Question

[Mai Hà Anh]

Tìm số nguyên dương m và n, biết: \(2^m+2^n=2^{m+n}\)

Answer 1

[Nguyễn Việt Lâm]

\(2^m+2^n=2^{m+n}\Leftrightarrow2^{m+n}-2^m-2^n=0\)

\(\Leftrightarrow2^m\left(2^n-1\right)-2^n+1-1=0\Leftrightarrow2^m\left(2^n-1\right)-\left(2^n-1\right)-1=0\)

\(\Leftrightarrow\left(2^m-1\right)\left(2^n-1\right)=1=1.1\)

\(\Rightarrow\left\{{}\begin{matrix}2^m-1=1\\2^n-1=1\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}2^m=2\\2^n=2\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}m=1\\n=1\end{matrix}\right.\)

Answer 2

[Nguyễn Ngọc Thái]

=>(2^n-1).2^m

=>2^n=2^n.2^m-2^m

=>2^n=(2^n-1).2^m

=>(2^n-1).2^m-2^n=0

=>(2^n-1).2^m-(2^n-1)=1

=>(2^n-1).(2^m-1)=1

Vì m,n là số nguyên dương

=>2^m;2^n là 2 số nguyên dương và 2^m;2^n>=1

=>2^n-1=2^m-1=1

=>2^n=2^m=2

=>n=m=1