Ta co : 3.n^3+10.n^2-5

=3.n^3+9.n^2+3n-3n-1-4

=n^2.(3n+1)+3n(3n+1)-(3n+1)-4

=(3n+1)(n^2+3n-1)-4

De 3.=10.-5 chia het cho 3n+1

=> (3n+1)(3n-1)-4 chia het cho 3n+1

=> -4 chia het cho 3n+1

Ma U(-4)={-4;-2;-1;1;2;4}

=>3n+1={-4;-2;-1;1;2;4}

=>3n={-5;-3;-2;0;1;4}

=>n={5/3;-1;-2/3;0;1/3;1}

Ma n thuoc N

Vay n={-1;0;1}

lik e nhe

Ta có \(n^3+10n^2-5\)chia hết cho \(3n+1\)

Suy ra \(\frac{n^3+10n^2-5}{3n+1}\)là số nguyên

Ta thấy \(\frac{n^3+10n^2-5}{3n+1}\)

\(=\frac{n^3+9n^2+n^2-5}{3n+1}\)

\(=\frac{n^2.\left(1+3n\right)+n^2-5}{3n+1}\)

\(=n^2+\frac{n^2-5}{3n+1}\)

Tách tiếp cái phân số phía sau là ra nhé , lười @@

Ta có:  \(3n^3+10n^2-5=\left(3n+1\right).\left(n^2+3n-1\right)-4\)

để  \(3n^3+10n^2-5⋮3n+1\) thì  \(4⋮3n+1\)

Tức là \(3n+1\) là ước của 4

\(\Rightarrow\left(3n+1\right)\in\left\{-4;-2;-1;1;2;4\right\}\)

\(3n+1=-4\Rightarrow n=\frac{-5}{3}\left(loai\right)\)

\(3n+1=-2\Rightarrow n=-1\)

\(3n+1=-1\Rightarrow n=\frac{-2}{3}\left(loai\right)\)

\(3n+1=1\Rightarrow n=0\)

\(3n+1=2\Rightarrow n=\frac{1}{3}\left(loai\right)\)

\(3n+1=4\Rightarrow n=1\)

Vậy.....................

1: \(\Leftrightarrow3n^3+n^2+9n^2+3n-3n-1-4⋮3n+1\)

\(\Leftrightarrow3n+1\in\left\{1;4;2;-2;-1;-4\right\}\)

\(\Leftrightarrow3n\in\left\{0;3;-3\right\}\)

hay \(n\in\left\{0;1;-1\right\}\)

a: n lẻ nên n=2k+1

\(A=n^3+3n^2-n-3\)

\(=n^2\left(n+3\right)-\left(n+3\right)\)

\(=\left(n+3\right)\left(n^2-1\right)=\left(n+3\right)\left(n-1\right)\left(n+1\right)\)

\(=\left(2k+1+3\right)\left(2k+1-1\right)\left(2k+1+1\right)\)

\(=2k\left(2k+2\right)\left(2k+4\right)=2k\cdot2\left(k+1\right)\cdot2\left(k+2\right)=8k\left(k+1\right)\left(k+2\right)\)

Vì k;k+1;k+2 là ba số nguyên liên tiếp

nên k(k+1)(k+2)⋮3!=6

=>A=8k(k+1)(k+2)⋮8*6

=>A⋮48

c: n lẻ nên n=2k+1

\(C=n^4-10n^2+9\)

\(=n^4-n^2-9n^2+9\)

\(=\left(n^2-1\right)\left(n^2-9\right)=\left(n-1\right)\left(n+1\right)\left(n-3\right)\left(n+3\right)\)

=(2k+1-1)(2k+1+1)(2k+1-3)(2k+1+3)

\(=2k\cdot\left(2k+2\right)\left(2k-2\right)\left(2k+4\right)=16k\left(k+1\right)\left(k-1\right)\left(k+2\right)\)

Vì k-1;k;k+1;k+2 là bốn số nguyên liên tiếp

nên \(\left(k-1\right)\cdot k\cdot\left(k+1\right)\left(k+2\right)\) ⋮4!=24

=>C=16k(k+1)(k-1)(k+2)⋮16*24

=>C⋮384

Ta có: \(3n^3+10n^2-5⋮3n+1\)

\(\Rightarrow3n^3+n^2+9n^2+3n-3n-1-4⋮3n+1\)

\(\Rightarrow n^2\left(3n+1\right)+3n\left(3n+1\right)-\left(3n+1\right)-4⋮\left(3n+1\right)\)

\(\Rightarrow\left(3n+1\right)\left(n^2+3n-1\right)-4⋮3n+1\)

Vì \(3n+1⋮3n+1\) nên để \(\left(3n+1\right)\left(n^2+3n-1\right)-4⋮3n+1\) thì \(4⋮3n+1\)

\(\Rightarrow3n+1\inƯ\left(4\right)\)

\(\Rightarrow3n+1\in\left\{1;2;4;-1;-2;-4\right\}\)

\(\Rightarrow3n\in\left\{0;1;3;-2;-3;-5\right\}\)

\(\Rightarrow n\in\left\{0;\frac{1}{3};1;-\frac{2}{3};-1;-\frac{5}{3}\right\}\)

Mà \(n\in Z\Rightarrow n\in\left\{0;1;-1\right\}\)

Vậy \(n\in\left\{0;1;-1\right\}\)

a: =>\(n+2\in\left\{1;-1;7;-7\right\}\)

=>\(n\in\left\{-1;-3;5;-9\right\}\)

b: =>n-3+4 chia hết cho n-3

=>\(n-3\in\left\{1;-1;2;-2;4;-4\right\}\)

=>\(n\in\left\{4;2;5;1;7;-1\right\}\)

c: =>3n^3+n^2+9n^2-1-4 chia hết cho 3n+1

=>\(3n+1\in\left\{1;-1;2;-2;4;-4\right\}\)

=>\(n\in\left\{0;-\dfrac{2}{3};\dfrac{1}{3};-1;1;-\dfrac{5}{3}\right\}\)

d: =>10n^2-10n+11n-11+1 chia hết cho n-1

=>\(n-1\in\left\{1;-1\right\}\)

=>\(n\in\left\{2;0\right\}\)

= (3n^3 + 10n^2 - 5)/(3n + 1)
A = (3n^3 + n^2 + 9n^2 + 3n - 3n - 1 -4)/(3n+1)
A= n^2 + 3n - 1 - 4/(3n+1)
biểu thức 3n^3 + 10n^2 - 5 chia hết cho giá trị của biểu thức 3n + 1 khi:
3n+1 = ±1,±2, ±4
=> n = 0,-2/3,1/3,-1,1,-5/3
chọn giá trị nguyên: n = 0,-1,1