Question

Chứng minh rằng với mọi số nguyên dương n, thì:

3^n+2 - 2^n+2 + 3^n -2^n chia hết cho 10

Answer 1

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

\(=3^n.10-2^n.5=3^{10}.10-2^{n-1}.2.5=3^n.10-2^{n-1}.10\)

\(=\left(3^n-2^{n-1}\right).10\)chia hết cho 10

Answer 2

Đặt D=3^n+2 - 2^n+2 + 3^n + 2^n

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

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

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

=3^n . 10 - 2 ^n .5

=3^n .10 - 2^n-1 .10

=(3^n - 2^n-1) . 10 chia hết cho 10 (ĐPCM)

Chúc bạn học tốt!

Answer 3

3^n+2-2^n+2+3^n-2^n

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

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

=3^n*10-2^n*5

=3^n*10-2^n-1*10

Có 10 chia het cho 10 nen (1) dung 

Suy ra dieu phai chung minh

Answer 4

\(x = {-b \pm \sqrt{b^2-4ac} \over 2a}\)

Answer 5

\(3^{n+2}-2^{n+2}+3^n-2^n\left(n\inℕ\right)\)

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

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

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

\(=3^n.10-2^n.5\)

\(=3^n.10-2^{n-1}.2.5\)

\(=3^n.10-2^{n-1}.10\)

\(=10\left(3^n-2^{n-1}\right)\)

vì \(10⋮10\Rightarrow10\left(3^n-2^{n-1}\right)⋮10\Rightarrow3^{n+2}-2^{n+2}+3^n-2^n⋮10\)

vậy......................

Answer 6

3^n+2-2^n+2+3^n-2^n
=3^n*3^2-2^n*2^2+3^n*1-2^n*1 (1)
=3^n*(3^2+1)-2^n*(2^2+1)
=3^n*10-2^n*5
=3^n*10-2^n-1*10
Có 10 chia het cho 10 nen (1) dung 
Suy ra 3^n+2-2^n+2+3^n-2^n:10( dpcm)

:3

Answer 7

thì bạn cứ giải đi dễ mà 

Answer 8

Toán lớp 8 mà

sao chúng nóa học nhanh quá z trời !!!!!!!!!

Đây là toán nâng cao lớp 8 mới đúng chớ. huhu

Answer 9

\(3n^2+2-2^{n+2}+3^n-2^n\)

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

=\(3^n.10-2^n.5\)

=\(3^n.10-2^{n-1}.10⋮10\)

Vậy \(3n^2+2-2^{n+2}+3^n-2^n\)\(⋮\)10