HB
2
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Khách
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25 tháng 2 2017
2.
\(\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+...+\frac{1}{\left(2x+1\right).\left(2x+3\right)}=\frac{15}{93}\)
\(\Rightarrow\frac{1}{2}.\left(\frac{1}{3}-\frac{1}{5}\right)+\frac{1}{2}.\left(\frac{1}{5}-\frac{1}{7}\right)+\frac{1}{2}.\left(\frac{1}{7}-\frac{1}{9}\right)+...+\frac{1}{2}.\left(\frac{1}{2x+1}-\frac{1}{2x+3}\right)=\frac{15}{93}\)
\(\Rightarrow\frac{1}{2}.\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{2x+1}-\frac{1}{2x+3}\right)=\frac{15}{93}\)
\(\Rightarrow\frac{1}{2}.\left(\frac{1}{3}-\frac{1}{2x+3}\right)=\frac{15}{93}\)
\(\Rightarrow\frac{1}{3}-\frac{1}{2x+3}=\frac{15}{93}:\frac{1}{2}\)
\(\Rightarrow\frac{1}{3}-\frac{1}{2x+3}=\frac{10}{31}\)
\(\Rightarrow\frac{1}{2x+3}=\frac{1}{3}-\frac{10}{31}\)
\(\Rightarrow\frac{1}{2x+3}=\frac{1}{93}\)
\(\Rightarrow\)2x + 3 = 93
\(\Rightarrow\)2x = 93 - 3
\(\Rightarrow\)2x = 90
\(\Rightarrow\)x = 90 : 2 = 45
S
25 tháng 2 2017
\(H=\frac{3}{1.5}+\frac{3}{5.9}+\frac{3}{9.13}+...+\frac{3}{33.37}\)
= \(\frac{3}{4}\left(1-\frac{1}{5}+\frac{1}{5}-\frac{1}{9}+\frac{1}{9}-\frac{1}{13}+...+\frac{1}{33}-\frac{1}{37}\right)\)
= \(\frac{3}{4}\left(1-\frac{1}{37}\right)\)
= \(\frac{3}{4}.\frac{36}{37}=\frac{27}{37}\)
6 tháng 8 2016
\(A=\frac{1}{1.5}+\frac{1}{5.9}+\frac{1}{9.13}+...+\frac{1}{93.97}\)
\(A=\frac{1}{4}.\left(\frac{4}{1.5}+\frac{4}{5.9}+\frac{4}{9.13}+...+\frac{1}{93.97}\right)\)
\(A=\frac{1}{4}.\left(1-\frac{1}{5}+\frac{1}{5}-\frac{1}{9}+\frac{1}{9}-\frac{1}{13}+...+\frac{1}{93}-\frac{1}{97}\right)\)
\(A=\frac{1}{4}.\left(1-\frac{1}{97}\right)\)
\(A=\frac{1}{4}.\frac{96}{97}=\frac{24}{97}\)
6 tháng 8 2016
\(A=\frac{1}{1.5}+\frac{1}{5.9}+\frac{1}{9.13}+...+\frac{1}{93.97}\)
\(A=\frac{1}{4}.\left(\frac{4}{1.5}+\frac{4}{5.9}+\frac{4}{9.13}+...+\frac{1}{93.97}\right)\)
\(A=\frac{1}{4}.\left(1-\frac{1}{5}+\frac{1}{5}-\frac{1}{9}+\frac{1}{9}-\frac{1}{13}+...+\frac{1}{93}-\frac{1}{97}\right)\)
\(A=\frac{1}{4}.\left(1-\frac{1}{97}\right)\)
\(A=\frac{1}{4}.\frac{96}{97}=\frac{24}{97}\)
23 tháng 3 2016
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
6 tháng 4 2017
Ta có:\(\frac{1}{1.5}+\frac{1}{5.9}+\frac{1}{9.13}+......+\frac{1}{81.85}\)
\(=\frac{1}{4}\left(\frac{4}{1.5}+\frac{4}{5.9}+\frac{4}{9.13}+......+\frac{4}{81.85}\right)\)
\(=\frac{1}{4}\left(1-\frac{1}{5}+\frac{1}{5}-\frac{1}{9}+.......+\frac{1}{81}-\frac{1}{85}\right)\)
\(=\frac{1}{4}.\left(1-\frac{1}{85}\right)\)
\(=\frac{1}{4}.\frac{84}{85}=\frac{21}{85}\)
6 tháng 4 2017
\(A=\frac{1}{1.5}+\frac{1}{5.9}+...+\frac{1}{81.85}\)
Ta có công thức
