By reducing to absurdity, let a=√2+√3√5 be reasonable, then
a^2=2+ 15+2√30，
√ 30 = (a 2- 17)/2 is a rational number, which contradicts the known conditions of irrational numbers with root number 30, so
√2+√3√5 is an irrational number.
2.n is an integer greater than or equal to 1, and solve the equation (cosx) n-(sinx) n = 1.
When n is an even number, there are solutions x = 0 and x = π in [0,2π], and it is proved that there are no other solutions.
0<x< is at π, (cosx) n.
π& lt； X & lt2π, similarly, is not the solution of the equation.
When n is an odd number greater than 1, there is a solution at [0,2π] where X = 0 and X = 3π/2, and it is proved that there is no other solution.
0 & lt(cosx) n
π& lt； x & lt3π/2，(cosx) n
3π/2 & lt； x & lt2π，(cosx) n
N= 1 is, and the original equation is cosx-sinx= 1.
cos(x+π/4)=√2/2，
X=0, x=3π/2, which is the same as when n is an odd number greater than 1.
If it is a general solution, add 2kπ to the above solution. This equation is probably the only way, first find the solution through observation, and then prove that there is no other solution.
Supplementary question: The first question is wrong.
It is proved that root 2+ root 3+ root 5 is also irrational.
The proof method is similar: Let a=√2+√3+√5 be reasonable, then
a-√5 =√2+√3
a^2+5-2a√5 =2+3+2√6，
a^2=2a√5+2√6，
a^4=20a^2+24+8a√30，
√30=(a^4-20a^2-24)/(8a)，
......