Question

tìm n thuộc Z biết: 

a) n+8 chia hết cho n-9

b) 2^n * 4^12 = 8^30 : 16^10

c) 2^n+2 + 2^n+1 + 2^n = 112

Answer 1

a, \(n+8\)chia hết cho \(n-9\)

\(\Rightarrow n+8-n+9\) chia hết cho \(n-9\)

\(\Rightarrow17\)chia hết cho \(n-9\)

\(\Rightarrow n-9\inƯ\left(17\right)=1;-1;17;-17\)

TH1 : \(n-9=1\Rightarrow n=10\)

TH2 :\(n-9=-1\Rightarrow n=8\)

TH3: \(n-9=17\Rightarrow n=26\)

TH4: \(n-9=-17\Rightarrow n=-8\)

Vậy \(n\in10;8;26;-8\)

b, \(2^n.4^{12}=8^{30}:16^{10}\)

\(\Rightarrow2^n.2^{24}=2^{90}:2^{40}\)

\(\Rightarrow2^n.2^{24}=2^{50}\)

\(\Rightarrow2^n=2^{26}\)

\(\Rightarrow n=26\)

Vậy \(n=26\)

c, \(2^n+2+2^n+1+2^n=112\)

\(\Rightarrow3.2^n+3=112\)

\(\Rightarrow3.2^n=109\)

\(\Rightarrow2^n=\frac{109}{3}\)