Ta có;

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

\(=n^2+n^2+2n+1+n^2+6n+9\)

\(=3n^2+8n+10\)

Ta có:

\(\left[n^2+\left(n+1\right)^2+\left(n+3\right)^2\right]⋮5\)

\(\Leftrightarrow n^2+\left(n+1\right)^2+\left(n+3\right)^2\equiv0\left(mod5\right)\)

\(\Leftrightarrow3n^2+8n+10\equiv0\left(mod5\right)\)

\(\Leftrightarrow3n^2+3n\equiv0\left(mod5\right)\)

\(\Leftrightarrow n\left(n+1\right)\equiv0\left(mod5\right)\)

Do đó n phải có dạng \(5k\) hoặc \(5k+4\)(\(k\in N\))

làm câu

bài 2 phần a

x^3-0,25x = 0

x*(x² - 0,25)=0

=> TH1: x=0

TH2 : x² - 0.25=0

(x-0,5)(x+0,5)=0

=> x=0.5

     x=-0.5

Vậy x=0  , x=+ - 5

sai thì thông cảm

a: \(\Leftrightarrow2n^2+n-2n-1+3⋮2n+1\)

\(\Leftrightarrow2n+1\in\left\{1;-1;3;-3\right\}\)

hay \(n\in\left\{0;-1;1;-2\right\}\)

b: \(\Leftrightarrow2n^2-4n+5n-10+3⋮n-2\)

\(\Leftrightarrow n-2\in\left\{1;-1;3;-3\right\}\)

hay \(n\in\left\{3;1;5;-1\right\}\)

c: \(\Leftrightarrow10n^2-15n+8n-12+7⋮2n-3\)

\(\Leftrightarrow2n-3\in\left\{1;-1;7;-7\right\}\)

hay \(n\in\left\{2;1;5;-2\right\}\)

d: \(\Leftrightarrow2n^2-n+4n-2+5⋮2n-1\)

\(\Leftrightarrow2n-1\in\left\{1;-1;5;-5\right\}\)

hay \(n\in\left\{1;0;3;-2\right\}\)

6   \(n^5+5n=n^5-n+6n=n\left(n^4-1\right)+6n=n\left(n^2-1\right)\left(n^2+1\right)+6n\)

\(=n\left(n-1\right)\left(n+1\right)\left(n^2+1\right)+6n\)

vì n,n-1 là 2 số nguyên lien tiếp  \(\Rightarrow n\left(n-1\right)⋮2\Rightarrow n\left(n-1\right)\left(n+1\right)\left(n^2+1\right)⋮2\)

  n,n-1,n+1 là 3 sô nguyên liên tiếp \(\Rightarrow n\left(n-1\right)\left(n+1\right)⋮3\Rightarrow n\left(n-1\right)\left(n+1\right)\left(n^2+1\right)⋮3\)

\(\Rightarrow n\left(n-1\right)\left(n+1\right)\left(n^2+1\right)⋮2\cdot3=6\)

\(6⋮6\Rightarrow6n⋮6\Rightarrow n\left(n-1\right)\left(n+1\right)\left(n^2+1\right)-6n⋮6\Rightarrow n^5+5n⋮6\)(đpcm)

7   \(n\left(2n+7\right)\left(7n+1\right)=n\left(2n+7\right)\left(7n+7-6\right)=7n\left(n+1\right)\left(2n+7\right)-6n\left(2n+7\right)\)

\(=7n\left(n+1\right)\left(2n+4+3\right)-6n\left(2n+7\right)\)

\(=7n\left(n+1\right)\left(2n+4\right)+21n\left(n+1\right)-6n\left(2n+7\right)\)

\(=14n\left(n+1\right)\left(n+2\right)+21n\left(n+1\right)-6n\left(2n+7\right)\)

n,n+1,n+2 là 3 sô nguyên liên tiếp dựa vào bài 6 \(\Rightarrow n\left(n+1\right)\left(n+2\right)⋮6\Rightarrow14n\left(n+1\right)\left(n+2\right)⋮6\)

\(21⋮3;n\left(n+1\right)⋮2\Rightarrow21n\left(n+1\right)⋮3\cdot2=6\)

\(6⋮6\Rightarrow6n\left(2n+7\right)⋮6\)

\(\Rightarrow14n\left(n+1\right)\left(n+2\right)+21n\left(n+1\right)-6n\left(2n+7\right)⋮6\)

\(\Rightarrow n\left(2n+7\right)\left(7n+1\right)⋮6\)(đpcm)

......................?

mik ko biết

mong bn thông cảm 

nha ................