Tìm x thuộc N sao (x khác 0 ý). Giúp mk vs! 2 tick nhé!
\(\frac{1}{1.3}+\frac{1}{3.5}+...+\frac{1}{x\left(x+2\right)}=\frac{50}{102}\)
K
Khách
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29 tháng 12 2016
A\(A=\frac{1}{1.3}+..+\frac{1}{x\left(x+1\right)}\)
\(2A=\frac{1}{1}-\frac{1}{\left(x+1\right)}\)
\(A=\frac{x}{2.\left(x+1\right)}=\frac{8}{17}=\frac{16}{2.17}\)
X=16
XO
20 tháng 7 2021
ĐK : 51x \(\ge0\Rightarrow x\ge0\)
Với \(x\ge0\)thì \(x+\frac{1}{1.3}>0;x+\frac{1}{3.5}>0;...;x+\frac{1}{99.101}>0\)
Khi đó : \(\left|x+\frac{1}{1.3}\right|+\left|x+\frac{1}{3.5}\right|+\left|x+\frac{1}{5.7}\right|+...+\left|x+\frac{1}{99.101}\right|=51x\)
<=> \(x+\frac{1}{1.3}+x+\frac{1}{3.5}+x+\frac{1}{5.7}+....+x+\frac{1}{99.101}=51x\)(50 hạng tử x ở VT)
<=> \(50x+\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{99.101}=51x\)
<=> \(x=\frac{1}{2}.\left(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{1}{99.101}\right)\)
<=> \(x=\frac{1}{2}\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{99}-\frac{1}{101}\right)\)
<=> \(x=\frac{1}{2}\left(1-\frac{1}{101}\right)=\frac{50}{101}\)
Vậy x = 50/101
LD
6
13 tháng 8 2019
\(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{x\left(x+2\right)}=\frac{20}{41}\)
\(\Leftrightarrow1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{x}-\frac{1}{x+2}=\frac{20}{41}\)
\(\Leftrightarrow1-\frac{1}{x+2}=\frac{21}{41}\)
\(\Leftrightarrow\frac{1}{x+2}=1-\frac{21}{41}\)
\(\Leftrightarrow\frac{1}{x+2}=\frac{20}{41}\)
\(\Leftrightarrow20\left(x+2\right)=41\)
\(\Leftrightarrow x-2=\frac{41}{20}\)
\(\Leftrightarrow x=\frac{41}{20}+2\)
\(\Leftrightarrow x=\frac{81}{20}\)
S
13 tháng 8 2019
\(\frac{1}{1.3}+...+\frac{1}{a\left(a+2\right)}=\frac{1}{2}\left(\frac{2}{1.3}+\frac{2}{3.5}+....+\frac{2}{a\left(a+2\right)}\right)=\frac{1}{2}\left(1-\frac{1}{3}+....-\frac{1}{a+2}\right)\)
\(=\frac{1}{2}\left(1-\frac{1}{a+2}\right)=\frac{20}{41}\Rightarrow a+2=41\Leftrightarrow a=39\)
6 tháng 2 2022
a: \(\Leftrightarrow\dfrac{x-214}{86}-1+\dfrac{x-132}{84}-2+\dfrac{x-54}{82}-3=0\)
=>x-300=0
hay x=300
Y
26 tháng 3 2019
Xinloi, t ghi thiếu đề
\(\left|x+\frac{1}{1.3}\right|+\left|x+\frac{1}{3.5}\right|+...+\left|x+\frac{1}{97.99}\right|=50x\)
NH
26 tháng 3 2019
\(\left|x+\frac{1}{1.3}\right|+\left|x+\frac{1}{3.5}\right|+...+\left|x+\frac{1}{97.99}\right|=50x\)
Vì \(\left|x+\frac{1}{1.3}\right|\ge0\forall x\)
\(\left|x+\frac{1}{3.5}\right|\ge0\forall x\)
................
\(\left|x+\frac{1}{97.99}\right|\ge0\forall x\)
(VT: Vế trái; VP: Vế phải)
\(\Rightarrow VT\ge0\Rightarrow VP=50x\ge0\)mà \(50>0\)
\(\Rightarrow x>0\)
\(\Rightarrow x+\frac{1}{1.3}>0\forall x\)
..............
\(x+\frac{1}{97.99}>0\forall x\)(1)
(1) \(\Leftrightarrow x+\frac{1}{1.3}+x+\frac{1}{3.5}+...+x+\frac{1}{97.99}=50x\)
\(\Leftrightarrow49x+\left(\frac{1}{1.3}+...+\frac{1}{97.99}\right)=50x\)
\(\Leftrightarrow50x-49x=\frac{1}{2}\left(\frac{2}{1.3}+...+\frac{2}{97.99}\right)\)
\(\Leftrightarrow x=\frac{1}{2}\left(1-\frac{1}{3}+...+\frac{1}{97}-\frac{1}{99}\right)\)
\(\Leftrightarrow x=\frac{1}{2}\left(1-\frac{1}{99}\right)\)
\(\Leftrightarrow x=\frac{1}{2}\cdot\frac{98}{99}=\frac{49}{99}\)
Vậy....
P/s: Làm bừa :) Ko chắc đúng nhé
1/1.3+1/3.5+............+1/x.(x+2) = 50/102
2/1.3+2/3.3+...........+2/x.(x+2) =50/51
1-1/x+2=50/51
x+1/x+2=50/51
(x+1).51=(x+2).50
51x+51=50x+100
51x-50x=100-51
x=49
vay x=49