A=\(2^{n-1}+2.2^n+3-8.2^{n-4}-16.2^n=\)\(\frac{2^n}{2}+2.2^n-8.\frac{2^n}{2^4}-16.2^n+3\)

=\(2^n\left(\frac{1}{2}+2-\frac{8}{16}-16\right)+3\)=\(-14.2^n+3\)

Bạn làm như vầy nèe

A = (3^{n + 1} - 2.2ⁿ)(3^{n + 1} + 2.2ⁿ) - 3^{2n + 2} + (8.2^{n - 2})²

= (3^{n + 1} - 2^{n + 1})(3^{n + 1} + 2^{n + 1}) - 3^{2n + 2} + (2³.2^{n - 2})²

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

= 0

a) \(A=2^{n-1}+2.2^{n+3}-8.2^{n-4}-16.2^n\)

\(=2^{n-1}+2^{n+3+1}-2^{n-4+3}-2^{n+4}\)

\(=2^{n-1}+2^{n+4}-2^{n-1}-2^{n+4}\)

\(=0\)

b) \(B=\left(3^{n+1}-2.2^n\right)\left(3^{n+1}+2.2^n\right)-3^{2n+2}+\left(8.2^{n-2}\right)^2\)

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

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

\(=0\)

a,

\(A=2^{n-1}+2.2^{n+3}-8.2^{n-4}-16.2^n\)

\(=2^{n-1}+2^{n+3+1}-2^{n-4+3}-2^{n+4}\)

\(=2.2^{n-1}+2.2^{n+4}=2^n+2^{n+5}\)

b,

\(B=\left(3^{n+1}-2.2^n\right)\left(3^{n+1}+2.2^n\right)-3^{2n+2}+\left(8.2^{n-2}\right)^2\)

\(=\left(3^{n+1}\right)^2-\left(2.2^n\right)^2-\left(3^{n+1}\right)^2+\left(2^{n-2+3}\right)^2\)

\(=-2^{n+1}+2^{n+1}=0\)

làm câu

n-2 chia het cho n+3 

nen n+3-5 chia het cho n+3

5 chia het cho n+3

n+3 =cong tru1 cong tru 5

roi tim n

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

ĐKXĐ : \(n\ne-1\)

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

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

Với n nguyên, đặt ƯC( n² + n - 1 ; n² + n + 1 ) = d

=> n² + n - 1 ⋮ d và n² + n + 1 ⋮ d

=> ( n² + n + 1 ) - ( n² + n - 1 ) ⋮ d

=> n² + n + 1 - n² - n + 1 ⋮ d

=> 2 ⋮ d => d = 1 hoặc d = 2

Dễ thấy n² + n + 1 ⋮/ 2 ∀ n ∈ Z ( bạn tự chứng minh )

=> loại d = 2

=> d = 1

=> ƯCLN( n² + n - 1 ; n² + n + 1 ) = 1

hay P tối giản ( đpcm )