High school math problems, high scores!! Eager to solve math high school problems!!

Because the math format is wrong, I made a truncation of **It looks better, you can click on** to see the detailed process! I'm so tired! It's a pain to play those formats!

Solution: a(n+2)=3a(n+1)-2a(n)a(n+2)-a(n+1)=2[a(n+1)-a(n)]a(n+2)-a(n+1)] [a(n+1)-a(n)]=2[a(n+1)-a(n)] is a proportional series with a common ratio of 2.

From (1), it can be seen that:

a3-a2=2(a2-a1)

a4-a3=2(a3-a2)

an-an-1=2(an-1-an-2)

The above equation is added together.

an-a2=2a(n-1)-2a1

an-3=2a(n-1)-2

an=2a(n-1)+1

an+x=2a(n-1)+1+x=2[a(n-1)+(1+x) 2] let x=(1+x) 2, and the solution is x=1

an+1=2[a(n-1)+1]

an+1)/[a(n-1)+1]=2

Then the series is proportional to the common ratio = 2, and the first term is a proportional series of a(2-1) + 1 = a1 + 1 = 2.

an+1=2*2^（n-1)=2^n

The general formula for the series is: an=2 n-1

1. Prove: Sorted out by the formula, an+2-a n+1=2an+1-2a nan+2-a n+1) (an+2-a n+1- a n)=2 substantiation is worth a3=7, then (a3-a2) (a2-a1)=2, so the sequence (an+2-a n+1) (an+2-a n+1- a n) is a proportional series. q=2

2, by the proven proportional series, a2-a1=2, a3-a2=4....an-an-1=2^(n-1)

All kinds of additions can be observed, and finally an-a1=2+4+.2 (n-1) = 2 n-2, an=2 n-1

It should suit your taste...

It's very detailed...

The answer to the 5 questions is:

The answer to question 4 is: y=1 2sin(3x+6).

Solution: (1) Knowing from the question:

f(x)＝cos²(x＋π/12)＝1/2[1＋cos(2x＋π/6)]

x x0 is an axis of symmetry of the function y f(x) image.

2x0＝kπ－π/6(k∈z)

g(x0)＝1＋1/2sin2x0＝1＋1/2sin(kπ－π/6)

When k is even, g(x0) 1 1 2sin( 6) 1 1 4 3 4

When k is odd, g(x0) 1 1 2sin 6 1 1 4 5 4

2)h(x)＝f(x)＋g(x)＝1/2[1＋cos(2x＋π/6)]＋1＋1/2sin2x

1/2[cos(2x＋π/6)＋sin2x]＋3/2＝1/2(√3/2 cos2x＋1/2 sin2x)＋3/2

1/2sin(2x＋π/3)＋3/2

When 2k 2 x 3 2k 2

That is, when k 5 12 x k 12(k z), the function h(x) 1 2sin(2x 3) 3 2 is an increasing function.

When 2k2 2x 3 2k 3 2k3

That is: when k 12 x k 7 12(k z), the function h(x) 1 2sin(2x 3) 3 2 is the subtraction function.

Therefore, the monotonically increasing interval of the function h(x) 1 2sin(2x 3) 3 2 is: [k 5 12,k 12] (k z).

The monotonically decreasing interval of the function h(x) 1 2sin(2x 3) 3 2 is: [k 12, k 7 12] (k z).

Let me help you, here is the first to explain that s(n) represents n is the subscript, and the rest is the same!

There is a formula that goes like this, let the sum of the first n terms of the difference series a(n) and b(n) be s(n) and t(n) respectively

There is s(2n-1) t(2n-1)=(2n-1)*(a1+a(2n-1)) 2 (2n-1)*(b1+b(2n-1)) 2

a1+a(2n-1) b1+b(2n-1)=a(n) b(n), the last basic formula you should know, not much to say!

There is, s(2n-1) t(2n-1)=a(n) b(n).

That is: s(n) t(n)=a((n+1) 2) b((n+1) 2).

sn tn an 1 2n 7, and a5 b5 2, bring n=9 into the above equation, you can:

s(9) t(9)=a((9+1) 2) b((9+1) 2)=a5 b5 2 5, then:

a*9 1 2*9 7=2 5, then a=1

then sn tn n n 1 2n 7, and s(2n-1) t(2n-1)=a(n) b(n).

2n/(4n+5)

Here replace n with 2n-1 and it will come out)

then a(n) is a multiple of 2n, which is obtained according to the proportional formula.

