Toán Lớp 8: tìm số nguyên n sao cho n^3-3n^2-3n-1 chia hết cho n^2+n+1
giúp mik vs , cần gấp ạ :<
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Toán Lớp 8: tìm số nguyên n sao cho n^3-3n^2-3n-1 chia hết cho n^2+n+1
giúp mik vs , cần gấp ạ :<
Comments ( 2 )
n^3-3n^2-3n-1\vdots n^2+n+1
->n^3+n^2+n-4n^2-4n-4+3\vdots n^2+n+1
->n(n^2+n+1)-4(n^2+n+1)+3\vdots n^2+n+1
->(n^2+n+1)(n-4)+3\vdots n^2+n+1
->3\vdots n^2+n+1
->n^2+n+1\in Ư (3)={1;-1;3;-3}
Do n^2+n+1=(n+1/2)^2+3/4>=3/4>0
->n^2+n+1\in {1;3}
$\bullet$ n^2+n+1=1
->n^2+n=0
->n(n+1)=0
->n=0 (Tm) hoặc n=-1 (Tm)
$\bullet$ n^2+n+1=3
->n^2+n-2=0
->n^2+2n-n-2=0
->n(n+2)-(n+2)=0
->(n-1)(n+2)=0
->n=1 (Tm) hoặc n=-2 (Tm)
Vậy n\in{0;-1;1;-2} để n^3-3n^2-3n-1\vdots n^2+n+1
n^3-3n^2-3n-1\vdotsn^2+n+1
=(n^3-1)-(3n^2+3n+3)+3
=(n-1)(n^2+n+1)-3(n^2+n+1)+3
=(n^2+n+1)(n-1-3)+3
=(n^2+n+1)(n-4)+3
Để n^3-3n^2-3n-1\vdotsn^2+n+1
->3\vdotsn^2+n+1
->n^2+n+1\inƯ_((3))={-3;-1;1;3}
Mà n^2+n+1=(n+1/2)^2+3/4>=3/4>0
Nên n^2+n+1=1 hoặc n^2+n+1=3
$\quad\quad$n^2+n+1=1
$\quad\quad$->n^2+n=0
$\quad\quad$->n(n+1)=0
$\quad\quad$->n=0 hoặc n+1
$\quad\quad$->n=0(\text{Thõa mãn}) hoặc n=-1(\text{Thõa mãn})
$\bullet\quad$n^2+n+1=3
$\quad\quad$->n^2+n=2
$\quad\quad$->x^2+n-2=0
$\quad\quad$->n^2+2n-n-2=0
$\quad\quad$->n(n+2)-(n+2)=0
$\quad\quad$->(n+2)(n-1)=0
$\quad\quad$->n+2=0 hoặc n-1=0
$\quad\quad$->n=-2(\text{thõa mãn}) hoặc n=1(\text{thõa mãn})
Vậy n=0;n=1;n=-1;n=-2
$\quad$Để n^3-3n^2-3n-1\vdotsn^2+n+1