High School Mathematics Who knows, please teach with addition .

The first step is to use the derivative y'=nx^(n-1)-(n+1)x^n

So the slope of the tangent at x=2 is k=y'(x=2)=n*2^(n-1)-(n+1)*2^n

When x = 2 and y = -2 n, the equation for the tangent (using the point slope) can be obtained

y=[n*2^(n-1)-(n+1)*2^n]x+(n+1)*[2^(n+1)-2^n]

So the intersection of the tangent and the y-axis is (0,(n+1)[2 (n+1)-2 n]), so an=(n+1)[2 (n+1)-2 n].

an/(n+1)=2^(n+1)-2^n

Let bn=an (n+1), so bn=2 (n+1)-2 n, b(n-1)=2 n-2 (n-1).

So the former item and.

sn=b1+b2+b3+..bn

2^(n+1)-2^n+2^n-2^(n-1)+2^(n-1)

2^(n+1)-2

A few more points are added below.

It is necessary to prove that an is a positive number, that is, to prove that the intersection of the tangent and the y-axis must fall on the positive semi-axis of the y-axis, that is, the slope of the tangent must be more than y=-2 (n-1)x of the straight line (this line passes through two points (0,0), (2,-2 n), and the latter point is the tangent point).

y'-[-2^(n-1)]=n*2^(n-1)-(n+1)2^n+2^(n-1)

n+1)2^(n-1)-(n+1)2^n

n+1)[2^(n-1)-2^n]<0

Another one is the problem of the number series, which is similar to the sum or product of the problem type, and often uses the commutation method and the superposition method or the superposition method, so pay more attention.

y=x n(1-x), find the derivative y'=nx (n-1)-(n+1)x n and x=2, the tangent slope is k=-(n+2)*2 (n-1) and the tangent coordinates are (2, -2 n).

Write the tangent equation as y=-(n+2)*2 (n-1)x+(n+1)*2 n and the intersection of y is so that x=0

So an=(n+1)*2 n

Then = Okay, then you should have a plenary meeting.

(1) Take out two unknowns in the definition domain Assuming that x1, x2 are respectively, and x1 >x2, then f(x1)-f(x2)=a*[2 (x1)-2 (x2)]+b*[3 (x1)-3 (x2)], assuming that it is greater than zero, i.e., a*[2 (x1)-2 (x2)]>b*[3 (x1)-3 (x2)], obviously[2 (x1)-2 (x2)].

and [3 (x1)-3 (x2)] are both positive, then when a and b are also positive, the assumption is true, and the function is an increasing function; When both a and b are negative, the assumption is not true, i.e., f(x1)-f(x2) is less than zero, then the function is a subtraction function.

2) If ab<0 and f(x+1)>f(x), this sentence is translated into mathematical language, i.e. a*2 x+2b*3 x is greater than zero. i.e. a*2 x>-2b*3 xIf a is less than zero, then b is greater than zero, obviously the inequality will never hold, then, it must be a greater than zero, b less than zero, and then the inequality will be processed...

It's too troublesome to do math problems with a computer.,Seeing that it's almost done.,The results of the operation error are all gone.,The rest is easier The landlord will do it himself.。

1.If ab>0, then A and B have the same sign.

1) When A and B are both positive, set x1 > x2

f（x1）-f（x2）=a*2^x1+b*3^x1-（a*2^x2+b*3^x1）=a（2^x1-2^x2）+b（3^x1-3^x1）

Since 2 x and 3 x are increments, 2 x1-2 x2 and 3 x1-3 x1 are both greater than 0.

f(x1)-f(x2)>0 So f(x) is an increasing function (2) When a and b are both negative, the same can prove that f(x) is a base function 2 Sort it out.

f(x+1) > f(x) are.

a*2 (x+1)+b*3 (x+1)>a*2 x+b*3 x can be obtained.

a*2 x+2b*3 x>0 divided by 2 x to get a+2b*(3 2) x>0.

2b*(3/2)^x>-a

Because ab<0, A and B have different names.

