=11,4215495

\(=\frac{39}{100}=0,39\)

\(A=\frac{6}{5.8}+\frac{22}{8.19}+\frac{24}{19.31}+\frac{140}{31.101}+\frac{198}{101.200}\)

\(\Rightarrow A=\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{19}+\frac{1}{19}-\frac{1}{31}+\frac{1}{31}-\frac{1}{101}+\frac{1}{101}-\frac{1}{200}\)

\(\Rightarrow A=\frac{1}{5}-\frac{1}{200}\)

\(\Rightarrow A=\frac{39}{200}\)

A=2.(3\5.8+11\8.19+12\19.31+70\31.101+99\101\200)

A=2.(1\5-18+1\8-1\19+....+1\101-1\200)

A=2.(1\5-1\200)

A=2.39\200

A=39\100

Chúc bạn học tốt

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

A=39/100

100% đúng

Ta co:

1/2A= 3/5×8 + 11/8×19 + 12/19×31 +70 /31×101+ 99/101×200

1/2A= 1/5 - 1/8 + 1/8 - 1/19 + 1/19 - 1/31 + 1/31 -1/101 + 1/101 - 1/200

1/2A= 1/5 - 1/200

1/2A= 39/200

A= 39/200 ÷ 1/2

A= 39/100

\(=2\left(\dfrac{3}{5\cdot8}+\dfrac{11}{8\cdot19}+\dfrac{12}{19\cdot31}+\dfrac{70}{31\cdot101}+\dfrac{99}{101\cdot200}\right)\)

\(=2\left(\dfrac{1}{5}-\dfrac{1}{8}+\dfrac{1}{8}-\dfrac{1}{19}+...+\dfrac{1}{101}-\dfrac{1}{200}\right)\)

\(=2\cdot\dfrac{39}{200}=\dfrac{39}{100}\)

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 :

\(\dfrac{1}{2^2}< \dfrac{1}{1.2}\)

\(\dfrac{1}{4^2}< \dfrac{1}{3.4}\)

...........

\(\dfrac{1}{100^2}< \dfrac{1}{99.100}\)

\(\Leftrightarrow A< \dfrac{1}{1.2}+\dfrac{1}{2.3}+....+\dfrac{1}{99.100}\)

\(\Leftrightarrow A< 1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+.....+\dfrac{1}{99}-\dfrac{1}{100}\)

\(\Leftrightarrow A< 1-\dfrac{1}{100}< 1\left(đpcm\right)\)

Giải:

a) \(A=\dfrac{6}{5.8}+\dfrac{22}{8.19}+\dfrac{24}{19.31}+\dfrac{140}{31.101}+\dfrac{198}{101.200}\) 

\(A=2.\left(\dfrac{3}{5.8}+\dfrac{11}{8.19}+\dfrac{12}{19.31}+\dfrac{70}{31.101}+\dfrac{99}{101.200}\right)\) 

\(A=2.\left(\dfrac{1}{5}-\dfrac{1}{8}+\dfrac{1}{8}-\dfrac{1}{19}+\dfrac{1}{19}-\dfrac{1}{31}+\dfrac{1}{31}-\dfrac{1}{101}+\dfrac{1}{101}-\dfrac{1}{200}\right)\) 

\(A=2.\left(\dfrac{1}{5}-\dfrac{1}{200}\right)\) 

\(A=2.\dfrac{39}{200}\) 

\(A=\dfrac{39}{100}\) 

b) \(\dfrac{1}{2^2}+\dfrac{1}{4^2}+\dfrac{1}{6^2}+...+\dfrac{1}{100^2}\) 

Ta có:

\(\dfrac{1}{2^2}< \dfrac{1}{1.2}\) 

\(\dfrac{1}{4^2}< \dfrac{1}{3.4}\) 

\(\dfrac{1}{6^2}< \dfrac{1}{5.6}\) 

\(...\) 

\(\dfrac{1}{100^2}< \dfrac{1}{99.100}\) 

\(\Rightarrow\dfrac{1}{1.2}+\dfrac{1}{3.4}+\dfrac{1}{5.6}+...+\dfrac{1}{99.100}\) 

\(\Rightarrow\dfrac{1}{1}-\dfrac{1}{2}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{5}-\dfrac{1}{6}+...+\dfrac{1}{99}-\dfrac{1}{100}\) 

\(\Rightarrow\dfrac{1}{1}+\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{99}+\dfrac{1}{100}-2.\left(\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{1}{6}+...+\dfrac{1}{100}\right)\) 

\(\Rightarrow1-\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+\dfrac{1}{5}+\dfrac{1}{6}...+\dfrac{1}{99}+\dfrac{1}{100}-\left(1+\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{50}\right)\) 

\(\Rightarrow\dfrac{1}{51}+\dfrac{1}{52}+\dfrac{1}{53}+...+\dfrac{1}{99}+\dfrac{1}{100}\) 

Bạn tự lm theo đề bài của bạn nhé vì đề bài chỉ thế này thôi!

\(\frac{3}{5.8}+\frac{11}{8.19}+\frac{12}{19.31}+\frac{70}{31.101}+\frac{99}{101.200}\)

\(=\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{19}+\frac{1}{19}-\frac{1}{31}+\frac{1}{31}-\frac{1}{101}+\frac{1}{101}-\frac{1}{200}\)

\(=\frac{1}{5}-\frac{1}{200}\)

\(=\frac{39}{200}\)

1/5-1/8+1/8-1/19+1/19-1/31+1/31-1/101+1/200=1/5-1/200=195/1000=39/200

\(\Rightarrow A=\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{19}+\frac{1}{19}-\frac{1}{31}+\frac{1}{31}-\frac{1}{101}+\frac{1}{101}-\frac{1}{200}\)

\(\Rightarrow A=\frac{1}{5}-\frac{1}{200}\)

\(\Rightarrow A=\frac{39}{200}\)

vì \(\frac{39}{200}< 1\) nên A < 1

\(A=\frac{3}{5.8}+\frac{11}{8.19}+\frac{12}{19.31}+\frac{70}{31.101}+\frac{99}{101.200}\)

Áp dụng công thức \(\frac{b-a}{a.b}=\frac{1}{a}-\frac{1}{b}\) với a < b và a khác b khác 0, ta có:

\(A=\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{19}+...+\frac{1}{101}-\frac{1}{200}\\ =\frac{1}{5}-\frac{1}{200}\\ =\frac{40-1}{200}\\ =\frac{39}{200}\\ \frac{39}{200}< 1\\\Rightarrow A< 1\left(đpcm\right)\)

Chúc bạn học tốt!

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