Urgently ask for a high school math tutor!! The high school math god comes in!!

I very much agree with the point of view upstairs, I think you must first have confidence and believe that you have the ability to learn mathematics well, since you can get into other subjects in key high schools and it is very good, there is no reason why you can't learn mathematics well. The key is to calm down and see that the problems are all from the **, one by one, and the person you should trust the most is your math teacher, tell him what you think, after all, he sees you every day, knows best what you need, if he can tutor you should have twice the result with half the effort.

If this is not possible, then ask a high school mathematics teacher to tutor it, for example, a foreign mathematics teacher who is studying in Wuhan has a certain teaching experience and time, and the effect is generally good. I have a colleague's child who asked a teacher from the school to tutor in junior high school mathematics, and the results were very good. You can try it.

Good luck!

Whether or not to find a tutor is secondary, the important thing is that you can be ruthless to overcome your fear of mathematics. If you can focus on math for a week, you will have a breakthrough in your grades, and the important thing is that at this time, the teachers and classmates are impressed with you and you have confidence and interest in yourself, and your grades will skyrocket. You should seem smart, your grades are good, all you need is a breakthrough to gain confidence and interest.

If you still feel unsure, you can go up and see some teaching**, which is very good. Learn to use all the resources around you. The above words are my heartfelt words, and they are my personal feelings in high school.

I hope it can be of some help to you, and I wish you all the best in the college entrance examination!!

It is not difficult to improve your score by 20-30, just imagine that you can score more than 10 points in a big math problem. So you don't necessarily have to be a math teacher, in fact, you can find a college student with good math scores or other math scores that are about your age.

(1) According to the formula of double angle, we can get the formula given in the answer (2) change tan7° to x, then 2x 1-x 2=1 4 dislocation multiplication obtains: 4*2x=1-x 2, and finally get 0=1-8x-x 2, and then use the formula of the solution of the unary quadratic equation to obtain.

a = -1 (coefficient of x square term), b = -8 (coefficient of x square term), c = 1 (constant term).

Then the formula is x=[-b (root number B 2-4ac)] 2a gets: 8 root number 68 -2, 68 = 4*17, so: root number 68 = root number 4 * root number 17 = 2 * root number 17 then 8 root number 68 -2 = 8 2 * root number 17 (-2)=-4 root number 17 Then, because tan7° must be a positive number, so -4-root number 17 is not in place, then the rest is: root number 17-4

The function Qi grips the grinding co-conductance, and there is 3x 2+2ax-9=f'(x)

Roll the upper blind boy into the form of (a+b) 2+c=0.

Let c=-12, and find it.

Solution: f(x)+2f(1 x)=3x ......Style.

Replace x with 1 x.

f（1/x）+2f（x）=3/x ……Style.

Eq. 2- Eq. gets:

3f（x）=6/x-3x

f（x）=2/x-x

The question is f(x)+2(1 x)=3x?

That's not very simple, and the transfer is not just there.

f（x）=3x-2（1/x）

The method on the first floor can be as follows if it is a multiple-choice question, that is, if p is the center of the circle (x-2) 2+(y-3) 2=1, and at this time a=2, b=3, then q(0,1) is the center of the circle (x-2) 2+(y-3) 2=1 about the circle symmetrical with a straight line l.

Therefore, the equation is derived.

But if it's a problem, how can you make a=2, b=3?

To solve the problem, you can set the center point of pq to w(x',y'), then x'=(3+a-b)/2, y'=(3+b-a) 2, the slope of l is -1, and after the point w, let its equation be y=-x+m, and substitute the coordinates of w to obtain, m=3.

The equation for l is: y=-x+3

Let the center of the circle be n(x'',y''Then n is symmetrical with the point (2,3) with respect to l, using y''+3=-(x''+2)+6 and (y''-3)/(x''-2) = 1 to get x''=0,y''=1, the radius of the circle is also 1, so the equation is: x 2+(y-1) 2=1.

Use substitution: here let a=2, b=3

Then the coordinates of the symmetrical point (3-b,3-a) = (0,1) Since l bisects the line segment pq, the perpendicular bisector between (2,3) and (0,1) must also be l

So, the equation for the resulting circle is x 2+(y-1) 2=1

If the coordinates of two different points p and q are (a,b),(3-b,3-a), then the slope of the perpendicular bisector l of the line segment pq is -1, and the equation for the circle (x-2) 2+(y-3) 2=1 for a circle symmetrical with a straight line l is .

