2^3+3.(-1/2)^2-2^2.4+[(-2)^2:1/2].8 help me
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3/12.22 + 5/22.32 + 7/32.42 + ... + 19/92.102
= 3/1.4 + 5/4.9 + 7/9.16 + ... + 19/81.100
= 1 - 1/4 + 1/4 - 1/9 + 1/9 - 1/16 + ... + 1/81 - 1/100
= 1 - 1/100
= 99/100
\(A=\frac{3}{1.4}+\frac{5}{4.9}+\frac{7}{9.16}+..+\frac{17}{64.81}+\frac{19}{81.100}\)
\(A=\frac{4-1}{1.4}+\frac{9-4}{4.9}+\frac{16-9}{9.16}+...+\frac{81-64}{64.81}+\frac{100-81}{81.100}\)
\(A=1-\frac{1}{4}+\frac{1}{4}-\frac{1}{9}+\frac{1}{9}-\frac{1}{16}+...+\frac{1}{61}-\frac{1}{81}+\frac{1}{81}-\frac{1}{100}\)
\(A=1-\frac{1}{100}=\frac{99}{100}\)
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\(\dfrac{4}{3}\times\dfrac{9}{8}\times\dfrac{16}{15}\times\dfrac{25}{24}=\dfrac{5}{3}\)
`(2^2)/(1 . 3) . (3^2)/(2 . 4) . (4^2)/(3 . 5) . (5^2)/(4 . 6)`
`= 4/3 . 9/8 . 16/15 . 25/24 = 5/3`
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( 1/2 + 1/3 + 1/4 + ... + 1/10 ) . x = 1/9 + 2/8 + ... + 9/1
=> x = ( 1/9 + 2/8 + ... + 9/1 ) : ( 1/2 + 1/3 + ... 1/10 )
=> x = ( 9/1 + 8/2 + ... + 2/8 + 1/9 ) : ( 1/2 + 1/3 + 1/4 + ... + 1/10 )
=>x = [ ( 9 - 1 - 1 -... - 1 ) +( 8/2 + 1 ) + ( 7/3 + 1 ) + ... + ( 1/9 + 1 ) ] : ( 1/2 + 1/3 + 1/4 + ... + 1/10 )
=> x = ( 1 + 10/2 + 10/3 + ... + 10/9 ) : ( 1/2 + 1/3 + 1/4 + ... + 1/10 )
=> x = [10 . ( 1/2 + 1/3 + ... + 1/9 ) ] : ( 1/2 + 1/3 + 1/4 + ... + 1/10 )
=> x = 10
Chúc Bạn Học Tốt
#𝗝𝘂𝗻𝗻
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\(2A=\frac{2}{1.3}+\frac{2}{2.4}+\frac{2}{3.5}+......+\frac{2}{7.9}+\frac{2}{8.10}\)
\(2A=\left(\frac{2}{1.3}+\frac{2}{3.5}+.....+\frac{2}{7.9}\right)+\left(\frac{2}{2.4}+\frac{2}{4.6}+...+\frac{2}{8.10}\right)\)
\(2A=\left(\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+....+\frac{1}{7}-\frac{1}{9}\right)+\left(\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+.....+\frac{1}{8}-\frac{1}{10}\right)\)
\(2A=\left(\frac{1}{1}-\frac{1}{9}\right)+\left(\frac{1}{2}-\frac{1}{10}\right)\)
\(2A=\frac{8}{9}+\frac{2}{5}=\frac{58}{45}\)
\(A=\frac{58}{45}.\frac{1}{2}=\frac{29}{45}\)
Dễ thế mà cũng phải hỏi, ở trong SBT toán 6 có hết mà! còn bài 3 là ở trong SGK toán 6, bọn cậu học qua chương trình rùi mà? nếu mà học lớp 6
