Ta có :

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

\(\Leftrightarrow2^m.\left(2^n-1\right)-\left(2^n-1\right)=1\Leftrightarrow\left(2^n-1\right).\left(2^m-1\right)=1\)

\(\Leftrightarrow\hept{\begin{cases}2^n-1=1\\2^m-1=1\end{cases}}\Leftrightarrow m=n=1\)

Vậy m = 1 ; n = 1

<br class="Apple-interchange-newline"><div id="inner-editor"></div>2m+2n=2m+n⇔2m+n−2m−2n=0

⇔2m.(2n−1)−(2n−1)=1⇔(2n−1).(2m−1)=1

⇔{

Vậy m = 1 ; n = 1

\(\text{(m,n) = }\left\{\left(0;0\right);\left(1;1\right)\right\}\)

(m,n) = {(1,1)} (vi m, n la so nguyen duong)

\(2^m+2^n=2^{m+n}\)--->Chia 2 vế cho 2ⁿ

\(\Rightarrow2^{m-n}+1=2^m\Leftrightarrow2^m-2^{m-n}=1\)

\(\Leftrightarrow2^{m-n}\left(2^n-1\right)=1\)---> Các lũy thừa số mũ tự nhiên của 2 không thể bé hơn 1 nên pt chỉ có nghiệm khi:

\(\hept{\begin{cases}2^{m-n}=1\\2^n-1=1\end{cases}\Leftrightarrow\hept{\begin{cases}2^{m-n}=2^0\\2^n=2^1\end{cases}\Leftrightarrow}\hept{\begin{cases}m-n=0\\n=1\end{cases}\Rightarrow}m=n=1}\)

\(2^m+2^n=2^{m+n}\Leftrightarrow2^m.2^n-2^m-2^n+1=1\)

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

Vì \(2^m-1\)và \(2^n-1\)đều lớn hơn 0 nên ta chỉ có một trường hợp \(\hept{\begin{cases}2^m-1=1\\2^n-1=1\end{cases}\Leftrightarrow\hept{\begin{cases}m=1\\n=1\end{cases}}}\)

a)
x² - 4x + 3 = x² - x - 3x + 3
= x(x - 1) - 3(x - 1) = (x -1)(x - 3)
b)
x² + 5x + 4 = x² + 4x + x + 4
= x(x + 4) + (x + 4)
= (x + 4)(x + 1)

a)5/6=1/2+1/3

a)\(\frac{5}{6}=\frac{1}{2}+\frac{1}{3}\)

b)\(\frac{1}{m}+\frac{n}{6}=\frac{1}{2}\)

\(\frac{1}{m}=\frac{1}{2}-\frac{n}{6}\)

\(\frac{1}{m}=\frac{3}{6}-\frac{n}{6}\)

\(\frac{1}{m}=\frac{3-n}{6}\)

\(\Rightarrow m.\left(3-n\right)=1.6\)

\(m.\left(3-n\right)=6\)

\(\Rightarrow3-n\in\text{Ư}\left(6\right)=\left\{1;-1;2;-2;3;-3;6;-6\right\}\)

Nếu 2^m + 2ⁿ = 2^m+n 

thì:  2^m + 2ⁿ  = 2^m.2ⁿ

=>  2^m = 2^m.2ⁿ - 2ⁿ

=>  2^m = 2ⁿ.(2^m-1)

=>  1   = (2ⁿ - 1).(2^m-1)

 còn lại bạn lập bảng tự làm nhé