\(\dfrac{9}{9.19}+\dfrac{9}{19.29}+...+\dfrac{1}{89.99}\)
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1
19 tháng 9 2021
\(\dfrac{1}{19}+\dfrac{9}{19\cdot29}+...+\dfrac{9}{1999\cdot2009}\)
\(=\dfrac{1}{19}+\dfrac{9}{10}\left(\dfrac{10}{19\cdot29}+...+\dfrac{10}{1999\cdot2009}\right)\)
\(=\dfrac{1}{19}+\dfrac{9}{10}\left(\dfrac{1}{19}-\dfrac{1}{2009}\right)\)
\(=\dfrac{1}{19}+\dfrac{1791}{38171}=\dfrac{200}{2009}\)
TT
21 tháng 5 2017
Lưu An
\(A=\dfrac{1}{19}+\left(\dfrac{9}{19\cdot29}+\dfrac{9}{29\cdot39}+...+\dfrac{9}{1999\cdot2009}\right)\)
\(A=\dfrac{1}{19}+\dfrac{9}{10}\left(\dfrac{10}{19\cdot29}+\dfrac{10}{29\cdot39}+...+\dfrac{10}{1999\cdot2009}\right)\)
\(A=\dfrac{1}{19}+\dfrac{9}{10}\left(\dfrac{1}{19}-\dfrac{1}{29}+\dfrac{1}{29}-\dfrac{1}{39}+...+\dfrac{1}{1999}-\dfrac{1}{2009}\right)\)
\(A=\dfrac{1}{19}+\dfrac{9}{10}\left(\dfrac{1}{19}-\dfrac{1}{2009}\right)\)
\(A=\dfrac{1}{19}+\dfrac{9}{10}\cdot\dfrac{1990}{38171}\)
\(A=\dfrac{1}{19}+\dfrac{1791}{38171}\)
\(A=\dfrac{200}{2009}\)
21 tháng 5 2017
B=1/19+(9/19.29+9/29.39+...+9/1999.2009)
B=1/19+9/10+(10/19.29+10/29.39+.....+10/1999.2009
B=1/19+9/10+(1/19-1/29+1/29-1/39+....+1/1999-1/2009)
B=1/19+9/10+(1/19-1/2009)
B=1/19+9/10.1990/38171
B=1/19+1791/38171
B=200/2009
Vậy B= 200/2009
29 tháng 12 2020
=(54⋅9+59⋅14+514⋅19+...+544⋅49)⋅1−(3+5+7+...+49)89=(54⋅9+59⋅14+514⋅19+...+544⋅49)⋅1−(3+5+7+...+49)89=15⋅(14−19+19−114+114−119+...+144−149)⋅(1−(52⋅24)289)=15⋅(14−19+19−114+114−119+...+144−149)⋅(1−(52⋅24)289)=15⋅(14−149)⋅1−62489=15⋅(14−149)⋅1−62489=15⋅45196⋅−62389=15⋅45196⋅−62389=−928
S
29 tháng 12 2020
\(\dfrac{1}{4\cdot9}+\dfrac{1}{9\cdot14}+\dfrac{1}{14\cdot19}+...+\dfrac{1}{44\cdot49}\)
\(=\dfrac{1}{5}\left(\dfrac{9-4}{4\cdot9}+\dfrac{14-9}{9\cdot14}+\dfrac{19-14}{14\cdot19}+...+\dfrac{49-44}{44\cdot49}\right)\)
\(=\dfrac{1}{5}\cdot\left(\dfrac{1}{4}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{14}+\dfrac{1}{14}-\dfrac{1}{19}+....+\dfrac{1}{44}-\dfrac{1}{49}\right)\)
\(=\dfrac{1}{5}\cdot\left(\dfrac{1}{4}-\dfrac{1}{49}\right)\)
\(=\dfrac{1}{5}\cdot\dfrac{45}{196}\)
\(=\dfrac{9}{196}\)
AH
Akai Haruma
Giáo viên
12 tháng 5 2021
Lời giải:
\(2A=\frac{4}{1.5}+\frac{6}{5.11}+\frac{8}{11.19}+\frac{10}{19.29}+\frac{12}{29.41}\)
\(=1-\frac{1}{5}+\frac{1}{5}-\frac{1}{11}+\frac{1}{11}-\frac{1}{19}+...+\frac{1}{29}-\frac{1}{41}=1-\frac{1}{41}=\frac{40}{41}\)
\(\Rightarrow A=\frac{20}{21}\)
\(3B=\frac{3}{1.4}+\frac{6}{4.10}+\frac{9}{10.19}+\frac{12}{19.31}=1-\frac{1}{4}+\frac{1}{4}-\frac{1}{10}+\frac{1}{10}-\frac{1}{19}+\frac{1}{19}-\frac{1}{31}\)
\(=1-\frac{1}{31}=\frac{30}{31}\)
\(\Rightarrow B=\frac{10}{31}=\frac{20}{62}<\frac{20}{41}\)
Do đó $A>B$
OY
11 tháng 5 2021
A.2=4/1.5+6/5.11+...+12/29.41
A.2=1-1/5+1/5-1/11+...+1/29-1/41
A.2=1-1/41
A.2=40/41
A=20/41
