Open the roots first

Because 7 = 4+3 = 2 2+(root number 3) 2.

And the number of 4 roots is 3=2*2* 3.

So 7+4 radical number 3 = 2 2+(radical number 3) 2+2 * 2 * radical number 3=(2+ radical number 3) 2.

therefore

The 4th root number (7+4 root number 3)= root number (2+ root number 3)

2+ radical number 3=(4+2 radical number 3)/2

4+2 radical number 3= (radical number 3) 2+ 1 2+2 * radical number 3* 1= (radical number 3+ 1) 2

therefore

The fourth root number (7+4 root number 3)

= root number (2+ root number 3)

= root number [(root number 3+ 1) 2/2]

= (root number 3+ 1)/ root number 2

= (square root of 6+radical sign 2)/2

second

With abc on both sides

a(x-a)+b(x-b)+c(x-c)= 2(BC+ca+ab)

(a+b+c)x=a^2+b^2+c^2+2ab+2ac+2bc

(a+b+c)x=(a+b+c)^2

x=(a+b+c)

third

One set of solutions is m=4 and n= 13.