If n
2. One of the solutions X 1+X2 = A, X 1 * X2 = B, X3+X4 = B, X3 * X4 = A, X 1, X2, X3, X4 must be equal to 1, otherwise b = x/x4. X 1+x2=a, a = x3 * x4 & gtX3+x4=b, that is, b>a, a>b, which is impossible. Without loss of generality, let x 1= 1, a = x2+ 1, and b = x2, then a = b+ 1, x3 * x4 = x3+x4+ 1, x3 * x4-. (x3- 1)(x4- 1) =2, so x3- 1= 1 = 2, or x3- 1=2, x4-/kloc
So a=6, b=5 or a=5, b=6.
3. If loga(c)+logb(c)=0, what is the equivalence relation that a b c satisfies?
Solution lg(c)/lg(a)+ lg(c)/lg(b)=0,
lg(c)(lg(a)+ lg(b))=0，
Lg(c)=0 or lg(a)+ lg(b)=0, and lg(ab) =0 from lg(a)+ lg(b)=0, that is, c= 1 or ab= 1.
4. The solution consists of f (x) >; 0 de
[(k+ 1)x^2+(k+3)x+(2k-8)]/[(2k- 1)x^2+(k+ 1)x+(k-4)]& gt； 0,
There are two situations as follows:
(1) (k+1) x 2+(k+3) x+(2k-8) > 0 and (2k-1) x2+(k+1) x+(k-4).
(2)(k+ 1)x^2+(k+3)x+(2k-8)& lt； 0 and (2k-1) x2+(k+1) x+(k-4) < 0;
Case (1): the quadratic term coefficients of both quadratic polynomials are greater than zero, that is, k+1>; 0，2k- 1 & gt； 0, so k> 1/2, and the discriminant of quadratic polynomial is less than zero, that is
(k+3)^2-4(k+ 1)(2k-8)<； 0，(k+ 1)^2-4(2k- 1)(k-4)<； 0,
7k^2-30k-4 1>； 0，7k^2-38k+ 15<； 0, from 7k2 to 30k-41>; 0 to get k
Considering k> 1/2, so K >；; 5;
Case (2): The quadratic term coefficients of two quadratic polynomials are both greater than zero, k+ 1
The set of real number k is (-∞,-1)∩(5,+∞).