If the real number x, y satisfies X-Y + 1 > = 0, y + 1 > = 0, x + y + 1

If the real number x, y satisfies X-Y + 1 > = 0, y + 1 > = 0, x + y + 1

Using the geometric meaning, first draw the feasible region, the objective function can be seen as the slope of the line between the point on the feasible region and (1,1), then you rotate around (1,1), pay attention to the feasible region, observe the range of the slope!

Given y = (2x-1) / (x-1) and x > 0, the value range of real number y is?

There are two methods: (1) separation constant method; (2) inverse function method
(method 1) original function deformation: y = (2x-1) / (x-1) = (2x-2 + 1) / (x-1)
=[2(x-1)+1]/(x-1)=2+1/(x-1)
Because x > 0, so (x-1) > - 1
Then: when (x-1) > 0, 1 / (x-1) > 0, so 2 + 1 / (x-1) > 2, that is, Y > 2
When - 1