Tìm x , biết :
1/1.2 + 1/2.3 + 1/3.4 + 1/4.5 + ....... + 1/x.(x + 1) = 2018/2019
Giúp em nha !
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\(\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+\frac{1}{4\cdot5}+...+\frac{1}{x(x+1)}=\frac{2019}{2020}\)
\(\Rightarrow1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{x}-\frac{1}{x+1}=\frac{2019}{2020}\)
\(\Rightarrow1-\frac{1}{x+1}=\frac{2019}{2020}\)
\(\Rightarrow\frac{1}{x+1}=1-\frac{2019}{2020}\)
\(\Rightarrow\frac{1}{x+1}=\frac{1}{2020}\)
\(\Rightarrow x+1=2020\Leftrightarrow x=2019\)
Vậy x = 2019
ta có 1/1.2+1/2.3+1/3.4+1/4.5+...+1/x.(x+1)=17/18
1-1/2+1/2-1/3+1/3-1/4+1/4-1/5+...+1/x-1/x+1=17/18
1-1/x+1=17/18
1/x+1=1-17/18
1/x+1=1/18
suy ra: x+1=18
x=18-1
x=17
\(\frac{1313}{1212}:x=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+\)\(\frac{1}{5.6}\)
\(\Leftrightarrow\frac{13}{12}:x=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}\)
\(\Leftrightarrow\frac{13}{12}:x=1-\frac{1}{6}\)
\(\Leftrightarrow\frac{13}{12}:x=\frac{5}{6}\)
\(\Leftrightarrow x=\frac{13}{12}:\frac{5}{6}\)
\(\Leftrightarrow x=\frac{13}{10}\)
Vậy \(x=\frac{13}{10}\)
~~~~~Hok tốt ~~~~~
a,\(\frac{1313}{1212}\div x=\frac{1}{1\times2}+\frac{1}{2\times3}+\frac{1}{3\times4}+\frac{1}{4\times5}+\frac{1}{5\times6}\)
\(\frac{13}{12}\div x=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}\)
\(\frac{13}{12}\div x=1-\frac{1}{6}\)
\(\frac{13}{12}\div x=\frac{5}{6}\)
\(x=\frac{13}{12}\div\frac{5}{6}\)
\(x=\frac{13}{12}\times\frac{6}{5}\)
\(x=\frac{13}{10}\)
Chúc bạn hok tốt !
\(\frac{1313}{1212}:x=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+\)\(\frac{1}{5.6}\)
\(\Leftrightarrow\frac{13}{12}:x=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}\)
\(\Leftrightarrow\frac{13}{12}:x=1-\frac{1}{6}\)
\(\Leftrightarrow\frac{13}{12}:x=\frac{5}{6}\)
\(\Leftrightarrow x=\frac{13}{12}:\frac{5}{6}\)
\(\Leftrightarrow x=\frac{13}{10}\)
Hok tốt
=> 1 - 1/2 + 1/2 - 1/3 + 1/3 - 1/4 + ........ + 1 /x - 1/x + 1 = 17 /18
=> 1 - 1/x+1 = 17/18
=> 1/x+1 = 1/18
=> x + 1 = 18
=> x = 17 (tm)
Vậy x = 17 nha!
Ai mk mk lại !!
1/1.2 + 1/2.3 + 1/3.4 +......+ 1/x(x+1) = 17/18
=> 1- 1/x+1 = 17/18
=> 1/x +1 = 1-17/18
=> 1/x+1 = 1/18
=> x= 17
đặt A = 1.2. + 2.3 + 3.4 + ... + 49.50
3A = 1.2.3 + 2.3.3 + 3.4.3 + ... + 49.50.3
3A = 1.2.3 + 2.3.(4 - 1) + 3.4.(5 - 2) + ... + 49.50.(51 - 48)
3A = 1.2.3 + 2.3.4 - 1.2.3 + 3.4.5 - 2.3.4 + ... + 49.50.51 - 48.49.50
3A = 49.50.51
A = 41650
Thay vào ta được
41650 + 1/2x = 40642
=> 1/2x = 1008
=> x = 2016
1/1-1/2+1.2-1/3+1/3-1/4+..+1/x-1/x+1=2018/2019
1-1/x+1=2018/2019
1-2018/2019=1/x+1
1/2019=1/x+1
=>x+1=2019
=>x=2018
vậy...
\(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{x\left(x+1\right)}=\frac{2018}{2019}.\)
\(\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{x}-\frac{1}{x+1}=\frac{2018}{2019}.\)
\(\frac{1}{1}-\frac{1}{x+1}=\frac{2018}{2019}\)
\(\frac{1}{1}-\frac{2018}{2019}=\frac{1}{x+1}\)
\(\frac{1}{2019}=\frac{1}{x+1}\)
=> \(2019=x+1\)
\(x+1=2019\)
\(x=2019-1\)
\(x=2018\)
Vậy x = 2018