1.Tìm tích x,y biết: 1/4.15+1/5.18+1/6.21+....+1/x.y=3/40 Làm nhanh giúp em vs ạ pls!!!

`1/(4.15)+1/(5.18)+1/(6.21)+.......+1/(x.y)=3/40`

`⇔ 1/(4.3.5)+1/(5.3.6)+1/(6.3.7)+........+1/(3x.y/3)=3/40`

 `⇔ 1/3( 1/(4.5)+1/(5.6)+1/(6.7)+........+1/(x.y/3))=3/40`

`⇔ 1/3( 1/4-1/5+1/5-1/6+1/6-1/7+........+1/x-3/y))=3/40`

`⇔ 1/3(1/4-3/y)=3/40`

`⇔ 1/12-1/y=3/40`

`⇔ (y-12)/(12y)=3/40`

`⇔ 40(y-12)=36y`

`⇔ 40y-480=36y`

`⇔ 40y-36y=480`

`⇔ 4y=480`

`⇔ y=120`

`=> x=39`

                    Vậy `x=39; y=120`

Đáp án:

`x = 39`, `y = 120`

Giải thích các bước giải:

Ta có:

`1/4.15 + 1/5.18 + 1/6.21 + ... + 1/(x.y) = 3/40`

`<=> 1/(3.4.5) + 1/(3.5.6) + 1/(3.6.7) + ... + 1/(3.x.y/3) = 3/40`

`<=> 1/3(1/(4.5) + 1/(5.6) + 1/(6.7) + ... + 1/(x.y/3)) = 3/40`

`<=> 1/4 - 1/5 + 1/5 - 1/6 + 1/6 - 1/7 + ... + 1/x - 3/y = 3/40 : 1/3`

`<=> 1/4 - 3/y = 9/40`

`<=> (y - 12)/(4y) = 9/40`

`<=> 36y = 40y - 480`

`<=> y = 120`

`=> x = 120 : 3 - 1 = 39`

Vậy `x = 39`, `y = 120`