1, divide both sides by an x to get x-5- 1/x=0, that is, x- 1/x=5. Then, make a formula for the latter formula, subtract a 2 from it and add a 2, that is, X 2-2+ 1.

2, we first decompose the factor AB-A = 13 = A (B-1) = 13, and then we know that13 is a prime number, and its factors are only1and13, then A. And b- 1 must be 1 and 13, so it is possible that a= 1, b= 12, or a= 13, and b=0. Obviously, the latter is not valid, because ab-a=0 at this time.

3. I made a picture of this problem myself, which is not very good, so don't take it personally. Like the trapezoidal abcd in the picture, I can tell you that their two green diagonals are vertical, and the triangle divided into upper and lower sides is an isosceles right triangle. I won't write the specific proof process because it is too complicated, so the data in the figure can be obtained according to Pythagorean theorem. So it is concluded that the perimeter is equal to 6+ twice the root number six, and the area is equal to the sum of the four teaching triangles, that is, 3+ twice the root number two.