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\(B=\dfrac{2^2}{1\cdot3}+\dfrac{3^2}{2\cdot4}+\dfrac{4^2}{3\cdot5}+...+\dfrac{99^2}{98\cdot100}\\ =\dfrac{1\cdot3+1}{1\cdot3}+\dfrac{2\cdot4+1}{2\cdot4}+\dfrac{3\cdot5+1}{3\cdot5}+...+\dfrac{98\cdot100+1}{98\cdot100}\\ =\dfrac{1\cdot3}{1\cdot3}+\dfrac{1}{1\cdot3}+\dfrac{2\cdot4}{2\cdot4}+\dfrac{1}{2\cdot4}+\dfrac{3\cdot5}{3\cdot5}+\dfrac{1}{3\cdot5}+...+\dfrac{98\cdot100}{98\cdot100}+\dfrac{1}{98\cdot100}\\ =1+\dfrac{1}{1\cdot3}+1+\dfrac{1}{2\cdot4}+1+\dfrac{1}{3\cdot5}+...+1+\dfrac{1}{98\cdot100}\\ =\left(1+1+1+...+1\right)+\left(\dfrac{1}{1\cdot3}+\dfrac{1}{2\cdot4}+\dfrac{1}{3\cdot5}+...+\dfrac{1}{98\cdot100}\right)\\ =98+\left(\dfrac{1}{1\cdot3}+\dfrac{1}{2\cdot4}+\dfrac{1}{3\cdot5}+...+\dfrac{1}{98\cdot100}\right)\\ \)Gọi \(\dfrac{1}{1\cdot3}+\dfrac{1}{2\cdot4}+\dfrac{1}{3\cdot5}+...+\dfrac{1}{98\cdot100}\) là A
\(A=\dfrac{1}{1\cdot3}+\dfrac{1}{2\cdot4}+\dfrac{1}{3\cdot5}+...+\dfrac{1}{98\cdot100}\\ =\dfrac{1}{2}\cdot\left(\dfrac{2}{1\cdot3}+\dfrac{2}{2\cdot4}+\dfrac{2}{3\cdot5}+...+\dfrac{2}{98\cdot100}\right)\\ =\dfrac{1}{2}\cdot\left(\dfrac{1}{1}-\dfrac{1}{3}+\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{3}-\dfrac{1}{5}+...+\dfrac{1}{98}-\dfrac{1}{100}\right)\\ =\dfrac{1}{2}\cdot\left(\dfrac{1}{1}+\dfrac{1}{2}-\dfrac{1}{99}-\dfrac{1}{100}\right)\\ =\dfrac{1}{2}\cdot\left(\dfrac{3}{2}-\dfrac{1}{99}-\dfrac{1}{100}\right)\\ =\dfrac{1}{2}\cdot\left(\dfrac{295}{198}-\dfrac{1}{100}\right)\\ =\dfrac{1}{2}\cdot\dfrac{14651}{9900}=\dfrac{14651}{19800}\)
\(B=98+A=98+\dfrac{14651}{19800}=98\dfrac{14651}{19800}\)
Dễ thấy phần nguyên của B là 98
Vậy phần nguyên của B là 98
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mk ko ghi đb nhé
\(=\frac{1\cdot3+1}{1\cdot3}+\frac{2\cdot4+1}{2\cdot4}+...+\frac{99\cdot101+1}{99\cdot101}.\)
\(=1+\frac{1}{1\cdot3}+1+\frac{1}{2\cdot4}+...+1+\frac{1}{99\cdot101}\)
\(=99+\frac{1}{2}\left(\frac{2}{1\cdot3}+\frac{2}{2\cdot4}+...+\frac{2}{99\cdot101}\right)\)
\(=99+\frac{1}{2}\left(1-\frac{1}{3}+\frac{1}{2}-\frac{1}{4}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{98}-\frac{1}{100}+\frac{1}{99}-\frac{1}{101}\right)\)
\(=99+\frac{1}{2}\left(1+\frac{1}{2}-\frac{1}{100}-\frac{1}{101}\right)\)
phần còn lại bn tự tính nha
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A=2^2/1.3+3^2/2.4+4^2/3.5+....+99^2/98.100
A=2^2/(2-1)(2+1)+3^2/(3-1)(3+1)+4^2/(4-1)(4+1)+...+99^2/(99-1)(99+1)
A=2^2/2^2-1+3^2/3^2-1+...+99^2/99^2-1
A=2^2-1+1/2^2-1+3^2-1+1/3^2-1+...+99^2-1+1/99^2-1
A=1+1/1.3+1+1/2.4+1+1/3.5+...+1+1/98.100
A=(1+1+1+....+1)+(1/1.3+1/2.4+...+1/98.100) (1)
Ta có:
Đặt B=(1+1+1+...+1)=98[vì (99-2):1+1=98 số] (2)
Đặt C=1/1.3+1/2.4+1/3.5+...+1/98.100
=>C=1/2.(1-1/3)+1/2.(1/2-1/4)+1/2.(1/3-1/5)+...+1/2.(1/98-1/100)
=>C=1/2.(1-1/3+1/2-1/4+1/3-1/5+...+1/97-1/99+1/98-1/100)
=>C=1/2.(1+1/2-1/99-1/100)
=>C=1/2.(3/2-1/99.100) (3)
Thay (2),(3) vào(1), được:
A=98+1/2.(3/2-1/99.100)