2/5.8 + 2/8.11 + .. + 2/x.(x+3) = 1/5
Tìm x
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NT
1
6 tháng 9 2020
a) 2/2.5 + 2/5.8 + 2/8.11 + ... + 2/x(x+3) = 7/23
3/2.5 + 3/5.8 + 3/8.11 + ... + 3/x(x+3) = 21/46
1/2 - 1/5 + 1/5 - 1/8 + 1/8 - 1/11 + ... + 1/x - 1/x+1 = 21/46
1/2 - 1/x+1 = 21/46
=> 1/x+1 = 1/23
=> x + 1 = 23
=> x = 22
Vậy x = 22.
b) 3/4 . x - 1/5 = 7/4 . x + 11/5
3/4 . x - 7/4 . x = 1/5 + 11/5
x (3/4 - 7/4) = 12/5
-x = 12/5
x = -12/5
Vậy x = -12/5.
TH
1
7 tháng 8 2017
Đặt \(A=\frac{3}{5.8}+\frac{3}{8.11}+\frac{3}{11.14}+...+\frac{3}{32.35}\)
\(A=\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+\frac{1}{11}-\frac{1}{14}+...+\frac{1}{32}-\frac{1}{35}\)
\(A=\frac{1}{5}-\frac{1}{35}=\frac{6}{35}\)
\(\Rightarrow x+\frac{6}{35}=-\frac{2}{7}\Rightarrow x=-\frac{2}{7}-\frac{6}{35}=-\frac{16}{35}\)
TN
13 tháng 3 2016
còn chi tiết đây
a)1/5.8+1/8.11+1/11.14+...+1/x.(x+3)=101/1540
1/(5.8)+1/(8.11)+1/(11.14)+...1/x.(… =101/1540
3/(5.8)+3/(8.11)+...+3/x(x+3)=3.(10…
1/5-1/8+1/8-1/11+...+1/x-1/(x+3)=30…
1/5-1/(x+3)=303/1540
1/(x+3)=1/5-303/1540=1/308
=>x=305
TN
13 tháng 3 2016
lời giải nè : ấn vô dòng đen đen ở dưới ấy nhé
Tìm x, biết:a) 1/5.8 + 1/8.11 + 1/11.14 + ... + 1/x.(x+3)= 101/1540b) 1+ 1/3 + 1/6 + 1/10 +...+ 1/x.(x+1):2 = $1\frac{1991}{1993}$119911993
SK
1
QD
26 tháng 4 2017
Đề hình như bị sai ban ơi sửa lại
\(A=\dfrac{3}{2.5}+\dfrac{3}{5.8}+\dfrac{3}{8.11}+...+\dfrac{3}{92.95}\)
\(A=3\left(\dfrac{1}{2.5}+\dfrac{1}{5.8}+\dfrac{1}{8.11}+...+\dfrac{1}{92.95}\right)\)
\(A=3.\dfrac{1}{3}\left(\dfrac{1}{2}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{8}+\dfrac{1}{8}-\dfrac{1}{11}+...+\dfrac{1}{92}-\dfrac{1}{95}\right)\)
\(A=\dfrac{1}{2}-\dfrac{1}{95}\)
\(A=\dfrac{93}{190}\)
\(B=\dfrac{2}{2.5}+\dfrac{2}{5.8}+\dfrac{2}{8.11}+...+\dfrac{2}{92.95}\)
\(3B=2\left(\dfrac{1}{2.5}+\dfrac{1}{5.8}+\dfrac{1}{8.11}+...+\dfrac{1}{92.95}\right)\)
\(3B=2.\left(\dfrac{1}{2}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{8}+\dfrac{1}{8}-\dfrac{1}{11}+...+\dfrac{1}{92}-\dfrac{1}{95}\right)\)
\(3B=2\left(\dfrac{1}{2}-\dfrac{1}{95}\right)\)
\(3B=2.\dfrac{93}{190}\)
\(3B=\dfrac{93}{95}\)
\(\Rightarrow B=\dfrac{31}{95}\)
DQ
1
NY
4
10 tháng 5 2019
\(A=\frac{2}{2.5}+\frac{2}{5.8}+...+\frac{2}{95.98}\)
\(A=\frac{2}{3}.\left(\frac{3}{2.5}+\frac{3}{5.8}+...+\frac{3}{95.98}\right)\)
\(A=\frac{2}{3}.\left(\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+...+\frac{1}{95}-\frac{1}{98}\right)\)
\(A=\frac{2}{3}.\left(\frac{1}{2}-\frac{1}{98}\right)\)
\(A=\frac{2}{3}.\frac{24}{49}\)
\(A=\frac{16}{49}\)
KN
11 tháng 5 2019
\(A=\frac{2}{2.5}+\frac{2}{5.8}+\frac{2}{8.11}+...+\frac{2}{95.98}\)
\(\Leftrightarrow\frac{3}{2}A=\frac{3}{2.5}+\frac{3}{5.8}+\frac{3}{8.11}+...+\frac{3}{95.98}\)
\(\Leftrightarrow\frac{3}{2}A=\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+...+\frac{1}{95}-\frac{1}{98}\)
\(\Leftrightarrow\frac{3}{2}A=\frac{1}{2}-\frac{1}{98}\)
\(\Leftrightarrow\frac{3}{2}A=\frac{48}{98}=\frac{24}{49}\)
\(\Leftrightarrow A=\frac{24}{49}\div\frac{3}{2}\)
\(\Leftrightarrow A=\frac{48}{147}\)
2/5.8 + 2/8.11 + .. + 2/x.(x+3) = 1/5
=> 2/3(3/5*8 + 3/8*11 + ... + 3/n(n+3)) = 1/5
=> 2/3(1/5 - 1/8 + 1/8 - 1/11 + ... + 1/n - 1/n + 3) = 1/5
=> 2/3(1/5 - 1/n+3) = 1/5
=> 2/15 - 2/n + 3 = 1/5
=> 2/n + 3 = -1/15
=> 2/n + 3 = 2/-30
=> n + 3 = -30
=> n = -33
Bài làm:
Ta có:\(\frac{2}{5.8}+\frac{2}{8.11}+...+\frac{2}{x\left(x+3\right)}=\frac{1}{5}\)
\(\Leftrightarrow\frac{2}{3}\left(\frac{3}{5.8}+\frac{3}{8.11}+...+\frac{3}{x\left(x+3\right)}\right)=\frac{1}{5}\)
\(\Leftrightarrow\frac{8-5}{5.8}+\frac{11-8}{8.11}+...+\frac{x+3-x}{x\left(x+3\right)}=\frac{1}{5}\div\frac{2}{3}\)
\(\Leftrightarrow\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+...+\frac{1}{x}-\frac{1}{x+3}=\frac{3}{10}\)
\(\Leftrightarrow\frac{1}{5}-\frac{1}{x+3}=\frac{3}{10}\)
\(\Leftrightarrow\frac{x+3-5}{5\left(x+3\right)}=\frac{3}{10}\)
\(\Leftrightarrow10\left(x-2\right)=15\left(x+3\right)\)
\(\Leftrightarrow10x-20=15x+45\)
\(\Leftrightarrow5x=-65\)
\(\Rightarrow x=-13\)
Vậy \(x=-13\)
Học tốt!!!!