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Ta có :
\(\frac{1}{2018x}=\left(1-\frac{1}{2}\right)\left(1-\frac{1}{3}\right)\left(1-\frac{1}{4}\right)...\left(1-\frac{1}{2017}\right)\left(1-\frac{1}{2018}\right)\)
\(\Rightarrow\frac{1}{2018x}=\left(\frac{2}{2}-\frac{1}{2}\right)\left(\frac{3}{3}-\frac{1}{3}\right)\left(\frac{4}{4}-\frac{1}{4}\right)...\left(\frac{2017}{2017}-\frac{1}{2017}\right)\left(\frac{2018}{2018}-\frac{1}{2018}\right)\)
\(\Rightarrow\frac{1}{2018x}=\frac{1}{2}.\frac{2}{3}.\frac{3}{4}...\frac{2016}{2017}.\frac{2017}{2018}\)
\(\Rightarrow\frac{1}{2018x}=\frac{1}{2018}\)
\(\Rightarrow2018x=2018\)
\(\Rightarrow x=2018:2018\)
\(\Rightarrow x=1\)
Vậy \(x=1\)
Chúc bạn học tốt !!!
1/2018 * x = ( 1 - 1/2 ) * ( 1 - 1/3 ) * ( 1 - 1/4 ) * ... ( 1 - 1/2018 )
1/2018 * x = 1/2 * 2/3 * 3/4 * ... * 2017/2018
1/2018 * x = 1/2018
x = 1/2018 : 1/2018
x = 1
\(\frac{2}{3}\cdot x+\frac{3}{4}=3\)
\(\frac{2}{3}\cdot x=3-\frac{3}{4}\)
\(\frac{2}{3}\cdot x=\frac{12}{4}-\frac{3}{4}\)
\(\frac{2}{3}x=\frac{9}{4}\)
\(x=\frac{9}{4}:\frac{2}{3}\)
\(x=\frac{27}{8}\)
\(\frac{2}{3}\)X+ \(\frac{3}{4}\)= 3.
=> \(\frac{2}{3}\)X= 3- \(\frac{3}{4}\).
=> \(\frac{2}{3}\)X= \(\frac{9}{4}\).
=> X= \(\frac{9}{4}\): \(\frac{2}{3}\).
=> X= \(\frac{27}{8}\).
Vậy X= \(\frac{27}{8}\).
`x + x +x + 91= ( - 2 )`
`=> 3x+91=(-2)`
`=> 3x=-2-91`
`=>3x=-93`
`=>x=-93:3`
`=>x=-31`
2017^2018 - 2017^2017 = 2017
2017^2019-2017^2018=2017
=>2017^2018-2017^2017 = 2017^2019-2017^2018
TH1: x<-2018 PT (1) <=>
-(x+2017)+2017x-(x+2018)+2018x=4040x
<=> -x-x+2017x+2018x-4040x=2017+2018
<=>-7x=4035
<=>x=(-4035/7)(loại).
TH2: -2018≤ x <-2017 PT (1) <=>
-(x+2017)+2017x+(x+2018)+2018x=4040x
<=> -x+x+2017x+2018x-4040x=2017-2018
<=>-5x=-1
<=>x=(1/5)(loại).
TH3: 2017≤x PT (1) <=>
(x+2017)+2017x+(x+2018)+2018x=4040x
<=> x+x+2017x+2018x-4040x=-2017-2018
<=>-3x=-4035
<=>x=1345
vậy PT (1) có một nghiệm duy nhất x=1345