tính hợp lý : 4/5x7 + 4/7x9 + 4/9x11 + ... + 4/50x61
Thanks you very much !
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GK
1
Những câu hỏi liên quan
GK
1
DD
Đoàn Đức Hà
Giáo viên
27 tháng 2 2021
\(A=\frac{4}{5.7}+\frac{4}{7.9}+\frac{4}{9.11}+...+\frac{4}{59.61}\)
\(A=2\left(\frac{2}{5.7}+\frac{2}{7.9}+\frac{2}{9.11}+...+\frac{2}{59.61}\right)\)
\(A=2\left(\frac{7-5}{5.7}+\frac{9-7}{7.9}+\frac{11-9}{9.11}+...+\frac{61-59}{59.61}\right)\)
\(A=2\left(\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11}+...+\frac{1}{59}-\frac{1}{61}\right)\)
\(A=2\left(\frac{1}{5}-\frac{1}{61}\right)=\frac{112}{305}\).
T
1
1 tháng 5 2018
\(\frac{4}{5.7}+\frac{4}{7.9}+\frac{4}{9.11}+...+\frac{4}{99.101}\)
\(=2.\left(\frac{2}{5.7}+\frac{2}{7.9}+\frac{2}{9.11}+...+\frac{2}{99.101}\right)\)
\(=2.\left(\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11}+...+\frac{1}{99}-\frac{1}{101}\right)\)
\(=2.\left(\frac{1}{5}-\frac{1}{101}\right)\)
\(=2.\frac{96}{505}\)
\(=\frac{192}{505}\)
T
1
AK
1 tháng 5 2018
\(\frac{4}{5.7}+\frac{4}{7.9}+...+\frac{4}{99.101}\)
\(=2.\left(\frac{2}{5.7}+\frac{2}{7.9}+...+\frac{2}{99.101}\right)\)
\(=2.\left(\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{99}-\frac{1}{101}\right)\)
\(=2.\left(\frac{1}{5}-\frac{1}{101}\right)\)
\(=2.\left(\frac{101}{505}-\frac{5}{505}\right)\)
\(=2.\frac{96}{505}\)
\(=\frac{192}{505}\)
Chúc bạn học tốt !!!
NT
Nguyễn Thị Thương Hoài
Giáo viên
VIP
11 tháng 3 2023
A = \(\dfrac{4}{1\times3}\) - \(\dfrac{8}{3\times5}\) + \(\dfrac{12}{5\times7}\) - \(\dfrac{16}{7\times9}\) + \(\dfrac{20}{9\times11}\) - \(\dfrac{24}{11\times13}\)
A = ( \(\dfrac{1}{1}+\dfrac{1}{3}\)) - ( \(\dfrac{1}{3}\) + \(\dfrac{1}{5}\)) + (\(\dfrac{1}{5}\)+ \(\dfrac{1}{7}\)) - ( \(\dfrac{1}{7}\) + \(\dfrac{1}{9}\)) +( \(\dfrac{1}{9}\)+ \(\dfrac{1}{11}\)) - (\(\dfrac{1}{11}\)+\(\dfrac{1}{13}\))
A = \(\dfrac{1}{1}+\dfrac{1}{3}\) - \(\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}+\dfrac{1}{7}\) - \(\dfrac{1}{7}\) - \(\dfrac{1}{9}\) + \(\dfrac{1}{9}\) + \(\dfrac{1}{11}\) - \(\dfrac{1}{11}\) - \(\dfrac{1}{13}\)
A = \(\dfrac{1}{1}\) - \(\dfrac{1}{13}\)
A = \(\dfrac{12}{13}\)
XO
6 tháng 8 2020
Ta có :\(\frac{4}{5.7}+\frac{4}{7.9}+...+\frac{4}{59.61}=2.\left(\frac{2}{5.7}+\frac{2}{7.9}+...+\frac{2}{59.61}\right)\)
\(=2\left(\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{59}-\frac{1}{61}\right)=2\left(\frac{1}{5}-\frac{1}{61}\right)=2.\frac{56}{305}=\frac{112}{305}\)
P/S : Dấu "." là dấu "x"
NC
6 tháng 8 2020
Bài làm:
Ta có: \(\frac{4}{5.7}+\frac{4}{7.9}+...+\frac{4}{59.61}\)
\(=2\left(\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{59}-\frac{1}{61}\right)\)
\(=2\left(\frac{1}{5}-\frac{1}{61}\right)\)
\(=2.\frac{56}{305}=\frac{112}{305}\)
DT
2
BG
10 tháng 3 2018
Ta có:
A = \(\frac{2}{1x3}+\frac{2}{3x5}+\frac{2}{5x7}+\frac{2}{7x9}+\frac{2}{9x11}\)
= \(\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11}\)
= \(\frac{1}{1}-\frac{1}{11}\)
=\(\frac{10}{11}\)
AK
24 tháng 7 2020
3/2 . x + ( 5/3 - 3/2) : 2/3 = 5/3
3/2.x + 1/6 : 2/3 = 5/3
3/2.x + 1/4 = 5/3
3/2.x = 5/3 - 1/4
3/2.x=17/12
x= 17/12 : 3/2
x= 17/18
Vậy...
Bài 2:
4/5x7 + 4/7x9 + 4/9x11 +...+4/17x19
= 2(2/5.7 + 2/7.9 + 2/9.11+...+ 2/17/19)
= 2( 1/5 - 1/7 + 1/7 -1/9 + 1/9 -1/11 +...+ 1/17 - 1/19)
= 2( 1/5- 1/19)
= 2 . 14/95
= 28/95
24 tháng 7 2020
Trả lời:
Bài 1
\(\frac{3}{2}\times x+\left(\frac{5}{3}-\frac{3}{2}\right)\div\frac{2}{3}=\frac{5}{3}\)
\(\Leftrightarrow\frac{3}{2}\times x+\frac{1}{6}\div\frac{2}{3}=\frac{5}{3}\)
\(\Leftrightarrow\frac{3}{2}\times x+\frac{1}{6}\times\frac{3}{2}=\frac{5}{3}\)
\(\Leftrightarrow\frac{3}{2}\times x+\frac{1}{4}=\frac{5}{3}\)
\(\Leftrightarrow\frac{1}{6}\times x=\frac{17}{12}\)
\(\Leftrightarrow x=\frac{17}{2}\)
Vậy \(x=\frac{17}{2}\)
LQ
1
26 tháng 4 2015
S=1/5-1/7+1/7-1/9+1/9-1/11+...+1/93-1/95
S=1/5-1/95
S=18/95
4/5x7 + 4/7x9 + 4/9x11 + ... + 4/50x61
= 2x(2/5x7 + 2/7x9 + 2/9x11 + ... + 2/59x61)
= 2x [(1/5-1/7)+(1/7-1/9)+(1/9-1/11)+...+(1/59-1/61)]
= 2x(1/5-1/61)
= 2x 56/305
= 112/305