\(a.\)

\(8^7-2^{18}\)

\(=\left(2^3\right)^7-2^{18}\)

\(=2^{21}-2^{18}\)

\(=2^{18}.2^3-2^{18}\)

\(=2^{18}\left(2^3-1\right)\)

\(=2^{18}.7\)

\(=2^{17}.7.2⋮14\)

Vậy \(8^7-2^{18}⋮14\)

\(b.\)

\(5^5-5^4+5^3\)

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

\(=5^3.21\)

\(=5^3.7.3⋮7\)

Vậy \(5^5-5^4+5^3⋮7\)

\(c.\)

\(7^6+7^5-7^4\)

\(=7^4\left(7^2+7-1\right)\)

\(=7^4.55\)

\(=7^4.5.11⋮11\)

Vậy \(7^6+7^5-7^4⋮11\)

mk chỉ bt làm phần b với c thui xin lỗi bn nha

a)

b)

c)

Ta có :

3n+2−2n+2+3n−2n3n+2−2n+2+3n−2n =3n.32−2n.22+3n−2n3n.32−2n.22+3n−2n

=3n.9−2n.4+3n−2n3n.9−2n.4+3n−2n =3n.(9+1)−2n.(4+1)3n.(9+1)−2n.(4+1)

=3n.10−2n.5=3n.10−2n−1.2.53n.10−2n.5=3n.10−2n−1.2.5 = 3n.10−2n−1.103n.10−2n−1.10

=10.(3n−2n−1)⋮1010.(3n−2n−1)⋮10

⇒3n+2−2n+2+3n−2n⋮10⇒3n+2−2n+2+3n−2n⋮10 (ĐPCM) 

TK NHA

3n + 2−2n + 2 + 3n−2n3n + 2−2n + 2 + 3n−2n =3n.32−2n.22 + 3n−2n3n.32−2n.22 + 3n−2n
=3n.9−2n.4 + 3n−2n3n.9−2n.4 + 3n−2n =3n.(9 + 1)−2n.(4 + 1)3n.(9 + 1)−2n.(4 + 1)
=3n.10−2n.5 = 3n.10−2n−1.2.53n.10−2n.5 = 3n.10−2n−1.2.5 = 3n.10−2n−1.103n.10−2n−1.10
=10.(3n−2n−1)⋮1010.(3n−2n−1)⋮10
⇒3n + 2−2n + 2 + 3n−2n⋮10⇒3n + 2−2n + 2 + 3n−2n⋮10 (ĐPCM)
TK NHA. chúc bn hok tốt @_@

\(A=\frac{2n-1}{n+8}-\frac{n-14}{n+8}=\frac{2n-1-\left(n-14\right)}{n+8}=\frac{n+13}{n+8}\)

Để A thuộc Z thì \(n+13⋮n+8\Rightarrow n+13-\left(n+8\right)⋮n+8\)

\(\Rightarrow5⋮n+8\Rightarrow n+8\inƯ\left(5\right)=\left\{1;5;-1;-5\right\}\)

\(\Leftrightarrow n\in\left\{-7;-3;-9;-13\right\}\)

OK

hi lily

\(25-y^2=8\left(x-2009\right)^2.\)

\(8\left(x-2009\right)^2\ge0\text{ nên } 25-y^2\ge0\)

\(\text{Mặt khác: }8\left(x-2009\right)^2⋮2\text{ nên }25-y^2\text{ luôn chẵn}\Rightarrow y^2\text{ luôn lẻ }\)

\(\text{Nên }y^2\text{ luôn tồn tại ở các giá trị :}y^2=1;y^2=9;y^2=25\)

+\(y^2=1\Rightarrow8\left(x-2009\right)^2=24\Leftrightarrow\left(x-2009\right)^2=3\left(\text{loại vì x thuộc N}\right)\)

+\(y^2=9\Rightarrow8\left(x-2009\right)^2=16\Leftrightarrow\left(x-2009\right)^2=2\left(\text{Loại}\right)\)

+\(y^2=25\Rightarrow8\left(x-2009\right)^2=0\Leftrightarrow\left(x-2009\right)^2=0\Rightarrow x=2009\)

Vậy...

tích mình đi rùi mình giải

mk chi tim dc 1 nghiem la x=0

k biet con nghiem nao k

a)\(=8.2^n+2^n.2=2^n\left(8+2\right)=2^n\cdot10\)

do đó \(8\cdot2^n+2^{n+1}\)có tận cùng là 0

b)\(=3^n\cdot3 ^3-2\cdot3^n+2^n\cdot2^5-7\cdot2^n\)

\(=3^n\left(3^2-2\right)+2^n\left(2^5-7\right)\)

\(3^n\cdot7+2^n\cdot25⋮25\)

do đó biểu thức b) chia hết cho 25

câu a)

\(8.2^n+2^{n+1}=8.2^n+2.2^n=10.2^n\)

chia hết cho 10 nên có tận cùng là 0

câu b)

\(3^{n+3}-2.3^n+2^{n+5}-7.2^n=3^3.3^n-3^n+2^5.2^n-7.2^n\)

\(=3^n\left(3^2-2\right)+2^n\left(2^5-7\right)=25.3^n+25.2^n=25\left(3^n+2^n\right)\)

chia hết cho 5

chúc bạn học tốt