\(\frac{a}{b.c}=\frac{a}{c-b}.\left(\frac{1}{b}-\frac{1}{c}\right)\)
\(\Rightarrow A=\frac{1}{4}.\left(1-\frac{1}{5}+\frac{1}{5}-\frac{1}{9}+..+\frac{1}{81}-\frac{1}{85}\right)\)
\(A=\frac{1}{4}.\left(1-\frac{1}{85}\right)\)
\(A=\frac{84}{340}\)
S
10 tháng 5 2017
\(A=8400\left(\frac{1}{1.5}+\frac{1}{5.9}+\frac{1}{9.13}+\frac{1}{13.17}+\frac{1}{17.21}+\frac{1}{21.25}\right)\)
\(=\frac{8400}{4}.\left(\frac{4}{1.5}+\frac{4}{5.9}+\frac{4}{9.13}+\frac{4}{13.17}+\frac{4}{17.21}+\frac{4}{21.25}\right)\)
\(=2100\left(1-\frac{1}{5}+\frac{1}{5}-\frac{1}{9}+\frac{1}{9}-\frac{1}{13}+\frac{1}{13}-\frac{1}{17}+\frac{1}{17}-\frac{1}{21}+\frac{1}{21}-\frac{1}{25}\right)\)
\(=2100\left(1-\frac{1}{25}\right)\)
\(=2100\cdot\frac{24}{25}\)
\(=2016\)
10 tháng 5 2017
\(A=8400.\left(\frac{1}{1.5}+\frac{1}{5.9}+\frac{1}{9.13}+\frac{1}{13.17}+\frac{1}{17.21}+\frac{1}{21.25}\right)\)
\(A=8400.\left(\frac{1.4}{1.5.4}+\frac{1.4}{5.9.4}+\frac{1.4}{9.13.4}+\frac{1.4}{13.17.4}+\frac{1.4}{17.21.4}+\frac{1.4}{21.25.4}\right)\)
\(A=8400.\frac{1}{4}.\left(\frac{1}{1.5}+\frac{1}{5.9}+\frac{1}{9.13}+\frac{1}{13.17}+\frac{1}{17.21}+\frac{1}{21.25}\right)\)
\(A=8400.\frac{1}{4}.\left(\frac{1}{1}-\frac{1}{5}+\frac{1}{5}-\frac{1}{9}+\frac{1}{9}-\frac{1}{13}+\frac{1}{13}-\frac{1}{17}+\frac{1}{17}-\frac{1}{21}+\frac{1}{21}-\frac{1}{25}\right)\)
\(A=8400.\frac{1}{4}.\left(\frac{1}{1}-\frac{1}{25}\right)\)
\(A=8400.\frac{1}{4}.\frac{24}{25}\)
\(A=2016\)
NP
4
D
11 tháng 4 2017
\(4S=4.\left(\frac{1}{5.9}+\frac{1}{9.13}+...+\frac{1}{21.25}\right)\)
=\(\frac{4}{5.9}+\frac{4}{9.13}+....+\frac{4}{21.25}_{ }\)
=\(\frac{1}{5}-\frac{1}{9}+\frac{1}{9}-\frac{1}{13}+....+\frac{1}{21}-\frac{1}{23}\)
=\(\frac{1}{5}-\frac{1}{25}=\frac{5}{25}-\frac{1}{25}=\frac{4}{25}\)
=> \(S=\frac{4}{25}:4=\frac{4}{25}.\frac{1}{4}=\frac{1}{25}\)
11 tháng 4 2017
\(S=\frac{1}{5\times9}+\frac{1}{9\times13}+...+\frac{1}{21\times25}\)
\(S\times4=\frac{4}{5\times9}=\frac{4}{9\times13}+...+\frac{4}{21\times25}\)
\(S\times4=\frac{1}{5}-\frac{1}{9}+\frac{1}{9}-\frac{1}{13}+...+\frac{1}{21}-\frac{1}{25}\)
\(S\times4=\frac{1}{5}-\frac{1}{25}\)
\(S\times4=\frac{4}{25}\)
\(S=\frac{1}{25}\)
NP
2
25 tháng 2 2023
\(\dfrac{1}{5.9}+\dfrac{1}{9.13}+...+\dfrac{1}{21.25}\\ =\dfrac{4\cdot\dfrac{1}{4}}{5.9}+\dfrac{4\cdot\dfrac{1}{4}}{9.13}+...+\dfrac{4\cdot\dfrac{1}{4}}{21.25}\\ =\dfrac{1}{4}\left(\dfrac{4}{5.9}+\dfrac{4}{9.13}+...+\dfrac{4}{21.25}\right)\\ =\dfrac{1}{4}\cdot\left(\dfrac{1}{5}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{13}+...+\dfrac{1}{21}-\dfrac{1}{25}\right)\\ =\dfrac{1}{4}\left(\dfrac{1}{5}-\dfrac{1}{25}\right)=\dfrac{1}{4}\left(\dfrac{5}{25}-\dfrac{1}{25}\right)\\ =\dfrac{1}{4}\cdot\dfrac{4}{25}=\dfrac{1}{25}\)