B.3=3/1.4+6/4.10+...+12/29.31
B.3=1-1/4+1/4-1/10+...+1/29-1/31
B.3=1-1/31
B.3=30/31
B=10/31
Vì 20/41.10/31 nên A>B
11 tháng 5 2021
\(A=\dfrac{2}{1.5}+\dfrac{3}{5.11}+\dfrac{4}{11.19}+\dfrac{5}{19.29}+\dfrac{6}{29.41}\)
\(\Rightarrow2A=\dfrac{4}{1.5}+\dfrac{6}{5.11}+\dfrac{8}{11.19}+\dfrac{10}{19.29}+\dfrac{12}{29.41}\)
\(\Rightarrow2A=1-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{11}+\dfrac{1}{11}-\dfrac{1}{19}+\dfrac{1}{19}-\dfrac{1}{29}+\dfrac{1}{29}-\dfrac{1}{41}\)
\(\Rightarrow2A=1-\dfrac{1}{41}=\dfrac{40}{41}\)
\(\Rightarrow A=\dfrac{40}{41}:2=\dfrac{20}{41}\)(1)
\(B=\dfrac{1}{1.4}+\dfrac{2}{4.10}+\dfrac{3}{10.19}+\dfrac{4}{19.31}\)
\(\Rightarrow3B=\dfrac{3}{1.4}+\dfrac{6}{4.10}+\dfrac{9}{10.19}+\dfrac{12}{19.31}\)
\(\Rightarrow3B=\dfrac{1}{1}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{10}+\dfrac{1}{10}-\dfrac{1}{19}+\dfrac{1}{19}-\dfrac{1}{31}\)
\(\Rightarrow3B=\dfrac{1}{1}-\dfrac{1}{31}=\dfrac{30}{31}\)
\(\Rightarrow B=\dfrac{30}{31}:3=\dfrac{10}{31}\)
\(\Rightarrow B=\dfrac{2}{2}.\dfrac{10}{31}=\dfrac{20}{62}\)
+)Ta có:\(\dfrac{20}{62}< \dfrac{20}{41}\Rightarrow B< A\)
Hay A>B(ĐPCM)
Chúc bn học tốt
16 tháng 2 2022
a)\(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+\frac{1}{9.11}\)
\(=\frac{1}{2}.\left(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+\frac{2}{9.11}\right)\)
\(=\frac{1}{2}.\left(\frac{3-1}{1.3}+\frac{5-3}{3.5}+\frac{7-5}{5.7}+\frac{9-7}{7.9}+\frac{11-9}{9.11}\right)\)
\(=\frac{1}{2}.\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11}\right)\)
\(=\frac{1}{2}.\left(1-\frac{1}{11}\right)\)
\(\frac{10}{22}\)
2 tháng 7 2023
a: =-21/36-3/36=-24/36=-2/3
b: =43/12*1/2+5/24=43/24+5/24=2
c: =8/9+1/9=1
e: =1-1/4+1/4-1/7+...+1/97-1/100
=1-1/100=99/100
P
1
19 tháng 6 2023
`@` `\text {Ans}`
`\downarrow`
\(S=\sqrt{0,49}+\sqrt{\dfrac{1}{9}}-\sqrt{\dfrac{25}{4}}\)
`S=0,7 + 1/3 - 5/2`
`S=31/30 - 5/2 = -22/15`
\(\dfrac{9}{9.19}+\dfrac{9}{19.29}+...+\dfrac{1}{89.99}\)
\(=9.\left(\dfrac{1}{9.19}+\dfrac{1}{19.29}+...+\dfrac{1}{79.89}\right)+\dfrac{1}{89.99}\)
\(=\dfrac{9}{10}.\left(\dfrac{10}{9.19}+\dfrac{10}{19.29}+...+\dfrac{10}{79.89}\right)+\dfrac{1}{89.99}\)
\(=\dfrac{9}{10}.\left(\dfrac{1}{9}-\dfrac{1}{19}+\dfrac{1}{19}-\dfrac{1}{29}+...+\dfrac{1}{79}-\dfrac{1}{89}\right)+\left(\dfrac{1}{89}-\dfrac{1}{99}\right)\)
\(=\dfrac{9}{10}.\left(\dfrac{1}{9}-\dfrac{1}{89}\right)+\dfrac{10}{8811}\)
\(=\dfrac{9}{10}.\dfrac{80}{801}+\dfrac{10}{8811}\)
\(=\dfrac{8}{89}+\dfrac{10}{8811}\)
\(=\dfrac{802}{8811}\)
Đề là `9/[89.99]` thay vì là `1/[89.99]` chứ nhỉ? Vì nếu bạn để đề vậy thì không thấy quy luật là như thế nào.
*Sửa: `9/[9.19]+9/[19.29]+....+9/[89.99]`
`=9/10(10/[9.19]+10/[19.29]+....+10/[89.99])`
`=9/10(1/9-1/19+1/19-1/29+.....+1/89-1/99)`
`=9/10(1/9-1/99)`
`=9/10 . 10/99`
`=1/11`