a) Tính tổng A=6/5.8+22/8.19+24/19.31+140/31.101+198/101.200 b) Chứng minh : 1/2^2+1/4^2+1/6^2+...+1/100^2
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Những câu hỏi liên quan
26 tháng 6 2022
\(=2\left(\dfrac{3}{5\cdot8}+\dfrac{11}{8\cdot19}+\dfrac{12}{19\cdot31}+\dfrac{70}{31\cdot101}+\dfrac{99}{101\cdot200}\right)\)
\(=2\left(\dfrac{1}{5}-\dfrac{1}{8}+\dfrac{1}{8}-\dfrac{1}{19}+...+\dfrac{1}{101}-\dfrac{1}{200}\right)\)
\(=2\cdot\dfrac{39}{200}=\dfrac{39}{100}\)
10 tháng 4 2019
\(=2.\left(\frac{3}{5.8}+\frac{11}{8.19}+...+\frac{99}{101.200}\right)\)
\(=2.\left(\frac{1}{5}-\frac{1}{200}\right)\)
\(=2.\frac{39}{200}=\frac{39}{100}\)
TH
10 tháng 4 2019
\(E=\frac{6}{5.8}+\frac{22}{8.19}+\frac{24}{19.31}+\frac{140}{31.101}+\frac{198}{101.200}\)
\(=2.\left(\frac{3}{5.8}+\frac{11}{8.19}+\frac{12}{19.31}+\frac{70}{31.101}+\frac{99}{101.200}\right)\)
\(=2\left(\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{19}+\frac{1}{19}-\frac{1}{31}+\frac{1}{31}-\frac{1}{101}+\frac{1}{101}-\frac{1}{200}\right)\)
\(=2\left(\frac{1}{5}-\frac{1}{200}\right)\)
\(=2.\frac{39}{200}\)
\(=\frac{39}{100}\)
G
23 tháng 3 2019
giữ lời ko làm chó
gọi A = 6 / ( 5*8 ) + ... + 198 / ( 101 * 200 )
=> A / 2 = 3 / ( 5*8 ) + 11 / ( 8 * 19 ) + ... + 99 / ( 101*200 )
A / 2 = 1/5 - 1/8 + 1/8 - 1/11 + ... + 1 / 101 - 1 / 200
A / 2 = 1/ 5 -1 / 200
A / 2 = 39 /200
A = 39 / 100
đã làm bài này rồi , đúng, giờ thì k hộ cái , ko giết đấy
23 tháng 3 2019
\(A=\frac{6}{5.8}+\frac{22}{8.19}+\frac{24}{19.31}+\frac{140}{31.101}+\frac{198}{101.200}\)
\(\Rightarrow A=\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{19}+\frac{1}{19}-\frac{1}{31}+\frac{1}{31}-\frac{1}{101}+\frac{1}{101}-\frac{1}{200}\)
\(\Rightarrow A=\frac{1}{5}-\frac{1}{200}\)
\(\Rightarrow A=\frac{39}{200}\)
BT
3
S
11 tháng 5 2017
\(B=\frac{3}{5.8}+\frac{11}{8.19}+\frac{12}{19.31}+\frac{70}{31.101}+\frac{99}{101.200}\)
\(=\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{19}+\frac{1}{19}-\frac{1}{31}+\frac{1}{31}-\frac{1}{101}+\frac{1}{101}-\frac{1}{200}\)
\(=\frac{1}{5}-\frac{1}{200}\)
\(=\frac{39}{200}\)
\(\frac{x+2}{3}=\frac{x-2}{2}\)
=> \(\left(x+2\right)2=3\left(x-2\right)\)
2x + 4 = 3x - 6
2x - 3x = -6 - 4
-x = -10
x = 10
YN
27 tháng 4 2022
\(Q=\dfrac{3}{5.8}+\dfrac{11}{8.19}+\dfrac{12}{19.31}+\dfrac{70}{31.101}+\dfrac{90}{101.200}\)
