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\(n^2+3⋮n-1\)

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

\(\Rightarrow n+3⋮n-1\)

\(\Rightarrow\left(n-1\right)+4⋮n-1\)

\(\Rightarrow4⋮n-1\)

\(\Rightarrow n-1\inƯ\left(4\right)=\left\{\pm1;\pm2;\pm4\right\}\)

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

Vậy.......................................

a) n^2 + 3n - 13 chia hết cho n + 3

n(n + 3) - 13 chia hết cho n + 3

n(n + 3) chia hết cho n + 3

Nên 13 chia hết cho n + 3

Tự tìm nhé!

chịu.bo tay.com
 

mk chỉ biết làm phần a thôi

a) ta có: n²+3n-13 chia hết cho n+3

n(n+3) -13 chia hết cho n+3

ta thấy n(n+3) chia hết cho n+3 => 13 cũng phải chia hết cho n+3

=> n+3 E Ư₍₁₃₎={ 1;13;-1;-13}

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

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

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

a)n=5

b)X=16;-10;2;4

c)x=113;39;5;3;1;-1;-35;-109

Answer:

a) \(\left(n+2\right)⋮\left(n-3\right)\)

\(\Rightarrow\left(n-3+5\right)⋮\left(n-3\right)\)

\(\Rightarrow5⋮\left(n-3\right)\)

\(\Rightarrow n-3\) là ước của \(5\), ta có:

Trường hợp 1: \(n-3=-1\Rightarrow n=2\)

Trường hợp 2: \(n-3=1\Rightarrow n=4\)

Trường hợp 3: \(n-3=5\Rightarrow n=8\)

Trường hợp 4: \(n-3=-5\Rightarrow n=-2\)

b) Ta có: \(x-3\inƯ\left(13\right)=\left\{\pm1;\pm13\right\}\)

\(\Rightarrow x\in\left\{4;16;2;-10\right\}\)

Vậy để \(x-3\inƯ\left(13\right)\Rightarrow x\in\left\{4;16;2;-10\right\}\)

c) Ta có: \(x-2\inƯ\left(111\right)\)

\(\Rightarrow x-2\in\left\{\pm111;\pm37;\pm3;\pm1\right\}\)

\(\Rightarrow x\in\left\{-99;-35;1;1;3;5;39;113\right\}\)

d) \(5⋮n+15\Rightarrow n+15\inƯ\left(5\right)=\left\{\pm1;\pm5\right\}\)

Trường hợp 1: \(n+15=-1\Rightarrow n=-16\)

Trường hợp 2: \(n+15=1\Rightarrow n=-14\)

Trường hợp 3: \(n+15=5\Rightarrow n=-10\)

Trường hợp 4: \(n+15=-5\Rightarrow n=-20\)

Vậy \(n\in\left\{-14;-16;-10;-20\right\}\)

e) \(3⋮n+24\)

\(\Rightarrow n+24\inƯ\left(3\right)=\left\{\pm1;\pm3\right\}\)

\(\Rightarrow n\in\left\{-23;-25;-21;-27\right\}\)

f) Ta có:  \(x-2⋮x-2\)

\(\Rightarrow4\left(x-2\right)⋮x-2\)

\(\Rightarrow4x-8⋮x-2\)

\(\Rightarrow\left(4x+3\right)-\left(4x-8\right)⋮x-2\)

\(\Rightarrow11⋮x-2\)

\(\Rightarrow x-2\inƯ\left(11\right)=\left\{\pm1;\pm11\right\}\)

\(\Rightarrow x\in\left\{3;13;1;-9\right\}\)

4x-3⋮x-2

--> 4(x-2)+5⋮x-2

--> 5⋮x-2   (vì 4(x-2)⋮ x-2)

-->x-2⋴Ư(5) =⩲1;⩲5

ta có bảng

vậy x=1;3;7;-3 thì 4x-3⫶x-2

a) Đặt \(A=\frac{3n-13}{n+3}=\frac{3\left(n+3\right)-22}{n+3}=3-\frac{22}{n+3}\)

=> 22 \(⋮\)n + 3 => n + 3 \(\in\)Ư(22) = { \(\pm1;\pm2;\pm11;\pm22\)}

b) Đặt \(B=\frac{2n+3}{n-1}=\frac{2\left(n-1\right)+5}{n-1}=2+\frac{5}{n-1}\)

=> 5 \(⋮\)n - 1 => n - 1 \(\inƯ\left(5\right)=\left\{\pm1;\pm5\right\}\)

\(\left(a\right)3n-13⋮n+3\)

\(3n-13=3\left(n+3\right)-22\)

\(=>n+3=Ư\left(22\right)\)

\(n+3=\left\{-22;-11;-2;-1;1;2;11;22\right\}\)

\(=>n=\left\{-25;-14;-5;-4;-2;-1;8;19\right\}\)

\(\left(b\right)2n+3⋮n-1\)

\(2n+3=2\left(n-1\right)+5\)

\(=>n-1=Ư\left(5\right)\)

\(n-1=\left\{-5;-1;1;5\right\}\)

\(=>n=\left\{-4;0;2;6\right\}\)

a) Để \(3n-13⋮n+3\)

=> \(3n+9-22⋮n+3\)

=> \(3\left(n+3\right)-22⋮n+3\)

Vì \(3\left(n+3\right)⋮n+3\)

=> \(-22⋮n+3\)

=> \(n+3\inƯ\left(-22\right)\)

=> \(n+3\in\left\{1;-1;2;-2;11;-11;22;-22\right\}\)

=> \(n\in\left\{-2;-4;-1;-5;8;-14;19;-25\right\}\)

b) \(2n+3⋮n-1\)

=> \(2n-2+5⋮n-1\)

=> \(2\left(n-1\right)+5⋮n-1\)

=> \(2\left(n-1\right)⋮n-1\)

=> \(5⋮n-1\)

=> \(n-1\inƯ\left(5\right)\)

=> \(n-1\in\left\{1;-1;5;-5\right\}\)

=> \(n\in\left\{2;0;6;-4\right\}\)

\(\frac{2n+1}{n-5}=\frac{2n-10+11}{n-5}=\frac{2n-10}{n-5}+\frac{11}{n-5}=2+\frac{11}{n-5}\)

=> 11 chia hết cho n-5

n-5 thuộc Ư (11) = { -11; -1; 1; 11}

( rồi bạn thế vô rồi tính nha ^^ ... tương tự đối với b và c)