\(a.\) theo đề thì :
\(\dfrac{a+10}{a+1}=1\)
=> a = 8
b. \(\dfrac{2a+6}{a+2}=1\)
2a + 6 = a + 2
a - 4 =0
=> a = 4
tìm a \(\in\) N để a+10 \(⋮\) a + 1
A = \(\dfrac{a+10}{a+1}\) = \(\dfrac{a+1}{a+1}\)+ \(\dfrac{9}{a+1}\) = 1 + \(\dfrac{9}{a+1}\)
để a+10 \(⋮\)a+1 thì A \(\in\) N vì a \(\in\) N
A \(\in\)N \(\Leftrightarrow\) \(\dfrac{9}{a+1}\)\(\in\)N \(\Leftrightarrow\) a+ 1 \(\in\){1; 3; 9} \(\Leftrightarrow\) a \(\in\){0; 2; 8}
b, 2a + 6 \(⋮\)a +2
`a,`
`(a+10)/(a+1)=(a+1+9)/(a+1)=1+9/(a+1)`
Để `(a+10)/(a+1) in NN` thì `9/(a+1) in NN `
`=>9 vdots (a+1)`
`=>a+1 in Ư(9)={1;3;9}`
`a+1=1`
`=>a=0`
`a+1=3`
`=>a=2`
`a+1=9`
`=>a=8`
Vậy `a in {0;2;8}`
2a + 6 \(⋮\) a + 2
B = \(\dfrac{2a+6}{a+2}\)
để 2a + 6 \(⋮\) a + 2
thì B \(\in\) N \(\Leftrightarrow\) \(\dfrac{2a+6}{a+2}\)\(\in\)N \(\Leftrightarrow\) \(\dfrac{2a+6}{a+2}\) = \(\dfrac{2a+4}{a+2}\) + \(\dfrac{2}{a+2}\) \(\in\) N
\(\Leftrightarrow\) \(\dfrac{2\left(a+2\right)}{a+2}\) + \(\dfrac{2}{a+2}\) = 2 + \(\dfrac{2}{a+2}\) \(\in\) N
\(\Leftrightarrow\) a + 2 \(\in\) { 1; 2} \(\Leftrightarrow\) a \(\in\) {-1; 0} \(\Leftrightarrow\) a \(\in\) {0} vì a là số tự nhiên nên a = -1 loại