c, \(\frac{-32}{-2^n}=4\)

\(\Rightarrow-2^n=-32:4\)

\(\Rightarrow-2^n=-8\)

\(\Rightarrow-2^n=-2^3\Rightarrow n=3\)

d, \(\frac{8}{2^n}=2\)

\(\Rightarrow2^n=8:2\)

\(\Rightarrow2^n=4\)

\(\Rightarrow2^n=2^2\Rightarrow n=2\)

e, \(\frac{25^3}{5^n}=25\)

\(\Rightarrow5^n=25^3:25\)

\(\Rightarrow5^n=25^2\)

\(\Rightarrow5^n=5^4\Rightarrow n=4\)

i , \(8^{10}:2^n=4^5\)

\(\Rightarrow2^n=8^{10}:4^5\)

\(\Rightarrow2^n=\left(2^3\right)^{10}:\left(2^2\right)^5\)

\(\Rightarrow2^n=2^{30}:2^{10}\)

\(\Rightarrow2^n=2^{20}\Rightarrow n=20\)

k, \(2^n.81^4=27^{10}\)

\(\Rightarrow2^n=27^{10}:81^4\)

\(\Rightarrow2^n=\left(3^3\right)^{10}:\left(3^4\right)^4\)

\(\Rightarrow2^n=3^{30}:3^{16}\)

\(\Rightarrow2^n=3^{14}\)

\(\Rightarrow2^n=4782969\)Không chia hết cho 2 nên ko có Gt n thỏa mãn 

Bài 1

Ta có:\(\left(x^2-x+a\right)\left(x+1\right)=x^3+x^2-x^2-x+ax+a=x^3-x\left(a-1\right)+a\)

Khi đó:

\(x^3+x\left(1-a\right)+a=bx^2+cx+2\)

Do đó \(1-a=c;a=2;b=0\Rightarrow a=2;b=0;c=-1\)

Bài 2:

\(A=\left(n^2+2n-5\right)\left(n+2\right)-2n^3+n+10\)

\(=n^3+2n^2+2n^2+4n-5n-10-2n^3+n+10\)

\(=-n^3+4n^2\)

\(=n^2\left(4-n\right)\)

Lập luận với n chẵn thì cái trên luôn chia hết cho 8

1. ( x² - x + a )( x + 1 ) = x³ + bx² + cx + 2

<=> x³ + x² - x² - x + ax + a = x³ + bx² + cx + 2

<=> x³ + 0x² + ( a - 1 )x + a = x³ + bx² + cx + 2

<=> \(\hept{\begin{cases}b=0\\a-1=c\\a=2\end{cases}}\Leftrightarrow\hept{\begin{cases}a=2\\b=0\\c=1\end{cases}}\)

2. n chẵn => n có dạng 2k ( \(k\inℕ^∗\))

Thế vào ta được :

A = [ ( 2k )² + 2.2k - 5 )( 2k + 2 ) - 2(2k)³ + 2k + 10 

A = ( 4k² + 4k - 5 )( 2k + 2 ) - 16k³ + 2k + 10

A = 8k³ + 16k² - 2k - 10 - 16k³ + 2k + 10

A = -8k³ + 16k² = -8k²(k-2) \(⋮\)8

=> A chia hết cho 8 với mọi n chẵn ( đpcm )