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5 tháng 9 2021
Ta có: \(\left(\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+\dfrac{1}{3\cdot4}+...+\dfrac{1}{8\cdot9}\right)\cdot x=\dfrac{23}{45}\)
\(\Leftrightarrow x\left(1-\dfrac{1}{9}\right)=\dfrac{23}{45}\)
\(\Leftrightarrow x=\dfrac{23}{45}\cdot\dfrac{9}{8}=\dfrac{23}{40}\)
5 tháng 9 2021
Đặt A bằng biểu thức trong ngoặc
\(2A=\dfrac{3-1}{1.2.3}+\dfrac{4-2}{2.3.4}+\dfrac{5-3}{3.4.5}+...+\dfrac{10-8}{8.9.10}\)
\(2A=\dfrac{1}{2}-\dfrac{1}{2.3}+\dfrac{1}{2.3}-\dfrac{1}{3.4}+\dfrac{1}{3.4}-\dfrac{1}{4.5}+...+\dfrac{1}{8.9}-\dfrac{1}{9.10}=\dfrac{1}{2}-\dfrac{1}{9.10}\)
\(2A=\dfrac{44}{90}\)
\(A=\dfrac{22}{90}\)
VN
4 tháng 2 2018
Ta có: 1/1.2+1/2.3+1/3.4+...+1/x(x+1)=2/3
=> 1-1/2+1/2-1/3+1/3-1/4+...+1/x-1/x+1=2/3
=>1-1/x+1=2/3
=>1/x+1=1/3
=>3=x+1
=>x=2
4 tháng 2 2018
Ta có\(\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+...+\frac{1}{x\left(x+1\right)}=\frac{2}{3}\)
=>\(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{x}-\frac{1}{x+1}=\frac{2}{3}\)
=>\(1-\frac{1}{x+1}=\frac{2}{3}\)
=>\(\frac{1}{x+1}=1-\frac{2}{3}\)
=>\(\frac{1}{x+1}=\frac{1}{3}\)
=>\(x+1=3\)
=>\(x=2\)
PT
1
18 tháng 7 2018
\(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{x.\left(x+1\right)}=201\)
\(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{x}-\frac{1}{x+1}=201\)
\(1-\frac{1}{x+1}=201\)
\(\frac{1}{x+1}=1-201\)
\(\frac{1}{x+1}=-200\)
\(\Rightarrow x+1=-\frac{1}{200}\)
\(x=-\frac{1}{200}-1\)
\(x=-\frac{201}{200}\)
Vậy \(x=-\frac{201}{200}\)
8 tháng 3 2017
Gọi A = \(\dfrac{1}{1.2}+\dfrac{1}{2.3}+...+\dfrac{1}{x.\left(x+1\right)}=\dfrac{19}{20}\)
\(\Rightarrow\) A = \(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{x}-\dfrac{1}{x+1}\)
\(\Rightarrow\) A = 1 - \(\dfrac{1}{x+1}\)
\(\Rightarrow\) 1 - \(\dfrac{1}{x+1}\) = \(\dfrac{19}{20}\)
\(\Rightarrow1-\dfrac{19}{20}=\dfrac{1}{x+1}\Rightarrow\dfrac{1}{x+1}=\dfrac{1}{20}\)
\(\Rightarrow\) x + 1 = 20\(\Rightarrow\) x=19
6 tháng 8 2016
1/1.2 +1/2.3 +...+ 1/x(x+1) = 2015/2016
<=> 1-1/2 + 1/2 - 1/3 + ... + 1/x - 1/x+1 = 2015/2016
<=> 1 - 1/x+1 = 2015/2016
<=> 1/x+1 = 1/2016
<=> x + 1 = 2016
<=> x = 2015
LH
6 tháng 8 2016
\(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{x\left(x+1\right)}=\frac{2015}{2016}\)
\(\Leftrightarrow\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{x}-\frac{1}{x+1}=\frac{2015}{2016}\)
\(\Leftrightarrow1-\frac{1}{x+1}=\frac{2015}{2016}\)
\(\Leftrightarrow\frac{1}{x+1}=1-\frac{2015}{2016}=\frac{1}{2016}\)
\(\Leftrightarrow x+1=2016\Rightarrow x=2015\)
BT
9 tháng 3 2017
\(\frac{1}{1.2}=\frac{1}{1}-\frac{1}{2}\); \(\frac{1}{2.3}=\frac{1}{2}-\frac{1}{3}\);.....; \(\frac{1}{x.\left(x+1\right)}=\frac{1}{x}-\frac{1}{x+1}\)
=> \(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{x\left(x+1\right)}=1-\frac{1}{x+1}=\frac{x}{x+1}\)
=> \(\frac{x}{x+1}=\frac{19}{20}\)=> 20x=19x+19 => x=19
ĐS: x=19
9 tháng 3 2017
\(\frac{1}{1\times2}+\frac{1}{2\times3}+\frac{1}{3\times4}+...+\frac{1}{x\times\left(x+1\right)}=\frac{19}{20}\)\(\frac{19}{20}\)
\(\Rightarrow1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{x}-\frac{1}{x+1}=\frac{19}{20}\)
\(\Rightarrow1-\frac{1}{x+1}=\frac{19}{20}\)
\(\Rightarrow\frac{x}{x+1}=\frac{19}{20}\)
\(\Rightarrow20x=19x+19\)\(\Rightarrow x=19\)
Vậy \(x=19\)
27 tháng 6 2023
=>1-1/2+1/2-1/3+...+1/x-1/(x+1)=2022/2021
=>1-1/(x+1)=2022/2021
=>1/(x+1)=-1/2021=1/-2021
=>x+1=-2021
=>x=-2022
VK
12 tháng 6 2018
\(\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+..+\frac{1}{49.50}\right)x=\frac{49}{50}\)
\(\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{49}-\frac{1}{50}\right)x=\frac{49}{50}\)
\(\left(1-\frac{1}{50}\right)x=\frac{49}{50}\)
\(\frac{49}{50}x=\frac{49}{50}\)
\(x=\frac{\frac{49}{50}}{\frac{49}{50}}\)
\(x=1\)
Vậy \(x=1\)
Sửa đề : \(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{x\left(x+1\right)}=\frac{1}{8}\)
\(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{x}-\frac{1}{x+1}=\frac{1}{8}\)
\(1-\frac{1}{x+1}=\frac{1}{8}\)
\(\frac{1}{x+1}=\frac{7}{8}\Leftrightarrow8=7x+7\Leftrightarrow x=\frac{1}{7}\)