1 1. I just answered this question yesterday and copied it down for your reference. (My Baidu knows that the account number is 02606938)

12, as shown in figure,

( 1)S△AOB=S△AOP+S△BOP？ (The method is not unique)

= 1/2 oph 1+ 1/2 oph 2 = 1/2OP(h 1+H2)

1, where OP is the common base of two triangles. When y=2x-2 gives y=0, x= 1, that is, op =1;

2. The absolute values of heights h 1 and h2, that is, the ordinate of points A and B, are given by y=2x-2 and y = 4/x..

A (2,2) and b (- 1, 4) are obtained from the simultaneous equations, that is, H 1 = 2 and H2 = 4.

So s △ AOB =1/2×1× (2+4) = 3.

(2) 1. First, find the coordinates of point A and point B about the axisymmetrical point of X: C(2, -2) d (- 1, 4);

2. From these two points, the analytical formula of straight line CD is obtained: y=-2x+2, (or from symmetry);

3. Because of the intersection, substitute y=k/x into: k/x=-2x+2, and get-2x 2+2x-k = 0.

As long as the discriminant △ = b 2-4ac is greater than or equal to 0, there must be intersection.

= 4-8k & gt； =0 to get k

That is, when k is less than or equal to 1/2, y=k/x must intersect with the straight line CD.

13, these equations are solved by (1) direct dichotomy, (2) formula method and (3) factorization method respectively.

And (4) and (5) matching problem-solving methods, I said is the method applied in sequence, and some only need simple deformation and sorting.

You can copy textbook examples. I really won't ask again, but I won't be back until evening.

I hope it helps you!