High school math problems, which are not too difficult and require process explanation

Solution analysis: unary equations, shift terms, merge similar terms, make coefficients into one, and find the result.

On the plane vector Ab1 perpendicular vector Ab2, the modulus of vector ob1 is equal to the modulus of vector ob2 = 1, vector ap = vector ab1 + vector ab2, if the modulus of vector op is less than 1 2, then the modulus of vector oa takes the value range.

Multiply 2 on both sides of the equation at the same time

Get 1- 3+x=0

Move the item, got. x=√3-1

High school math problems are actually about the same level of difficulty, and each problem has a different difficulty for everyone. How to learn mathematics well, in fact, mathematics is not as difficult as everyone thinks, mathematics is mainly to practice more, in the process of repeated practice to constantly consolidate the knowledge points, for some of the wrong questions in the problem to think repeatedly, why I am wrong in this question, **wrong, why I will do this here, these must be clear. In high school, remember not to pile up problems, piling up problems will only pile up more and more, what you don't understand in class should be raised in time, don't pretend to understand, it doesn't matter if the teacher doesn't understand after putting it forward, you can find the teacher or classmates to tell you separately after class, until you understand this question, don't worry about asking too many questions when communicating with the teacher, the teacher will think that you are a student is stupid or something, causing a psychological burden on yourself, in fact, the teacher likes this kind of student very much, and he is also very happy to see his students so diligent and studious.

Mathematics is very active, you don't need to spend too much time to memorize some theorems, meanings, you just need to understand more, the only thing to memorize is the formula, but also to be able to use, the mutual conversion between formulas, there are more self-study classes in high school, you must use them reasonably, you can find some test papers to do, time yourself, treat them as tests, don't blindly practice questions, you have to find methods and some tricks from doing problems. It is recommended that you prepare a problem book, accumulate your wrong questions, and read more when you have nothing to do, there are many ways to solve each problem in high school mathematics, don't be too limited, try to solve the same problem with two or more methods, which is of great benefit to yourself. Learning is important, but to learn to combine work and rest, if you want to have a good study, you also need to have a strong physique, have a good body to have the capital to talk about learning, you can spend some time on physical exercise, for example, students who like to play basketball can use the afternoon break to play basketball and so on.

The figure obtained from (2a=4) >2c=2 3) is an ellipse.

a=2，c=√3，b^=a^-c^;

c=1, so c:x4 + y=1

2) Set p(2cos,sin) according to the parametric equation of the ellipse) (this should be possible, it is available in the book).

Vector (pf1) = 3 - 2cos, sin) vector (pf2) = 3 - 2cos, sin ) vector (pf1) * vector (pf2) =

√3 - 2cosα ，sinα ）3 - 2cosα ，sinα ）

4cos3 + sin 3cos2 belongs to (0,2) so maximum=1

Obtained when = 2 or 3 2.

p(0,1) or (0,-1).

Let's work hard, this kind of question is very basic, it is the key, and it is easy to set questions in the college entrance examination.

It's not difficult to take a look, it's just that it's a bit more computational.

I haven't done a question for a long time.,It's an ellipse.,a=3,b=2,c=with thirteen.。。 Nothing else will be. Hehe.

In the planar Cartesian coordinate system, three points a(a,b),b(c,d),c(e,f), and p are the points (x,y) in the triangle

then according to the formula for the distance between two points on the plane.

pa^2=(x-a)^2+(y-b)^2

pb^2=(x-c)^2+(y-d)^2

pc^2=(x-e)^2+(y-f)^2

pa^2+pb^2+pc^2=(x-a)^2+(y-b)^2+(x-c)^2+(y-d)^2+(x-e)^2+(y-f)^2

x^2-2ax+a^2)+(y^2-2by+y^2)+(x^2-2cx+c^2)+(y^2-2dy+y^2)+(x^2-2ex+x^2)+(y^2-2fx+f^2)

3x^2-2(a+c+e)x+a^2+c^2+e^2]+[3y^2-2(b+d+f)y+b^2+d^2+f^2]

This is because a, b, c, d, e, and f are six unrelated values.

Therefore, only when the minimum value is taken in the upper two middle brackets, there is a minimum value for pa 2 + pb 2 + pc 2.

Let f(x)=3x 2-2(a+c+e)x+a 2+c 2+e 2

f'(x)=6x-2(a+c+e)

Order f'(x)=0 gives x=(a+c+e) 3

Let g(y)=3y2-2(b+d+f)y+b2+d2+f2

g'(y)=6y-2(b+d+f)

Ream'(y)=0 gives y=(b+d+f) 3

So the coordinates of point p are p((a+c+e) 3,(b+d+f) 3).

The following proves that p is the center of gravity.

Let the center of gravity be O, then the proportion of O to the directed line segment cd is 2, and the abscissa of the center of gravity O of the fixed score point formula is [E+2*(A+C) 2] (1+2)=(A+C+E) 3, and the ordinate is (B+D+F) 3.

So p coincides with O, i.e., p is the center of gravity.

Solution: Let the triangle be in a plane Cartesian coordinate system, a(a,a1); b（b,b1); c(c,c1)；p（x，y）

then pa +pb +pc =(x-a) +y-a1) +x-b) +y-b1) +x-c) +y-c1).

3x²-2(a+b+c)x+a²+b²+c²+3y²-2(a1+b1+c1)x+a1²+b1²+c1²

3[x-(a+b+c)/3]²-3[(a+b+c)/3]²+a²+b²+c²+3[y-(a1+b1+c1)/3]²-3[(a1+b1+c1)/3]²+a1²+b1²+c1²

So when x=(a+b+c) 3 and y=(a1+b1+c1) 3, ap +bp +cp is minimum, and the point p is the center of gravity of abc.

Because Pa +Pb +PC =GA2+GB2+GC2+3GP2

When p is the center of gravity, gp=0, pa +pb +pc is the smallest.

and Pa +Pb +PC =GA2+GB2+GC2

Don't give points for such hard work?

The answer to the question of the second **.

The first ** answer.

1）f(0+a)=f(0)f(a)

So f(0)=1

Suppose a is greater than 0

f=f(a)f(-a)

f(0)=f(a)f(-a)

1=f（a）f（-a）

Because f(a) 0 and f(a) 1

So when -a 0, f(-a) > 1

So when x 0, f(x) > 1

2) Let n be any number. m is a positive number.

Because f(m+n) = f(m)f(n).

Because m is positive, 0 is > 0 because f(n).

So f(m+n).

Because m+n > n, and n is an arbitrary number.

So the function is monotonically reduced.

This is the answer to the third question of the first **.

It is recommended that the landlord separate the topics and offer a reward of 5 points for each question.

I believe that the landlord's question will be answered soon.

The slope of the straight line is 1 6, so the equation of the parasensitive straight line can be set as the absolute key y=1 6x+b, and the intersection and merger points of the coordinate axis are (-6b,0), (0,b), so the area is s=1 2*|-6b|*|b| =3

B=1 or -1 can be obtained

So the linear equation is y=1 6x+1 , or y=1 6x-1 .

Let the difference of the straight line be y=1 6x+b, and the coordinate axis encloses a triangle with an area of 3.

When x=0, y=b, when y=0, x=-6b

1 Missing 2*6*b 2=3, b=1 or -1

The equation for a straight line is y=1 6x+1 or y=1 6x-1

Because it is a straight line of Bino laughing, it must be a one-dimensional equation.

Let the equation y=1 6x+b

The point on the x-axis is (-6b,0).

The point on the y-axis is (0,b).

s=1 2*6b*b=3 (b has positive imitation and negative) to find the spine, only b=plus or minus 1