Đáp án:
Giải thích các bước giải:
a) Để B có giá trị nguyên thì `(x-2)/(x+3) in Z<=>x-2\vdotsx+3`
`=>x-2vdotsx+3`
`=>x+3-5vdotsx+3=>5vdots x+3`
`=>x+3 inƯ(5)`
`=>x+3 in{+-1;+-5}`
`=>`$x\in\{-4;-2;-8;2\}$
Vậy $x\in\{-4;-2;-8;2\}$.
b) `1/20+1/44+1/77+...+2/(x.(x+3))=101/770`
`=>1/40+1/88+1/154+...+1/(x.(x+3))=101/1540`
`=>1/5.8+1/8.11+1/11.14+...+1/(x.(x+3))=101/1540`
`=>3/5.8+3/8.11+3/11.14+...+3/(x.(x+3))=303/1540`
`=>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`
Vậy `x=305`.