Completely incomprehensible to this math problem? I don t know much about this math problem

This question is adapted from the real questions of the college entrance examination. The core is to simplify according to the equation, don't fall into the mistake of combining numbers and shapes, and write it by hand here:

If you still have doubts, ask them.

Point p is the point on the circle with (0,0) as the center and 2 as the radius, so the coordinates of the p point (a,b) satisfy a +b = 4

PA, PB, PC are the ...... of the application of the formula for the distance between two pointsThe formula is obtained: 3a +3b -4b+68, and substituting a +b = 4 to simplify: -4b+80

And because p is a circle, the range of a and b is in the range of -2,2 to get the result 72,88

This is mainly a matter of concept, that is, what can be expressed in terms of the absolute value of pa pb pc. The absolute value of Pa is the distance from point P to point A, and the two lines connected into line segments plus two lines perpendicular to the coordinate axis are a right triangle, and the absolute value of Pa is expressed by the Pythagorean theorem, and the same is true for PB PC.

Your step is the derivation process, do you see if there is such a relationship in the known? You didn't list the original question, and you asked how this step came about, of course, it was not clear. If you list the original question, there must be known conditions in it that are related to this.

This process is okay, you can draw a picture and see.

Untie; The coordinates of the point p satisfy x y = 4

The point p is in a circle with a radius of 2 with the center of the circle (0,0), and the values of a and b must be greater than or equal to -2 and less than or equal to 2

Establish; The coordinates of the point p are (a, b).

pa｜²＋pb｜²＋pc｜²=3a²＋3b²-4b＋68=3(a²＋b²)-4b＋68=-4b＋12＋68=-4b＋80

The value of b must be greater than or equal to -2 and less than or equal to 2

The value range is 72,88 closed.

Two straight lines have a maximum of 1 intersection point, that is: 1+0=1 (intersection point) Three straight lines have a maximum of 3 intersection points, that is: 1+2=3 (intersection point) Four straight lines have a maximum of 6 intersection points, that is:

1+2+3=6 (intersection) Five straight lines have a maximum of 10 intersections, that is: 1+2+3+4=10 (intersection) Six straight lines have a maximum of 15 intersections, that is: 1+2+3+4+5=15 (intersections) n straight lines have a maximum of:

1+2+3+4+5+..n-1) = n(n-1) 2 (intersection).

When n=15, n(n-1) 2=105 (intersection) So: 6 points make up 15 straight lines, and 15 straight lines have a maximum of 105 intersections.

Using the equivalent infinitesimal substitution theorem, when x->0, ln(1+x) and x are equivalent infinitesimal, i.e., when x->0, lim( ln(1+x) ) lim( x).

Intuitively, this is where ln(1+x) and x are interchangeable.

Although there is an equivalent infinitesimal substitution theorem, there is no equivalent infinity substitution theorem, but it doesn't matter, as long as you construct a quantity that tends to 0.

Since x-> infinity (you can't use equivalent infinitesimal anymore), -1 (x-1) -0 (you can use equivalent infinitesimal anymore).

Let -1 (x-1)=t, where t->0 can be used as equivalent infinitesimal when t->0, ln(1+t) and t are equivalent infinitesimal.

then the original formula can be expressed as lim (when t->0) (xln(1+t)) = lim(x*t) = lim(x* (1 (x-1)).

Suppose that when perpendicular, the coordinates of point p are (m,n) and the intersection point is in the first quadrant, i.e., m 0, n 0.

According to the projective theorem there is n = ( 3-m) ( 3 + m ) = 3-m and it is on the ellipse, so there is m 4 + n = 1

The simultaneous solution is m=(2 6) 3

Then when x (-m, m), are obtuse angles, i.e., pf1 pf2 0, so, the probability is 2m 2a = 6 3

Untie; Working hours = workload work efficiency. Set the workload to 1If the first year of secondary school students work hours, their productivity is 1

The working hours of the second year of junior high school students are 5 hours, and its work efficiency is 1 5

First find the remaining part of the time to roll Biqing, plus the time used for cooperation 1 hour, on Huiji is the required time.

Use 1-(1 is the amount of work remaining.

The time for the second year of junior high school students to complete alone is 2 3 1 5 = 10 3, and finally 10 3 + 1 = 13 3 is the shared big grip time.