tìm x biết a/1/2.4+1/4.6+…+1/(2x-2).2x= 11/48 b/ 1/5.8+1/8.11+1/11.14+…+1/x.(x+3)=101/1540 13/08/2021 Bởi Adeline tìm x biết a/1/2.4+1/4.6+…+1/(2x-2).2x= 11/48 b/ 1/5.8+1/8.11+1/11.14+…+1/x.(x+3)=101/1540
Đáp án + Giải thích các bước giải: `a//` `(1)/(2.4)+(1)/(4.6)+….+(1)/((2x-2).2x)=(11)/(48)` `(ĐK:x\ne{0;1})` `=>(2)/(2.4)+(2)/(4.6)+….+(2)/((2x-2).2x)=(11)/(48).2` `=>(1)/(2)-(1)/(4)+(1)/(4)-(1)/(6)+….+(1)/(2x-2)-(1)/(2x)=(11)/(24)` `=>(1)/(2)-(1)/(2x)=(11)/(24)` `=>(1)/(2x)=(1)/(24)` `=>2x=24` `=>x=12 (TM)` `b//` `(1)/(5.8)+(1)/(8.11)+(1)/(11.14)+….+(1)/(x.(x+3))=(101)/(1540)` `(ĐK:x\ne{0;-3})` `=>(3)/(5.8)+(3)/(8.11)+(3)/(11.14)+….+(3)/(x.(x+3))=(101)/(1540).3` `=>(1)/(5)-(1)/(8)+(1)/(8)-(1)/(11)+(1)/(11)-(1)/(14)+….+(1)/(x)-(1)/(x+3)=(303)/(1540)` `=>(1)/(5)-(1)/(x+3)=(303)/(1540)` `=>(1)/(x+3)=(1)/(308)` `=>x+3=308` `=>x=305 (TM)` Bình luận
Đáp án: $a/$ `1/(2 . 4) + 1/(4 . 6) +…+ 1/( (2x – 2) 2x) = 11/48` `⇔ 1/2 [1/2 – 1/4 + 1/4 – 1/6 + … + 1/(2x – 2) – 1/(2x)] = 11/48` `⇔ 1/2 + (- 1/4 + 1/4 – 1/6 + … + 1/(2x – 2) ) – 1/(2x) = 11/48 : 1/2` `⇔ 1/2 – 1/(2x) = 11/24` `⇔ 1/(2x) = 1/24` `⇔ 24 = 2x` `⇔ x=12` Vậy `x = 12` $b/$ `1/(5 . 8) + 1/(8 . 11) + 1/(11 . 14) + … + 1/(x (x + 3) ) = 101/1540` `⇔ 1/3 [1/5 – 1/8 + 1/8 – 1/11 + 1/11 – 1/14 + .. + 1/x – 1/(x + 3)] = 101/1540` `⇔ 1/5 + (- 1/8 + 1/8 – 1/11 + 1/11 – 1/14 + .. + 1/x) – 1/(x + 3) = 101/1540 : 1/3` `⇔ 1/5 – 1/(x + 3) = 303/1540` `⇔ 1/(x + 3) = 1/308` `⇔ 308 = x + 3` `⇔ x = 305` Vậy `x = 305` Bình luận
Đáp án + Giải thích các bước giải:
`a//`
`(1)/(2.4)+(1)/(4.6)+….+(1)/((2x-2).2x)=(11)/(48)` `(ĐK:x\ne{0;1})`
`=>(2)/(2.4)+(2)/(4.6)+….+(2)/((2x-2).2x)=(11)/(48).2`
`=>(1)/(2)-(1)/(4)+(1)/(4)-(1)/(6)+….+(1)/(2x-2)-(1)/(2x)=(11)/(24)`
`=>(1)/(2)-(1)/(2x)=(11)/(24)`
`=>(1)/(2x)=(1)/(24)`
`=>2x=24`
`=>x=12 (TM)`
`b//`
`(1)/(5.8)+(1)/(8.11)+(1)/(11.14)+….+(1)/(x.(x+3))=(101)/(1540)` `(ĐK:x\ne{0;-3})`
`=>(3)/(5.8)+(3)/(8.11)+(3)/(11.14)+….+(3)/(x.(x+3))=(101)/(1540).3`
`=>(1)/(5)-(1)/(8)+(1)/(8)-(1)/(11)+(1)/(11)-(1)/(14)+….+(1)/(x)-(1)/(x+3)=(303)/(1540)`
`=>(1)/(5)-(1)/(x+3)=(303)/(1540)`
`=>(1)/(x+3)=(1)/(308)`
`=>x+3=308`
`=>x=305 (TM)`
Đáp án:
$a/$ `1/(2 . 4) + 1/(4 . 6) +…+ 1/( (2x – 2) 2x) = 11/48`
`⇔ 1/2 [1/2 – 1/4 + 1/4 – 1/6 + … + 1/(2x – 2) – 1/(2x)] = 11/48`
`⇔ 1/2 + (- 1/4 + 1/4 – 1/6 + … + 1/(2x – 2) ) – 1/(2x) = 11/48 : 1/2`
`⇔ 1/2 – 1/(2x) = 11/24`
`⇔ 1/(2x) = 1/24`
`⇔ 24 = 2x`
`⇔ x=12`
Vậy `x = 12`
$b/$ `1/(5 . 8) + 1/(8 . 11) + 1/(11 . 14) + … + 1/(x (x + 3) ) = 101/1540`
`⇔ 1/3 [1/5 – 1/8 + 1/8 – 1/11 + 1/11 – 1/14 + .. + 1/x – 1/(x + 3)] = 101/1540`
`⇔ 1/5 + (- 1/8 + 1/8 – 1/11 + 1/11 – 1/14 + .. + 1/x) – 1/(x + 3) = 101/1540 : 1/3`
`⇔ 1/5 – 1/(x + 3) = 303/1540`
`⇔ 1/(x + 3) = 1/308`
`⇔ 308 = x + 3`
`⇔ x = 305`
Vậy `x = 305`