1) When a>0 b<0, divide both sides of the inequality by 2b to get (3 2) x<-a 2b

x0 is obtained by dividing both sides of the inequality by 2b.

3 2) x>-a 2b so.

x>log(3/2)(-a/2b)

1. If ab>0, judge the monotonicity of the function fx and prove it with a definition.

Categorical discussions. If a>0, b>0, a*2 x and b*3 x are increments, then f(x)=a*2 x + b*3 x are increments.

2. If a<0, b<0, a*2 x and b*3 x are subtractive functions, then f(x)=a*2 x + b*3 x are subtractive functions.

2. If ab<0, find the value range of x when f(x+1) > f(x).

f（x+1）-f（x）>0

a*2^(x+1)+b*3^(x+1)-(a*2^x+b*3^x)>0

a*2^x+2b*3^x>0

The first case.

If a>0, b<0

then a*2 x>-2b*3 x>0, so (a*2 x) (-2b*3 x)<1, and x>log(2 3,-a 2b) is solved

The second case.

If a>0, b<0

then a*2 x+2b*3 x>0

0<-a*2 x<2b*3 x, so (a*2 x) (-2b*3 x)<1, the solution is x

1. It is to first set the desired point (x,y), then find out the relationship between x,y and the corresponding curve point a of the known equation (the point on it is represented), and then bring the coordinates represented by x,y of the corresponding point a into the equation after the simplification of x,y The functional relationship is the trajectory of the found point.

m(x,y), then p(x2,y 2).

The point p satisfies the elliptic equation, and all (x 2) 2 25+(y 2) 2 9=1 then x 2 25+y 2 9=4, that is, the trajectory of the m point 2, and the maximum angle is when p is located on the y axis, then cosa=b a=3 5cos angle f1pf2=2(cosa) 2-1=-7 25

Take a look at the screenshot, it's written more clearly!

Select C for this question.

The analysis of this question is known to be obtainable from ab|= oa 2+ob 2=10, let the center of the circle (m, m) then 丨oa丨-m+丨ob丨-m=丨ab丨 i.e. 6-m+8-m=10, m=2, so the radius of the inscribed circle is also 2. So the circular equation (x-2) 2+(y-2) 2=2 2 i.e. x 2 + y 2-4x-4y+4 = 0

Let the radius of the inscribed circle be r, and according to the title, in RTΔABC, OA=6, ob=8, AB=10

Because (oa-r) + (ob-r) = ab

6-r)+(8-r)=10

2r=4r=2 so the inscribed circle equation is: (x-2) 2+(y-2) 2=4x 2+y 2-4x-4y+4=0

The answer is C

It's a very simple question! Wait for me to explain it to you. But now at work, I can't do the questions, I'm afraid that the manager will be scolded when he sees it, so I'll answer it for you after work.

I am engaged in high school mathematics, physics and chemistry tutoring, and I am in Shanghai Petrochemical District.

The first question is that x is greater than -1 and less than or equal to 0

The second and third questions are to solve the range of x when x+3 is greater than 0 and less than 0 in 2x-1 x+3, which is u, and then find the union of b and the intersection of a after that, and the third question is the same as above.

Hope it helps.

First, solve the definition domain of u, 2x-1 x+3<1

2x-1x<4a b: To draw a number line.

2<-1a∩b=

a (cub): number line.

cub)=a∩(cub)=

cua) b: number axis.

cua=cua)∪b=

a∩b=

Set C? You don't have it in the question!

In high school, the so-called trajectory is to draw a diagram by yourself, roughly determine what the figure is, and then purposefully set the equation.

The first little question is all What happened to the vector op???

(1) Simplification first.

f(x)=cosxsinx+√3cos²x-√3/2=1/2sin2x+√3/2cos2x

sin（2x+π/3)

So the minimum period is

2) Figures omitted. 3) Obtained from the question.

g(x)=f(x+π/12)=sin【2（x+π/12）+π/3】=sin（2x+π/2)=cos2x

Because cos2x=cos(-2x).

So the function g(x) is an even function.

In the middle of the night, I'm here for you with dark circles under my eyes, and I ask you to ......... a lot