\(\frac{1+\frac{1}{3}+\frac{1}{5}+...+\frac{1}{99}}{\frac{1}{1\times99}+\frac{1}{39\times7}+...+\frac{1}{97\times3}+\frac{1}{99\times1}}\)
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Ta xét riêng tử số:
\(1+\frac{1}{3}+\frac{1}{5}+\frac{1}{7}+......+\frac{1}{97}+\frac{1}{99}\)
\(=\left(1+\frac{1}{99}\right)+\left(\frac{1}{3}+\frac{1}{97}\right)+\left(\frac{1}{5}+\frac{1}{95}\right)+......+\left(\frac{1}{49}+\frac{1}{51}\right)\)
\(=\frac{100}{1\times99}+\frac{100}{3\times97}+\frac{100}{5\times95}+......+\frac{100}{49\times51}\)
\(=100\times\left(\frac{1}{1\times99}+\frac{1}{3\times97}+\frac{1}{5\times95}+......+\frac{1}{49\times51}\right)\)
Bây giờ xét đến mẫu số:
\(\frac{1}{1\times99}+\frac{1}{3\times97}+\frac{1}{5\times95}+......+\frac{1}{97\times3}+\frac{1}{99\times1}\)
\(=\frac{2}{1\times99}+\frac{2}{3\times97}+\frac{2}{5\times95}+......+\frac{2}{49\times51}\)
\(=2\times\left(\frac{1}{1\times99}+\frac{1}{3\times97}+\frac{1}{5\times95}+......+\frac{1}{49\times51}\right)\)
Vậy giá trị của biểu thức là: \(\frac{100}{2}=50\)
\(\frac{1}{1\times2}+\frac{1}{2\times3}+\frac{1}{3\times4}+....+\frac{1}{98\times99}+\frac{1}{99\times100}\)
\(=\frac{2-1}{1\times2}+\frac{3-2}{2\times3}+\frac{4-3}{3\times4}+....+\frac{99-98}{98\times99}+\frac{100-99}{99\times100}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+....+\frac{1}{98}-\frac{1}{99}+\frac{1}{99}-\frac{1}{100}\)
\(=1-\frac{1}{100}\)
\(=\frac{99}{100}\)
\(\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+...+\frac{1}{98\cdot99}+\frac{1}{99\cdot100}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}\)
\(=1-\frac{1}{100}\)
\(=\frac{99}{100}\)
Câu A
Ta có (1/2)A = 1/22 + 1/23 + ... + 1/2100 + 1/2101
=> (1/2)A - A = - (1/2)A = (1/22 + 1/23 + ... + 1/2100 + 1/2101) - (1/2 + 1/22 + ... + 1/2100 )
= 1/2101 - 1/2
=> A = 1 - 1/2100
Câu B
Ta có 1/(1x2) = 1/1 - 1/2
1/(2.3) = 1/2 - 1/3
.................................
1/(99.100) = 1/99 - 1/100
=> B = 1/1 - 1/2 + 1/2 - 1/3 +.... +1/99 - 1/100
= 1 - 1/100
=99/100
\(=\frac{1.2}{99.100}\)
\(=\frac{2}{9900}=\frac{1}{4950}\)
=>A=\(\frac{7}{2}\)(\(\frac{1}{1}\)-\(\frac{1}{3}\)+\(\frac{1}{3}\)-\(\frac{1}{5}\)+...+\(\frac{1}{99}\)-\(\frac{1}{101}\))
=>A=\(\frac{7}{2}\)(1-\(\frac{1}{101}\))
=>A=\(\frac{350}{101}\)
7/2 ( \(\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-....+\frac{1}{99}-\frac{1}{101}\))
7/2 ( 1 - 1/101 )
7/2 x 100/101
=350/101
1/1x2 + 1/2x3 +1/3x4 + ......+1/98x99+1/99x100
=1/1 - 1/2 + 1/2 - 1/3 + 1/3 - 1/4 +......+ 1/98 - 1/99 + 1/99 + 1/100
=(1-1/100)+(1/2 - 1/2 ) + ( 1/3 - 1/3 ) + ...... + (1/98 - 1/98 ) + ( 1/99 - 1/99 )
= 100/100 - 1/100 + 0 + 0 +.....+ 0 + 0
=99/100
vậy GTBT = 99/100
A = 5(1/1.2 + 1/2.3 +......+ 1/99.100)
A = 5( 1 - 1/2 + 1/2 - 1/3 +........+ 1/99 - 1/100)
A = 5( 1 - 1/100)
A = 5 . 99/100
A = 99/20
** k mk nha!
\(\frac{5}{1\times2}+\frac{5}{2\times3}+...+\frac{5}{99\times100}=5\left(\frac{1}{1\times2}+\frac{1}{2\times3}+...+\frac{1}{99\times100}\right)=5\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{99}-\frac{1}{100}\right)=5\left(1-\frac{1}{100}\right)=5\times\frac{99}{100}=\frac{99}{20}=4\frac{19}{20}\)