And since the question is s2 6, then a1+a2=6, obviously only when a(n)=2n is obtained, so a(n)=2n

The function y g(x) is the inverse of the function f(x) 2x 1

Finding the inverse function, this according to the definition of the inverse function, gets:

y=2x+1

x=(y-1) 2 and xy are swapped to obtain: y=(x-1) 2, which is the inverse function g(x)=(x-1) 2

Because the question is set in cn g (cn 1), bring in:

c(n)=(c(n-1)-1)/2

2c(n)=(c(n-1)-1), here we use a method of undetermined coefficients to find a proportional series.

Let 2(c(n)+x)=(c(n-1)+x), and compare with the above equation to obtain x=1

Then there is 2(c(n)+1)=(c(n-1)+1), then (c(n)+1) (c(n-1)+1)=1 2

then it is an equal proportional series, and the general term formula is written:

c(n)+1=(c1+1)*(1 2) to the n-1 power.

c(n)+1=2*(1 2) to the power n-1 = (1 2) (n-2) (this represents the power).

then c(n)==(1 2) (n-2)-1

This question is so complicated to write, remember to give points.

Solution: (1) The inverse function y=g(x)=(x)=(x-1) 2 is obtained from the function f(x)=2x+1

Let an=a1+(n-1)d1; bn=b1+(n-1)d2

sn=(a1+an)n/2=(nd1+2a1-d1)n/2;

In the same way, tn=(nd2+2b1-d2) 2

Therefore, SN tn=(nd1+2a1-d1) (nd2+2b1-d2)=an+1 2n+7

We get d2=2, b1= d1=a, 2a1-d1=1 and equation 1

So, b5=b1+4d2=

then a5=b5*2 5=5

i.e. a1+4d1=5 Equation 2

The simultaneous equation 1,2 gives d1=1, a1=1, a=1

So a=1 and the function y=g(x)=(x-1) 2

2) From the first question, an=a1+(n-1)d1=1+(n-1)*1=n

cn=[(cn-1)-1] 2 (n is an integer greater than 1).

c1=1 (special note: for phases that do not satisfy the formula of the series or the first term known, it should be listed separately).

……Are all high school math problems so hard now? Poor child.

I him, and I never learned it.

1.Let m=1, then sn+1=s1+q*sn (1)

When n=1, the above equation becomes s2=s1+q*s1 => a2=q*a1 => is a proportional series.

When n>1, then sn-1+1=s1+q*sn-1 (2)(1)-(2) yields: an+1=q*an=> is a proportional series.

2，3.There is a problem with the writing of the title. What about h?

Assuming that the ground is x long and y wide, then there is xy=25. Let the total cost be z, then there is.

z=6*（x+y）*400+500*25

Therefore, z>=2400*2 root number xy+12500=36500 yuan, at this time x=y=5

If p is a true proposition, then f'(x)=x 2-2a<=0 (x belongs to [0,1]), i.e., -2a<=0 and 1-2a<=0, so a>=1 2, if q is a true proposition, then the discriminant of the equation "=0, the solution gives a<=-2 or a>=1, (1)p and q is a true proposition, then a>=1;

2) p or q is a true proposition, then a<=-2 or a>=1 2

p is true:

Derivative f'(x) = x^2-2a

f(x) decrements at [0,1], then there is f'(1) 0 A 1 2q is true

4a^2-4(2-a)≥0

A 2 + A-2 0 gives a 1 or a -2(1)p and q is a true proposition, taking the intersection a 1

2) p or q is a true proposition, take the union set a 1 2 or a -2

It is not good to do it, and it is difficult to do it with a negative proposition. - Department of Mathematics, Xi'an Jiaotong University.

1. The average monthly temperature of a place in the first half of the year is , and it can be made into a line chart in order to show the change in temperature in that place

2. There are 15 students with excellent results in the physical education test of the 61 class, accounting for 25% of the class, and when making a fan-shaped statistical chart, the central angle of the circle is (90 degrees); If the central angle of the fan circle of the student who indicates or gets good is 72°, then there are (3) people who get good.

3. On a bar chart, the longitudinal axis is represented by 300,000 yuan with a length of 1 cm, and the vertical bar representing 1.5 million yuan should be drawn with a length of (5) cm; A straight strip of centimeters long, he said (750,000).

4. The discount chart can not only represent ( number ). And it can be clearly expressed (the law of change).

5. The median of the mean (48) of a set of data is (48) and the mode is (48).

1.Line charts.

3.5, 700,000.

4.The number of data and the trend of data changes.

1.Line charts.

4.The value of each point is the overall trend.

1. Line chart.

degree, 12 people.

750,000 yuan. 4. Changes in variables Specific values of variables 48 48

1.Line charts.

2.90 degrees, 8 people.

3.5 cm, 750,000 yuan.

4.The number of quantities, the increase and decrease of the quantity.

Line chart, 90, 3, 5, 750,000, number, law of change, 48, 48, 48