Tìm X ,biêt 1+1/3 +1/6+1/10+...+2/X.(x+1) =1 2025/2027
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TH
Thầy Hùng Olm
Manager
VIP
23 tháng 6 2023
2 x A = 1 - \(\dfrac{1}{2027}\)
\(A=\dfrac{1013}{2027}\)
MT
1
DN
18 tháng 5 2023
Đề có phải là:
\(\dfrac{x+1}{2024}+\dfrac{x+2}{2025}+\dfrac{x+3}{2026}+\dfrac{x+4}{2027}=4\text{ ?}\)
\(\Rightarrow\text{ }\dfrac{x+1}{2024}+\dfrac{x+2}{2025}+\dfrac{x+3}{2026}+\dfrac{x+4}{2027}-4=0\)
\(\Rightarrow\text{ }\dfrac{x+1}{2024}+\dfrac{x+2}{2025}+\dfrac{x+3}{2026}+\dfrac{x+4}{2027}-1-1-1-1=0\)
\(\Rightarrow\left(\dfrac{x+1}{2024}-1\right)+\left(\dfrac{x+2}{2025}-1\right)+\left(\dfrac{x+3}{2026}-1\right)+\left(\dfrac{x+4}{2027}-1\right)=0\)
\(\Rightarrow\left(\dfrac{x+1-2024}{2024}\right)+\left(\dfrac{x+2-2025}{2025}\right)+\left(\dfrac{x+3-2026}{2026}\right)+\left(\dfrac{x+4-2027}{2027}\right)=0\)
\(\Rightarrow\dfrac{x-2023}{2024}+\dfrac{x-2023}{2025}+\dfrac{x-2023}{2026}+\dfrac{x-2023}{2027}=0\)
\(\Rightarrow\left(x-2023\right)\left(\dfrac{1}{2024}+\dfrac{1}{2025}+\dfrac{1}{2026}+\dfrac{1}{2027}\right)=0\)
Mà \(\dfrac{1}{2024}+\dfrac{1}{2025}+\dfrac{1}{2026}+\dfrac{1}{2027}\ne0\)
\(\Rightarrow x-2023=0\)
\(\Rightarrow x=0+2023\)
\(\Rightarrow x=2023\)
Vậy, \(x=2023.\)
GN
0
NT
1
KM
0
14 tháng 4 2017
(1+2+3+4+5+6+7+8+9+...............................+2016+2025) x (24,2 - 24,2) = (1 + 2 +3+4+5+6+7+8+9+...............................+2016+2025) x 0 = 0
NA
9 tháng 1 2024
a, 2\(^3\) . x + 2005\(^0\) . x = 994-15:3+1\(^{2025}\)
8 .x + 1 . x = 990
x . [ 8 +1 ] = 990
x . 9 = 990
x = 990 : 9
x = 110
10 tháng 1 2024
a: \(\left(2^3\right)^{1^{2005}}\cdot x+2005^0\cdot x=9915:3+1^{2025}\)
=>\(8\cdot x+1\cdot x=3305+1\)
=>\(9x=3306\)
=>\(x=\dfrac{3306}{9}=\dfrac{1102}{3}\)
b: \(2^x+2^{x+1}+2^{x+2}+2^{x+3}=480\)
=>\(2^x+2^x\cdot2+2^x\cdot4+2^x\cdot8=480\)
=>\(2^x\left(1+2+4+8\right)=480\)
=>\(2^x\cdot15=480\)
=>\(2^x=32\)
=>\(2^x=2^5\)
=>x+5
NH
0
Q
26 tháng 10 2017
a) \(\frac{1}{2}+\frac{2}{3}x=\frac{4}{5}\)
\(x=\frac{\left(\frac{4}{5}-\frac{1}{2}\right)}{\frac{2}{3}}\)
\(x=\frac{9}{20}\)
b) \(\left|x+\frac{3}{4}\right|-\frac{1}{2}=0\)
\(\left|x+\frac{3}{4}\right|=0+\frac{1}{2}\)
\(\left|x+\frac{3}{4}\right|=\frac{1}{2}\)
\(\Rightarrow\hept{\begin{cases}x+\frac{3}{4}=\frac{1}{2}\\x+\frac{3}{4}=-\frac{1}{2}\end{cases}\Rightarrow\hept{\begin{cases}x=\frac{-1}{4}\\x=\frac{-5}{4}\end{cases}}}\)
Vậy x=-1/4 hoặc x=-5/4
c) \(\left(x+\frac{1}{3}\right)^3=\frac{-1}{8}\)
\(\Leftrightarrow x+\frac{1}{3}=\frac{-1}{8}=\frac{\left(-1\right)^3}{2^3}=\frac{-1}{2}\)
\(x=\frac{-1}{2}-\frac{1}{3}\)
\(x=\frac{-5}{6}\)
26 tháng 10 2017
\(\frac{1}{2}+\frac{2}{3}x=\frac{4}{5}\)
\(\frac{2}{3}x=\frac{4}{5}-\frac{1}{2}\)
\(\frac{2}{3}x=\frac{3}{10}\)
\(x=\frac{3}{10}:\frac{2}{3}\)
\(x=\frac{9}{20}\)
b) l x + 3/4 l - 1/2 = 0
l x + 3/4 l = 1/2
TH1 : \(x+\frac{3}{4}\le0\) TH2: \(x+\frac{3}{4}\ge0\)
=> \(x+\frac{3}{4}=-\frac{1}{2}\) => \(x+\frac{3}{4}=\frac{1}{2}\)
\(x=-\frac{1}{2}-\frac{3}{4}\) \(x=\frac{1}{2}-\frac{3}{4}\)
\(x=-\frac{5}{4}\) \(x=-\frac{1}{4}\)
c) ( x + 1/3 )3 = ( -1/8 )
( x + 1/3 ) 3 = ( -1/3 )3
=> x + 1/3 = -1/3
x = -1/3 - 1/3
x = -2/3
\(1+\dfrac{1}{3}+\dfrac{1}{6}+...+\dfrac{2}{x\left(x+1\right)}=1\dfrac{2025}{2027}\)
=>\(\dfrac{2}{2}+\dfrac{2}{6}+\dfrac{2}{12}+...+\dfrac{2}{x\left(x+1\right)}=\dfrac{4052}{2027}\)
=>\(2\left(\dfrac{1}{2}+\dfrac{1}{6}+...+\dfrac{1}{x\left(x+1\right)}\right)=\dfrac{4052}{2027}\)
=>\(\dfrac{1}{2}+\dfrac{1}{6}+...+\dfrac{1}{x\left(x+1\right)}=\dfrac{2026}{2027}\)
=>\(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{x}-\dfrac{1}{x+1}=\dfrac{2026}{2027}\)
=>\(1-\dfrac{1}{x+1}=\dfrac{2026}{2027}\)
=>\(\dfrac{1}{x+1}=\dfrac{1}{2027}\)
=>x+1=2027
=>x=2026