1
Áp dụng tính chất dãy tỉ số bằng nhau
`=>a/(b+c)=c/(a+b)=b/(a+c)=(a+b+c)/(2a+2b+2c)=1/2`
`=>b+c=2a`
`=>a+b+c=3a`
Hoàn toàn tương tự:
`a+b+c=3b`
`a+b+c=3c`
`=>a=b=c`
`=>A=1/2+1/2+1/2=3/2`
2
`A in Z`
`=>x+3 vdots x-2`
`=>x-2+5 vdots x-2`
`=>5 vdots x-2`
`=>x-2 in Ư(5)={1,-1,5,-5}`
`+)x-2=1=>x=3(TM)`
`+)x-2=-1=>x=1(TM)`
`+)x-2=5=>x=7(TM)`
`+)x-2=-5=>x=-3(TM)`
Vậy với `x in {1,3,-3,7}` thì `A in Z`
`A in Z`
`=>1-2x vdots x+3`
`=>-2(x+3)+1+6 vdots x+3`
`=>7 vdots x+3`
`=>x+3 in Ư(7)={1,-1,7,-7}`
`+)x+3=1=>x=-2(TM)`
`+)x+3=-1=>x=-4(TM)`
`+)x+3=-7=>x=-10(TM)`
`+)x+3=7=>x=4(TM)`
Vậy `x in {2,-4,4,10}` thì `A in Z`
Câu 2:
a) Để A nguyên thì \(x+3⋮x-2\)
\(\Leftrightarrow x-2+5⋮x-2\)
mà \(x-2⋮x-2\)
nên \(5⋮x-2\)
\(\Leftrightarrow x-2\inƯ\left(5\right)\)
\(\Leftrightarrow x-2\in\left\{1;-1;5;-5\right\}\)
hay \(x\in\left\{3;1;7;-3\right\}\)
Vậy: \(x\in\left\{3;1;7;-3\right\}\)
