tìm x biết 1/1.2+1/2.3+...+1/(x-1)x=2 (x khác 0; x khác 1; x thuộc Z)
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25 tháng 3 2021
\(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{x.\left(x+1\right)}=\frac{2014}{2015}\)
\((1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{x}-\frac{1}{x+1})=\frac{2014}{2015}\)
\(\Rightarrow1-\frac{1}{x+1}=\frac{2014}{2015}\)
\(\Rightarrow\frac{1}{x+1}=\frac{1}{2015}\)
\(\Rightarrow x+1=2015\)
\(\Leftrightarrow x=2014\)
Vậy x=2014
NA
20 tháng 3 2022
\(x\cdot\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{49}-\dfrac{1}{50}\right)=1\\ x\cdot\left(1-\dfrac{1}{50}\right)=1\\ \dfrac{49}{50}x=1\\ x=1:\dfrac{49}{50}\\ x=\dfrac{50}{49}\)
VD
20 tháng 3 2022
\(x.\left(\dfrac{1}{1.2}+\dfrac{1}{2.3}+...+\dfrac{1}{49.50}\right)=1\\ \Rightarrow x.\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{49}-\dfrac{1}{50}\right)=1\\ \Rightarrow x.\left(1-\dfrac{1}{50}\right)=1\\ \Rightarrow x.\dfrac{49}{50}=1\\ \Rightarrow x=1:\dfrac{49}{50}\\ \Rightarrow x=\dfrac{50}{49}\)
TQ
0
NC
2
S
16 tháng 4 2017
a, \(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{x\left(x+1\right)}=\frac{44}{45}\)
=> \(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{x}-\frac{1}{x+1}=\frac{44}{45}\)
=> \(1-\frac{1}{x+1}=\frac{44}{45}\)
=> \(\frac{x}{x+1}=\frac{44}{45}\)
=> x = 44
b, Ta có: \(\frac{1}{2^2}< \frac{1}{1.2}=1-\frac{1}{2}\)
\(\frac{1}{3^2}< \frac{1}{2.3}=\frac{1}{2}-\frac{1}{3}\)
.................
\(\frac{1}{45^2}< \frac{1}{44.45}=\frac{1}{44}-\frac{1}{45}\)
=> \(\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{45^2}< 1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{44}-\frac{1}{45}=1-\frac{1}{45}< 1\)
Vậy \(\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{45^2}< 1\)
BT
16 tháng 4 2017
a) 1/1.2+1/2.3+1/3.4+...+1/x(x+1)=1-1/2+1/2-1/3+1/3-1/4+....+1/x-1/(x+1)=1-1/(x+1)=x/(x+1)=44/45
=> x=44
b/ 1/22 < 1/1.2; 1/32 < 1/2.3; ....; 1/452 < 1/44.45
=> A < 1/1.2+1/2.3+...+1/44.45=1-1/45=44/45 < 1
=> A < 1
15 tháng 7 2019
a) x - 3/97 + x - 2/98 = x - 1/99 + x/100
<=> x + 1/99 + 1 + x + 2/98 + 1 + x + 3/97 + 1 + (x + 4/96 + 1 + x + 5/95 + 1 + x + 10/90 + 1) = 0
<=> x + 100/99 + x + 100/98 + x + 100/97 + (x + 100/96 + x + 100/95 + x + 100/90) = 0
<=> (x + 100)(1/99 + 1/98 + 1/97 + 1/96 + 1/95 + 1/90) = 0
Mà 1/99 + 1/98 + 1/97 + 1/96 + 1/95 + 1/90 khác 0
=> x + 100 = 0
=> x = -100
c) (1/1.2 + 1/2.3 + ... + 1/99.100) - 2x = 1/2
<=> (1 - 1/2 + 1/2 - 1/3 + ... + 1/99 - 1/100) - 2x = 1/2
<=> (1 - 1/100) - 2x = 1/2
<=> 99/100 - 2x = 1/2
<=> -2x = 1/2 - 99/100
<=> -2x = -49/100
<=> x = 49/200
=> x = 49/200
15 tháng 7 2019
\(\frac{x+2}{327}+\frac{x+3}{326}+\frac{x+4}{325}+\frac{x+5}{324}+\frac{x+349}{5}=0\)
\(\Rightarrow\left(\frac{x+2}{327}+1\right)+\left(\frac{x+3}{326}+1\right)+\left(\frac{x+4}{325}+1\right)+\left(\frac{x+5}{324}+1\right)+\left(\frac{x+349}{5}-4\right)=0\)
\(\Rightarrow\frac{x+329}{327}+\frac{x+329}{326}+\frac{x+329}{325}+\frac{x+329}{324}+\frac{x+329}{5}=0\)
\(\Rightarrow\left(x+329\right)\left(\frac{1}{327}+\frac{1}{326}+\frac{1}{325}+\frac{1}{324}+\frac{1}{5}\right)=0\)
Dễ thấy \(\frac{1}{327}+\frac{1}{326}+\frac{1}{325}+\frac{1}{324}+\frac{1}{5}>0\Rightarrow x+329=0\)
\(\Rightarrow x=-329\)
TV
13 tháng 8 2017
\(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{x\left(x+1\right)}=\frac{75}{76}\)
\(\frac{1}{1}-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{x}-\frac{1}{x+1}=\frac{75}{76}\)
\(\frac{1}{1}-\frac{1}{x+1}=\frac{75}{76}\)
\(\frac{1}{x+1}=1-\frac{75}{76}\)
\(\frac{1}{x+1}=\frac{1}{76}\)
\(\Rightarrow x+1=76\)
\(x=75\)
vậy \(x=75\)
\(\frac{1}{1.2}+\frac{1}{2.3}+....+\frac{1}{\left(x-1\right)x}=2\)
\(\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+......+\frac{1}{x-1}-\frac{1}{x}=2\)
suy ra \(1-\frac{1}{x}=2\)
hay \(\frac{x-1}{x}=2\) .suy ra x-1=2x .tính ra ta có x=-1