tính 1/1+1/23+1/33+...1/20253
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=\(\dfrac{1}{18}-\dfrac{1}{23}+\dfrac{1}{23}-\dfrac{1}{24}+\dfrac{1}{24}-\dfrac{1}{31}+\dfrac{1}{31}-\dfrac{1}{33}+\dfrac{1}{33}-\dfrac{1}{37}+\dfrac{1}{37}-\dfrac{1}{38}\)
=\(\dfrac{1}{18}-\dfrac{1}{38}=\dfrac{19-9}{38x9}=\dfrac{10}{342}=\dfrac{5}{171}\)
\(\dfrac{5}{18\times23}+\dfrac{1}{23\times24}+\dfrac{7}{24\times31}+\dfrac{2}{31\times33}+\dfrac{4}{33\times37}+\dfrac{1}{37\tímes38}\)
\(\dfrac{23-18}{18\times23}+\dfrac{24-23}{23\times24}+\dfrac{31-24}{24\times31}+\dfrac{33-31}{31\times33}+\dfrac{37-33}{33\times37}+\dfrac{38-37}{37\times38}\)
\(\dfrac{23}{18\times23}-\dfrac{18}{18\times23}+\dfrac{24}{23\times24}-\dfrac{23}{23\times24}+\dfrac{31}{24\times31}-\dfrac{24}{24\times31}+\dfrac{33}{31\times33}-\dfrac{31}{31\times33}+\dfrac{37}{33\times37}-\dfrac{33}{33\times37}+\dfrac{38}{37\times38}-\dfrac{37}{37\times38}\)
\(\dfrac{1}{18}-\dfrac{1}{23}+\dfrac{1}{23}-\dfrac{1}{24}+\dfrac{1}{24}-\dfrac{1}{31}+\dfrac{1}{31}-\dfrac{1}{33}+\dfrac{1}{33}-\dfrac{1}{37}+\dfrac{1}{37}-\dfrac{1}{38}\)
\(\dfrac{1}{18}-\dfrac{38}=\dfrac{5}{171}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a)
-37 +14 +26+37=(-37+37)+(14+26)=30
b) =(-24+24)+(6+10)=16
c)=(-23+23)+(25+15)=40
d)=(-33+33)+(-50+60)=10
e)=(-209+209)+[-14+(-16)]=-30
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1) (-37) + 14 + 26 + 37
= [(-37) + 37] + 14 + 26
= 0 + 40
= 40
2) (-24) + 6 + 10 + 24
= [(-24) + 24] + 6 + 10
= 0 + 16
= 16
3) 15 + 23 + (-25) + (-23)
= 15 + (-25) + [(-23) + 23]
= -10 + 0
= -10
4) 60 + 33 + (-50) + (-33)
= 60 + (-50) + [(-33) + 33]
= 10 + 0
= 10
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\(=\dfrac{20}{21}x\dfrac{21}{22}x\dfrac{22}{23}x...x\dfrac{1999}{2000}\)
\(=\dfrac{20}{2000}=\dfrac{1}{100}\)
=20/21x21/22x22/23x..............x1998/1999x1999/2000
=20x21x22x23x.....................x1998x1999/21x22x23x24x...............x1999x2000
=20/2000
1/100
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Bài 1
a) S = 1 + 2 + 2² + 2³ + ... + 2²⁰²³
2S = 2 + 2² + 2³ + 2⁴ + ... + 2²⁰²⁴
S = 2S - S = (2 + 2² + 2³ + ... + 2²⁰²⁴) - (1 + 2 + 2² + 2³)
= 2²⁰²⁴ - 1
b) B = 2²⁰²⁴
B - 1 = 2²⁰²⁴ - 1 = S
B = S + 1
Vậy B > S
a,
\(S=1+2+2^2+...+2^{2023}\)
\(2S=2+2^2+2^3+...+2^{2024}\)
\(\Rightarrow S=2^{2024}-1\)
b.
Do \(2^{2024}-1< 2^{2024}\)
\(\Rightarrow S< B\)
2.
\(H=3+3^2+...+3^{2022}\)
\(\Rightarrow3H=3^2+3^3+...+3^{2023}\)
\(\Rightarrow3H-H=3^{2023}-3\)
\(\Rightarrow2H=3^{2023}-3\)
\(\Rightarrow H=\dfrac{3^{2023}-3}{2}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a, A = 1 + 3 + 32 + 33 + ... + 32000
3.A = 3 + 32 + 33+ 33+... + 32001
3A - A = 3 + 32 + 33 + ... + 32001 - (1 + 3 + 32 + 33 + ... + 32000)
2A = 3 + 32 + 33 + ... + 32001 - 1 - 3 - 32 - 33 - ... - 32000
2A = 32001 - 1
A = \(\dfrac{3^{2001}-1}{2}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
mik làm 1 câu thôi các cau khác 1 chang lun chỉ khác số
A = 1 + (-2) + 3 + (-4) + 5 +(-6) + ... + 99 + (-100)
A=(1+(-2))+(3+(-4))+.....+(99+(-100))
A=(-1)+(-1)+......+(-1) CÓ 50 SỐ
A= -50