2 Senior 1 math problems The process is a little more detailed. Thank you all 5

The word vector is omitted.

The projection of a on b is a cos, a cos = (a·b) b =-7 5=-7 5 5

a+b｜=√（a+b）²=√（3，4）²=√25=5

Because a and c are parallel, 5 4 = 3m and m = 20 3

b and d are perpendicular, so -6 + n = 0 and n = 6

m+n=38/3

When a=2, x 2-2ax+3 0 is x -4x+3 0

Solve this inequality, x 1 or x 3

The outer function decreases, so the decreasing interval of the inner function is the increasing interval of the outer layer (negative infinity, 1).

The domain is defined as r, that is, x 2-2ax+3 is 0, =4a -12 0 on r, and the solution is -3 a 3

The value range is r, that is, the inner function can take all the positive numbers =4a -12 0 on the defined domain to get a - 3 or a 3

The outer layer decreases, while f(x) is an increasing function within -1, so the inner layer decreases at -1.

So 1 4a 1 0 a 1 4

The first one, (vector a vector b) |b|Projection, |Vector a vector b|＝|5-3，3 1）|Root number (2 2 4 2).

3×m=5×4，-2×3 1×n=0,m n=...

Second, the position of the true number should be greater than zero, discriminative, symmetry axis, combined.

This is the teacher's question for you, he wants to teach you to use the superposition method (the first question) and the cumulative multiplication method (the second question) to find the general formula of the number series, which is very common in high school and very important, so you must absorb it well!

1.Solution: According to the meaning of the question an+1-an=n

Then there is: a2-a1=1

a3-a2=2

an-an-1=n-1

Stacking the above equation (i.e., adding it all together) gives :

an-a1=1+2+3+……n-1）

n^2/2+n/2

and a1=3, so an=n 2 2+n 2+3

2.Solution: According to the topic:

a2/a1=2×1/2

a3/a2=4×1/3

a4/a3=6×1/4

a5/a4=8×1/5

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

Explanation: From the above, we can find the law, the adjacent two terms are multiplied, and the numerator and denominator can be reduced, and 2) can be obtained after dividing

Multiply the above equation (i.e., multiply it all) to get:

an/a1=2^(n-1)/n

and a1=1, so an=2 (n-1) n

The first line is added.

The second link.

You can call me for the specific process.

Let's give you the answer first.

Question 1: an=3+n(n-1) 2

Question 2: an=[2 (n-1)] n

At the suggestion, the first question uses superposition, and the second question uses multiplication.

One. a2-a1=1

a3-a2=2

a4-a3=3

an-a(n-1）=n-1

So the addition of the left and the addition of the right are:

an-a1=1+2+3+……n-1

an=n*(n-1) 2+a1

The second is the same as the multiplication on the left, eliminating the middle term.

The interval is in (- 1).

When x=1, f(x)min=0

amin=-1

And you can get it-

a＜0a∈[-1，0）

I mistook the length of the circle as the circumference, and I couldn't understand the title.

Make CE AB at point E, DF AB at point F, and set of=oe=kod=oc=oa=ob=r=1

of=oe=√(1-k2),af=be=1-kda2=df2+af2

x2=2-2k

k=(2-x2)/2

f(x)=-x2+2x+4

x∈（0，√2）

f(x)∈（4，5）

It's a hassle to write these questions online, did you know?

This question is a very classic question, it is a college entrance examination question in Beijing in a certain year, you can check it yourself, in fact, it is not difficult, many teaching aids are available, you can find it yourself.

(1)s= t=

2) A has the property p, then 0 does not belong to a

If (ai,aj) t,a has the property p, then ai,aj,ai-aj a and aj-ai does not belong to a and ai ≠ aj

aj, ai) does not belong to t and ai ≠ aj, and a total of k elements.

Then the number of elements in t n c(k,2) = k(k-1) 2

3) If there are elements in S or T, then there must be three elements xi, xj, and xr in a (they can all be the same, or they can all be different, or there are two of the same) so that xi+xj=xr

When xi≠xj, it corresponds to two elements in s (xi, xj), (xj, xi) and two elements in t (xr, xi), (xr, xj).

When xi=xj, it corresponds to an element in s (xi,xi) and an element in t (xr,xi).

s corresponds to the number of elements in t, i.e., m=n

The problem-solving process is too complicated.

: x 2-x-6<0 solution -20 solution x>2 or x<-4a b={xi2 in (x-a)(x-3a)<0 because the equation x 2-4ax+3a 2=0 has roots, the discriminant formula is greater than 0, the solution is a>0, so a because a b c so a"2 3a"3 3 3 3 a, 2 "2"

Please adopt the correct answer in time, and may help you next time, you adopt the correct answer, and you can also get wealth value, thank you.

1.a=b, x=2, and x=3 are the solutions of the equation ax -2x+6a=0; So: 2+3=2 a; 2×3=6a/a

So a=2 5

2.A is included in B, i.e. A is a subset of B;

1) a= (empty set), 4-4a 6a 0; and a>0;

Solution: a 6 6;

2) If a=0, a=(0,+ is not in place;

3) 00 hours, 4-24a >0; 00, -6/6a=(-∞1+√1-6a²)/a)∪(1-√1-6a²)/a,+∞

Only (1-1-6a) a 2; A 0, or A 2 5 So-6 6 (3) A - 6 6, obviously suitable.

4) When a=0, a=(0,+ suitable.

In summary, the range of a is: {2 5} (0].

1.If a=r, then the inequality holds when x takes a real number, so the function image is below the y-axis. So the function graph opening is downward, and the maximum value is less than 0

i.e., a<0, and when x takes the axis of symmetry - (2a b) = 1 a, f(x) < 0

So a<0,6a< (1 a).

Get a< - minus sixths of the root number.

2 If a belongs to b, then x is 2 or 3... Bring x=2 and x=3 into the original solution, respectively.

3 If b belongs to a, then the x solution must contain both 2 and 3, from which the set of solutions a can be obtained.