\frac{1}{13}+...+\frac{1}{37}-\frac{1}{41}\right)\)
\(=\frac{1}{4}.\left(\frac{1}{5}-\frac{1}{41}\right)\)
\(=\frac{1}{4}.\frac{36}{205}=\frac{9}{205}\)
\(\frac{1}{5.9}+\frac{1}{9.13}+\frac{1}{13.17}+...+\frac{1}{37.41}\)
\(=\frac{1}{4}\left(\frac{4}{5.9}+\frac{4}{9.13}+...+\frac{4}{37.41}\right)\)
\(=\frac{1}{4}\left(\frac{1}{5}-\frac{1}{9}+\frac{1}{9}-\frac{1}{13}+...+\frac{1}{37}-\frac{1}{41}\right)\)
\(=\frac{1}{4}\left(\frac{1}{5}-\frac{1}{41}\right)\)
\(=\frac{1}{4}.\frac{36}{205}=\frac{9}{205}\)

\(4S=4.\left(\frac{1}{5.9}+\frac{1}{9.13}+...+\frac{1}{21.25}\right)\)
=\(\frac{4}{5.9}+\frac{4}{9.13}+....+\frac{4}{21.25}_{ }\)
=\(\frac{1}{5}-\frac{1}{9}+\frac{1}{9}-\frac{1}{13}+....+\frac{1}{21}-\frac{1}{23}\)
=\(\frac{1}{5}-\frac{1}{25}=\frac{5}{25}-\frac{1}{25}=\frac{4}{25}\)
=> \(S=\frac{4}{25}:4=\frac{4}{25}.\frac{1}{4}=\frac{1}{25}\)
\(S=\frac{1}{5\times9}+\frac{1}{9\times13}+...+\frac{1}{21\times25}\)
\(S\times4=\frac{4}{5\times9}=\frac{4}{9\times13}+...+\frac{4}{21\times25}\)
\(S\times4=\frac{1}{5}-\frac{1}{9}+\frac{1}{9}-\frac{1}{13}+...+\frac{1}{21}-\frac{1}{25}\)
\(S\times4=\frac{1}{5}-\frac{1}{25}\)
\(S\times4=\frac{4}{25}\)
\(S=\frac{1}{25}\)

4/1.5+4/5.9+4/9.13+....+4/21.25
=1-1/5+1/5-1/9+1/9-1/13+......+1/21-1/25
=1-1/25
=24/25
Tích đúng cho mình nha
\(\frac{4}{1.5}+\frac{4}{5.9}+\frac{4}{9.13}+....+\frac{4}{21.25}\)
\(=1-\frac{1}{5}+\frac{1}{5}-\frac{1}{9}+....+\frac{1}{21}-\frac{1}{25}\)
\(=1-\frac{1}{25}=\frac{24}{25}\)

a)187 + [923 - (923 + 887)]
= 187 + [923 - 923 - 887]
= 187 + 923 - 923 - 887
= 187 + 0 - 887
= 187 - 887
= -700
b)39 . 187 + 302 - 78 . 39
= 39 . 187 + 900 - 78 . 39
= 39 . (187 - 78) + 900
= 39 . 109 + 900
= 4251 + 900
= 5151
a, 187+[923-(923+887)]
=187+[923-923-887]
=187-887
=-700
b, 39.187+302-78.39
=39.(187-78)+900
=39.109+900
=4251+900
=5151
Câu b bạn xem lại đầu bài là 78 hay 87 nha. Là 87 sẽ đúng hơn đó.

Ta có :
\(x+\frac{4}{5.9}+\frac{4}{9.13}+\frac{4}{13.17}+...+\frac{4}{41.45}\)\(=1\)
\(x+\left(\frac{4}{5.9}+\frac{4}{9.13}+\frac{4}{13.17}+...+\frac{4}{41.45}\right)=1\)
\(x+\left(\frac{1}{5}-\frac{1}{9}+\frac{1}{9}-\frac{1}{13}+\frac{1}{13}-\frac{1}{17}+...+\frac{1}{41}-\frac{1}{45}\right)=1\)
\(x+\left(\frac{1}{5}-\frac{1}{45}\right)=1\)
\(x+\frac{8}{45}=1\)
\(x=1-\frac{8}{45}=\frac{37}{45}\)
Ủng hộ mk nha !!! ^_^
giu minh voi
\(\frac{1}{5\cdot9}+\frac{1}{9\cdot13}+...+\frac{1}{49\cdot53}\)
\(=\frac{1}{4}\left(\frac{4}{5\cdot9}+\frac{4}{9\cdot13}+...+\frac{4}{49\cdot53}\right)\)
\(=\frac{1}{4}\left(\frac{1}{5}-\frac{1}{9}+\frac{1}{9}-\frac{1}{13}+...+\frac{1}{49}-\frac{1}{53}\right)\)
\(=\frac{1}{4}\left(\frac{1}{5}-\frac{1}{53}\right)\)
\(=\frac{1}{4}\cdot\frac{48}{265}\)
\(=\frac{12}{265}\)