4/2.5 + 4/5.8 +...+ 4/x(x+3) = 22/35
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Những câu hỏi liên quan
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.
KT
12 tháng 8 2018
mạo phép chỉnh đề:
\(\frac{3x}{2.5}+\frac{3x}{5.8}+\frac{3x}{8.11}+\frac{3x}{11.14}=\frac{1}{21}\)
<=> \(x\left(\frac{3}{2.5}+\frac{3}{5.8}+\frac{3}{8.11}+\frac{3}{11.14}\right)=\frac{1}{21}\)
<=> \(x\left(\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+\frac{1}{11}-\frac{1}{14}\right)=\frac{1}{21}\)
<=> \(x\left(\frac{1}{2}-\frac{1}{14}\right)=\frac{1}{21}\)
<=> \(x.\frac{3}{7}=\frac{1}{21}\)
<=> \(x=\frac{1}{9}\)
Vậy...
NT
1
S
7 tháng 4 2023
\(\left(\dfrac{3}{2.5}+\dfrac{3}{5.8}+...+\dfrac{3}{17.20}\right).x=\dfrac{45}{23}\)
\(\left(\dfrac{1}{2}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{8}+....+\dfrac{1}{17}-\dfrac{1}{20}\right).x=\dfrac{45}{23}\)
\(\left(\dfrac{1}{2}-\dfrac{1}{20}\right).x=\dfrac{45}{23}\)
\(\dfrac{9}{20}\cdot x=\dfrac{45}{23}\)
\(x=\dfrac{45}{23}:\dfrac{9}{20}\)
\(x=\dfrac{100}{23}\)
TA
7 tháng 4 2023
\(\left(\dfrac{3}{2.5}+\dfrac{3}{5.8}+...+\dfrac{1}{17.20}\right)x=\dfrac{45}{23}\Leftrightarrow\left(\dfrac{1}{2}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{8}+...+\dfrac{1}{17}-\dfrac{1}{20}\right)x=\dfrac{45}{23}\)\(\Leftrightarrow\left(\dfrac{1}{2}-\dfrac{1}{20}\right)x=\dfrac{45}{23}\)
\(\Rightarrow\dfrac{9}{20}.x=\dfrac{45}{23}\)
\(\Rightarrow x=\dfrac{100}{23}\)
24 tháng 2 2022
\(=\dfrac{4}{3}\left(\dfrac{1}{2}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{8}+...+\dfrac{1}{302}-\dfrac{1}{305}\right)=\dfrac{4}{3}\cdot\dfrac{303}{610}=\dfrac{202}{305}\)
24 tháng 2 2022
\(\dfrac{4}{2.5}+\dfrac{4}{5.8}+\dfrac{4}{8.11}+...+\dfrac{4}{302.305}\)
\(=4\left(\dfrac{1}{2.5}+\dfrac{1}{5.8}+\dfrac{1}{8.11}+...+\dfrac{1}{302.305}\right)\)
\(=\dfrac{4}{3}\left(\dfrac{3}{2.5}+\dfrac{3}{5.8}+\dfrac{3}{8.11}+...+\dfrac{3}{302.305}\right)\)
\(=\dfrac{4}{3}\left(\dfrac{1}{2}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{8}+...+\dfrac{1}{302}-\dfrac{1}{305}\right)\)
\(=\dfrac{4}{3}\left(\dfrac{1}{2}-\dfrac{1}{305}\right)\)
\(=\dfrac{4}{3}.\dfrac{303}{610}\\ =\dfrac{202}{305}\)
NT
Nguyễn Thị Thương Hoài
Giáo viên
VIP
30 tháng 7 2023
A = \(\dfrac{4}{2.5}\) + \(\dfrac{4}{5.8}\)+...+ \(\dfrac{4}{47.50}\)
A = \(\dfrac{4}{3}\).( \(\dfrac{3}{2.5}+\dfrac{3}{5.8}+...+\dfrac{3}{47.50}\))
A = \(\dfrac{4}{3}\).(\(\dfrac{1}{2}\) - \(\dfrac{1}{5}\) + \(\dfrac{1}{5}\) - \(\dfrac{1}{8}\)+...+ \(\dfrac{1}{47}\) - \(\dfrac{1}{50}\))
A = \(\dfrac{4}{3}\).( \(\dfrac{1}{2}\) - \(\dfrac{1}{50}\))
A = \(\dfrac{4}{3}\). \(\dfrac{24}{50}\)
A = \(\dfrac{16}{25}\)
\(\frac{4}{2\cdot5}+\frac{4}{5\cdot8}+...+\frac{4}{x\cdot\left(x+3\right)}=\frac{22}{35}\)
\(\frac{3}{2\cdot5}+\frac{3}{5\cdot8}+...+\frac{3}{x\cdot\left(x+3\right)}=\frac{33}{70}\)
\(\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+...+\frac{1}{x}-\frac{1}{x+3}=\frac{33}{70}\)
\(\frac{1}{2}-\frac{1}{x+3}=\frac{33}{70}\)
\(\frac{1}{x+3}=\frac{1}{35}\)
\(\Rightarrow x+3=35\)
\(\Rightarrow x=32\)
\(\frac{4}{2\cdot5}+\frac{4}{5\cdot8}+...+\frac{4}{x\left(x+3\right)}=\frac{22}{35}\)
\(\frac{3}{4}\left(\frac{4}{2\cdot5}+\frac{4}{5\cdot8}+...+\frac{4}{x\left(x+3\right)}\right)=\frac{3}{4}\cdot\frac{22}{35}\)
\(\frac{3}{2\cdot5}+\frac{3}{5\cdot8}+...+\frac{3}{x\left(x+3\right)}=\frac{33}{70}\)
\(\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+...+\frac{1}{x}-\frac{1}{x+3}=\frac{33}{70}\)
\(\frac{1}{2}-\frac{1}{x+3}=\frac{33}{70}\)
\(\frac{1}{x+3}=\frac{1}{35}\)
\(\Rightarrow x+3=35\)
\(\Rightarrow x=32\)
Vậy x=32