Just 1+1

$1 + 1 =\dfrac{2}{1}$ . So, $m = 2$ and $n = 1$ . Now, $2(m^2 + n^2 + 2mn) =2{(m + n)}^2 = 2×3^2$ . So, $a + b = \color{#69047E}{\boxed{3}}$ .

Hello,

as x= 1+1 =2, it also can be written as m/n= 2 / 2^0,

m = 2 , n= 2^0 = 1,substitute them into the equation,

2(4 + 1 + 2(2)(1) )= 2^a . 3^b,

2^a . 3^b = 18=2 x 9

2^a . 3^b = 2 x 3^2

a=1, b=2, therefore, a+b = 1+2 = 3...

thanks....

2=2/1 2=m/n 2(m^ 2 + n^2 + 2mn) = 2^a * 3^b m=b n=a 18=2^n * 3^m m+n=a+b a+b=3

$x = 1 + 1 = 2$

The only way to express 2 as $\frac{m}{n}$ where $m$ and $n$ are co-prime positive integers is $\frac{2}{1}$ . This implies that $m = 2$ and $n = 1$ .

Substituting the values of $m$ and $n$ in the expression $2( m^2 + n^2 + 2mn)$ , we get 18.

Now, $18 = 2^1 \times 3^2$ , which implies that $a = 1$ and $b = 2$ .

So, $a + b = 1 + 2 = \boxed{3}$