Solution: Let p(a,b); q(3-b, 3-a), p, q symmetrical with respect to l. There are many, many such p,q,According to the topic,Of course I can choose。

choose a=2, b=3In this way, the center of the symmetrical garden is quickly found to be (0,1), and the equation for the desired garden is quickly obtained:

x²+(y-1)²=1.Whether it's a multiple-choice or math-based question, you can do that! Friends on the second floor are a bit sloppy.

Center of the circle (3, 2).

The radius r=2 and let the centricity be d

Then d +(mn2) = r

d²=r²-(mn/2)²

mn≥2√3

mn/2)²≥3

So d 4-3 = 1

Straight line kx--y+3=0

So d=|3k-2+3|/√(k²+1)

d²=(9k²+6k+1)/(k²+1)≤19k²+6k+1≤k²+1

8k²+6k≤0

2k(4k+3)≤0

3/4≤k≤0

mn>=2√3

The radius of the circle is 2, and the center of the circle is on the point (3,2).

According to the Pythagorean theorem.

then the distance from the center of the circle to the straight line < = (2 2-( 3) 2)=1 then |3k+3-2|/√(1+k^2)<=1(3k+1)^2<=(k^2+1)

4k^2+3k<=0

3/4<=k<=0

Because the straight line is truncated by the circle and the chord is 2 times longer than the root number 3, the distance from the center of the circle to the straight line is 1, so |2k|Under the root number (k 2 + 1) < = 1, the solution is - root number 3 3< = k< = root number 3 3

The first big question.

f(1)=2

m=1f(-x)=-x-1 x=-(x+1 x)=-f(x) f(x) is an odd function.

Let a, b be any two values on (1,+ and a2 solve this inequality.

The second big question. f(1)]²g(1)]²=

f(x)]²g(x)]²==(2^x)×(

Let's teach you how to solve the problem.

Function parity proof.

If f(-x)=f(x), it is an even function, and if f(-x)=-f(x) it is an odd function.

The function increment and decrease proves that these two methods are more commonly used.

1.Subtracting the former from the latter term, it can be seen that it is exactly negative, and the positive number is the increasing function, and the negative number is the subtraction function.

2.Derivative of the function, the derivative is greater than zero, the function increases, and vice versa, decrementIn your first question, according to f(1)=2, bring the point (1,2) into the original function, you can find the value of m, the function is there, you can find anything.

The second problem is to bring the two functions into the equation in the problem and slowly simplify the solution.

1. sin +cos == 2 2 2 squared on both sides to get 2*sin *cos = 1 - = > 0 so there is (0, 2).

again sin( + 4) = 1 2 so 4 = 5 6 sin -cos = 2cos( + 4) = 6 2

2, x 10 = ((x - 1) +1) 10 according to the binomial formula knows a8 = 10!/(8!2!)=45

3. Let the tolerance be d then a = b -d; c = b + d;The equation of the straight line b(x + y + 1) +d(1 - x) = 0 so the straight line passes the fixed point d(1,2);

The distance between the point d and the center of the circle is [1 -1) 2 + 1 - 2) 2] = 1 < 2 so the point d is inside the circle, so when the line is perpendicular to the line between d and the center of the circle, the chord length is the smallest. The minimum is 2* [2) 2 - 1 2] = 2

a 2 = c 2 + b 2 and a =c 2+b 2 - 2*b*c*cos(a) So cos(a) = 2a 2 (b*c), based on the vector ab and vector ac, then the length of vector ab is c and the length of vector ac is b. The inner product of vector ab and vector ac is b*c*cos(a) = 2*a2

be = 1/2(ba + bc) = 1/2ac - ab

cf = 1/2(cb + ca)= 1/2ab - ac

The inner product of the vector be and vector cf is -1 2|ac|^2 - 1/2|ab|^2 + 5/4)*|ab|*|ac|*cos(a) = 0

So the vector be and vector cf are perpendicular to each other, and the cosine of the included angle is 0

sin +cos == 2 2, then square both sides, find the sin multiplied by cos, and you should know how to do it below.

Contemptible high school one, others will not.

1 sinα+cosα=√2sin(α+45)=√2/2 sin(α+45)=1/2,α=105 ,cos(α+45)=-√3/2<0

sinα-cosα=-√2cos(α+45)=√3/22x^10=[(x-1)+1]^10, a8=10!/(8!2!)=45

3ax+by+c=0, x 2+y 2-2x-2y=0 circle(x-1) 2+(y-1) 2=2, when the line is tangent to the circle, the minimum value of the chord is 04

Because (0, )sin +cos == 2 2,, according to the title, sin -cos =as(in-co).