BF
25 tháng 2 2023
`1/(5.9) + 1/(9.13) + ...+ 1/(21.25)`
`= 1/5 - 1/9 + 1/9 - 1/13 + ... + 1/21 - 1/25`
`= 1/5 - 1/25`
`= 4/25`
24 tháng 2 2022
bạn sửa số cuối tử là 4 nhé
\(=1-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{9}+...+\dfrac{1}{401}-\dfrac{1}{405}=1-\dfrac{1}{405}=\dfrac{404}{405}\)
24 tháng 2 2022
\(\dfrac{4}{1.5}+\dfrac{4}{5.9}+...+\dfrac{4}{401.405}\\ =1-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{9}+...+\dfrac{1}{401}-\dfrac{1}{405}\\ =1-\left(\dfrac{1}{5}-\dfrac{1}{5}\right)-\left(\dfrac{1}{9}-\dfrac{1}{9}\right)-...-\left(\dfrac{1}{401}-\dfrac{1}{401}\right)-\dfrac{1}{405}\\ =1-0-0-....-0-\dfrac{1}{405}\\ =1-\dfrac{1}{405}\\ =\dfrac{404}{405}\)
ND
2
5 tháng 5 2021
Sửa đề: 3x1⋅5+3x5⋅9+3x9⋅13+...+3x81⋅85=4153x1⋅5+3x5⋅9+3x9⋅13+...+3x81⋅85=415
a) Ta có: 3x1⋅5+3x5⋅9+3x9⋅13+...+3x81⋅85=4153x1⋅5+3x5⋅9+3x9⋅13+...+3x81⋅85=415
⇔3x4(41⋅5+45⋅9+49⋅13+...+481⋅85)=415⇔3x4(41⋅5+45⋅9+49⋅13+...+481⋅85)=415
⇔x⋅34(1−15+15−19+19−113+...+181−185)=415⇔x⋅34(1−15+15−19+19−113+...+181−185)=415
⇔x⋅34(1−185)=415⇔x⋅34(1−185)=415
⇔x⋅6385=415⇔x⋅6385=415
hay x=68189x=68189
Vậy: x=68189
OD
1
5 tháng 5 2021
Sửa đề: \(\dfrac{3x}{1\cdot5}+\dfrac{3x}{5\cdot9}+\dfrac{3x}{9\cdot13}+...+\dfrac{3x}{81\cdot85}=\dfrac{4}{15}\)
a) Ta có: \(\dfrac{3x}{1\cdot5}+\dfrac{3x}{5\cdot9}+\dfrac{3x}{9\cdot13}+...+\dfrac{3x}{81\cdot85}=\dfrac{4}{15}\)
\(\Leftrightarrow\dfrac{3x}{4}\left(\dfrac{4}{1\cdot5}+\dfrac{4}{5\cdot9}+\dfrac{4}{9\cdot13}+...+\dfrac{4}{81\cdot85}\right)=\dfrac{4}{15}\)
\(\Leftrightarrow x\cdot\dfrac{3}{4}\left(1-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{13}+...+\dfrac{1}{81}-\dfrac{1}{85}\right)=\dfrac{4}{15}\)
\(\Leftrightarrow x\cdot\dfrac{3}{4}\left(1-\dfrac{1}{85}\right)=\dfrac{4}{15}\)
\(\Leftrightarrow x\cdot\dfrac{63}{85}=\dfrac{4}{15}\)
hay \(x=\dfrac{68}{189}\)
Vậy: \(x=\dfrac{68}{189}\)
\(A=\dfrac{1}{1\cdot5}+\dfrac{1}{5\cdot9}+\dfrac{1}{9\cdot13}+...+\dfrac{1}{33\cdot37}\\ =\dfrac{1}{4}\cdot\left(\dfrac{4}{1\cdot5}+\dfrac{4}{5\cdot9}+\dfrac{4}{9\cdot13}+...+\dfrac{4}{33\cdot37}\right)\\ =\dfrac{1}{4}\cdot\left(1-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{13}+...+\dfrac{1}{33}-\dfrac{1}{37}\right)\\ =\dfrac{1}{4}\cdot\left(1-\dfrac{1}{37}\right)\\ =\dfrac{1}{4}\cdot\dfrac{36}{37}\\ =\dfrac{9}{37}\)
\(A=\dfrac{1}{1\cdot5}+\dfrac{1}{5\cdot9}+...+\dfrac{1}{33\cdot37}\)
\(=\dfrac{1}{4}\left(\dfrac{4}{1\cdot5}+\dfrac{4}{5\cdot9}+...+\dfrac{4}{33\cdot37}\right)\)
\(=\dfrac{1}{4}\left(1-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{9}+...+\dfrac{1}{33}-\dfrac{1}{37}\right)\)
\(=\dfrac{1}{4}\left(1-\dfrac{1}{37}\right)=\dfrac{1}{4}\cdot\dfrac{36}{37}=\dfrac{9}{37}\)