\(=\dfrac{1}{5}-\dfrac{1}{8}+\dfrac{1}{8}-\dfrac{1}{19}+\dfrac{1}{19}-\dfrac{1}{31}+\dfrac{1}{31}-\dfrac{1}{101}+\dfrac{1}{101}-\dfrac{1}{200}\)
\(=\dfrac{1}{5}-\dfrac{1}{200}\)
\(=\dfrac{39}{200}\)
YN
18 tháng 3 2022
`Answer:`
1. \(S=\frac{1}{41}+\frac{1}{42}+...+\frac{1}{80}\)
\(\Rightarrow S=\left(\frac{1}{41}+\frac{1}{42}+...+\frac{1}{60}\right)+\left(\frac{1}{61}+...+\frac{1}{80}\right)\)
\(\Rightarrow S>\left(\frac{1}{60}+\frac{1}{60}+...+\frac{1}{60}\right)+\left(\frac{1}{80}+...+\frac{1}{80}\right)\)
\(\Rightarrow S>20.\frac{1}{60}+20.\frac{1}{80}\)
\(\Rightarrow S>\frac{1}{3}+\frac{1}{4}\)
\(\Rightarrow S>\frac{7}{12}\)
2. \(S=\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{2009^2}\)
Ta có:
\(2^2< 1.2\Rightarrow\frac{1}{2^2}< \frac{1}{1.2}\)
\(3^2< 2.3\Rightarrow\frac{1}{3^2}< \frac{1}{2.3}\)
\(4^2< 3.4\Rightarrow\frac{1}{4^2}< \frac{1}{3.4}\)
...
\(2009^2< 2008.2009\Rightarrow\frac{1}{2009^2}< \frac{1}{2008.2009}\)
\(\Rightarrow S< \frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{2008.2009}\)
\(\Rightarrow S< 1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2008}-\frac{1}{2009}\)
\(\Rightarrow S< 1-\frac{1}{2009}< 1\)
\(\Rightarrow S< 1\)
3. \(\frac{3}{5.8}+\frac{11}{8.19}+\frac{12}{19.31}+\frac{70}{31.101}+\frac{99}{101.200}\)
\(=\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{19}+\frac{1}{19}-\frac{1}{31}+\frac{1}{31}-\frac{1}{101}+\frac{1}{101}-\frac{1}{200}\)
\(=\frac{1}{5}-\frac{1}{200}\)
\(=\frac{39}{200}\)
a/ \(A=\dfrac{6}{5.8}+\dfrac{22}{8.19}+\dfrac{24}{19.31}+\dfrac{198}{101.200}\)
\(=2\left(\dfrac{3}{5.8}+\dfrac{11}{8.19}+\dfrac{12}{19.31}+...+\dfrac{99}{101.200}\right)\)
\(=2\left(\dfrac{1}{5}-\dfrac{1}{8}+\dfrac{1}{8}-\dfrac{1}{19}+....+\dfrac{1}{101}-\dfrac{1}{200}\right)\)
\(=2\left(\dfrac{1}{5}-\dfrac{1}{200}\right)\)
\(=\dfrac{39}{100}\)
b/ \(A=\dfrac{1}{2^2}+\dfrac{1}{4^2}+.....+\dfrac{1}{100^2}\)
Ta có :
\(\dfrac{1}{2^2}< \dfrac{1}{1.2}\)
\(\dfrac{1}{4^2}< \dfrac{1}{3.4}\)
...........
\(\dfrac{1}{100^2}< \dfrac{1}{99.100}\)
\(\Leftrightarrow A< \dfrac{1}{1.2}+\dfrac{1}{2.3}+....+\dfrac{1}{99.100}\)
\(\Leftrightarrow A< 1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+.....+\dfrac{1}{99}-\dfrac{1}{100}\)
\(\Leftrightarrow A< 1-\dfrac{1}{100}< 1\left(đpcm\right)\)
câu b sai rồi chị ơi!