Question

[Đào Thị Mai Hà]

cứu huhu giúp tớ

Answer 1

[Yen Nhi]

`Answer:`

Câu 4:

a. Để cho `6a+4;a+2` đều là luỹ thừa của `2` thì \(\hept{\begin{cases}6a+2=2^m\\a+2=2^n\end{cases}}\left(m,n\inℤ\right)\)

Ta có: \(a\inℤ\Rightarrow6a+4>4,a^2>2\Rightarrow m>2,n>0\)

Mặt khác: \(6a+4>a+2\rightarrow2^m>2^n\rightarrow m>n\)

\(\Rightarrow\hept{\begin{cases}6a=2^m-4\\a=2^n-2\end{cases}}\left(m,n\inℤ\right)\)

\(\Rightarrow6.\left(2^n-2\right)=2^m-4\)

\(\Rightarrow6.2^n-12=2^m-4\)

\(\Rightarrow6.2^n-8=2^m\)

\(\Rightarrow6.2^n-2^m=8\)

\(\Rightarrow2^n.\left(6-2^{m-n}\right)=8\)

\(\Rightarrow\left(2^n,6-2^{m-n}\right)\) là cặp \(Ư\left(8\right)\)

Ta có: `2^n>2`

\(\Rightarrow\left(2^n,6-2^{m-n}\right)\in\left\{\left(2,4\right);\left(4,2\right);\left(8,1\right)\right\}\)

\(\Rightarrow\left(n,2^{m-n}\right)\in\left\{\left(1,2\right);\left(2,4\right);\left(3,7\right)\right\}\)

Mà `m,n\inZZ`

\(\Rightarrow\left(n,2^{m-n}\right)\in\left\{\left(1,2\right);\left(2,4\right)\right\}\)

\(\Rightarrow\left(n,m-n\right)\in\left\{\left(1,1\right);\left(2,2\right)\right\}\)

\(\Rightarrow\left(n,m\right)\in\left\{\left(1,2\right);\left(2,4\right)\right\}\)

Mà `n>1`

\(\Rightarrow\left(n,m\right)=\left(2,4\right)\)

`=>a=2`