$\frac{1}{3.5}$ + $\frac{1}{5.7}$ + $\frac{1}{7.9}$ + … + $\frac{1}{(2x+1)(2x+3)}$ = $\frac{15}{83}$ giải giùm đi,

Đáp án:

`x\in∅`

Giải thích các bước giải:

`1/3.5+1/5.7+1/7.9+…+1/((2x+1)(2x+3))=15/83`
`=>1/2.(2/3.5+2/5.7+2/7.9+…+2/((2x+1)(2x+3)))=15/83`
`=>1/2.(1/3-1/5+1/5-1/7+1/7-1/9+…+1/(2x+1)-1/(2x+3))=15/83`
`=>1/2.(1/3-1/(2x+3))=15/83`
`=>1/3-1/(2x+3)=15/83:1/2`
`=>1/3-1/(2x+3)=15/83 . 2/1`
`=>1/3-1/(2x+3)=30/83`
`=>1/(2x+3)=1/3-30/83`
`=>1/(2x+3)=83/249-90/249`
`=>1/(2x+3)=-7/249`
`=>(2x+3).(-7)=1.249`
`=>(2x).(-7)+3.(-7)=249`
`=>-14x-21=249`
`=>-14x=249+21`
`=>-14x=270`
`=>x=270:(-14)`
`=>x=-270/14`
`=>x=-135/7`

Vì `x\inNN=>x\in∅`
Vậy `x\in∅`