200 for senior one mathematics and 200 for senior one mathematics

Solution: Let the coordinates of the center of this circle be (x, y), then the equation for the perpendicular bisector of the line segment ac is:

x--5) 2+y 2=(x+3) 2+(y--3) 2, i.e.: 16x--6y--7=0 (1).

In the same way, the equation for the perpendicular bisector of the line segment cd is:

x=2 (2)

From (1), (2): the coordinates of the center of the circle are (2,25 6), and since the point (a, 0) is on this circle, (a--5) 2+(0--0) 2=(a+3) 2+(0--3) 2

a^2--10a+25=a^2+9a+9+919a=7a=7/19.

Solution: a(5,0),b,c(-3,3),d(7,3) concyclical, yc=ydxc+xd=2 x center of circle.

3+7=2 x center of the circle.

Solution to x-center = 2

xb+xa=2 x center of the circle.

a+5=2x2

a=-1 Good luck in your studies!

a=-1b(-1,0)

Let the center of the circle be p(a,b).

pa=pc=pd

Find p(2,25 6).

pa= pb

a=-1

First, the domain is defined as r, which satisfies symmetry.

f（1）=2+2=4

f(-1)=0+4=0

i.e. f(1) = f(-1).

f(x) is an even function! PS: This method is a big problem.

The big questions should be explained by f(x) and f(-x).

i.e. prove f(x) = f(-x).

If the landlord is from Guangdong, this question is generally not a big question in the college entrance examination.

It can be seen that C d is parallel to AB and their midpoint abscissa should coincide. The abscissa of the midpoint of cd is 2, so a+5=2*2, a=-1

Question 1: a={x 2-3x+2=0}=

b={2x 2-ax+2=0}, and there are only three possible compositions of set b:

a.The equation has no solution, i.e. =a 2-4*2*2=a 2-16<0, we get -40, we get a>4 or a<,x2}, x1,2=(1 4)*[a (a 2-16)].

1) The set a is a subset of the set b, then firstly, a=b is not satisfied, and the two solutions of 2x 2-ax+2=0 cannot be the same as the set a, excluded; In addition, set B has at most two elements, so no matter how A takes the value, set A cannot be a subset of set B, that is, the value of A is an empty set;

2) Set b is a subset of set a, then firstly, a=b is not satisfied, and the two solutions of 2x 2-ax+2=0 cannot be the same as set a, excluded; When the set b has a unique element, a=4, b={2x 2-4x+2=0}={1} satisfies the condition and is true; When set b is an empty set, the condition that set b is a subset of set a is also satisfied, and the value range of a is -40, a={x |Since 0 is a=b, it must satisfy both -1 a=-1 2 and 4 a=2 to get: a=2.

3) When a<0, a={x |0 Since the set a is the interval between the left and the right open, and the set b is the interval between the left and the right closed, no matter how a is valued, it is impossible to satisfy a=b, so the value of a is an empty set.

In summary, the value range of a is a=2.

Question 3: Set p={x |What is x2-1}?

a=, to a=b, then:

1 a=-1 2 and 4 a=2

The solution is a=2

When a<0:

a=, to a=b, then:

4 a=-1 2 and -1 a=2

The solution is: a=-8, and a=-1 2

So there is no solution to the system of equations, i.e., when a<0, a≠b

So when a=2, a=b

For your question, it is like this, when a=0, no matter what value x takes, the equation or inequality holds, then x r

When a=0, no matter what value x takes, the equation or inequality does not hold, then: x is an empty set.

Question 1: If a is less than 0, then the original formula is reduced to 4 a" x"-a 1, so 4 a = -1 2. -1 a=2, and the solution to a has two values, so it is impossible.

If a=0, then ax must be equal to 0, the set is about x, and the condition that x takes any real number is true, so the set of x is r.

What does the second question mean? There is no relational formula.

In the first problem, when a=0, no matter what value x is, there is ax=0, then there is ax+1=1 constant, then for the inequality 0< ax+1<5, the inequality holds regardless of the value of x, so the range of x is a real number. In the second question, it is impossible to make ax=0 no matter what value x takes when a=0, so there is no such x in the range of real numbers, so it is an empty set.

Depending on the problem, [0] contains elements that are multiples of all 5s, including negative numbers.

a-b belongs to [0], which means that the difference between a and b is a multiple of 5.

What is a class here? According to the meaning of the question, there are 5 categories in total, which are [0], 1], 2], 3], 4].These 5 class representations mean that the remainders of the integer elements in the class divided by 5 are 0, 1, 2, 3, 4

A b is in the same category, which means that the remainder of a divided by 5 is equal to the remainder of b divided by 5.

This conclusion is correct. Since the difference between a and b is a multiple of 5, then a = b+5*m, and m is some integer.

The remainder of the divisible b divided by 5 is p, p=one of the , then b can be expressed as b=p+5n, and n is also an integer.

Then a=b+5m=p+5n+5m= p+5(n+m), it is easy to know that 5(n+m) is divisible by 5, then the remainder of a divisible by 5 is the remaining p. That is, the remainders of a, b divisible by 5 are the same.

So A and B belong to the same category.

Quite verbose, hope that helps.

The integers a and b belong to the same class, and here I will explain to you the class mentioned in this question in layman's language. (A class can be thought of as numbers with the same properties grouped together to form a set).

For example, numbers. For these seven digits, the remainder of the first six digits divided by 5 is 1, according to the explanation of the title (the remainder obtained by dividing by 5 is k (where k is 1) is of the same class.

And the remainder after dividing the seventh number by 5 is 3 and does not equal 1, so it does not belong to the previous class. It's another category.

Yes, that's it. Next, we explain "a-b [0]", which means that a-b is divisible by 5 after that. It can be assumed that b is (the remainder is 1 after dividing by 5), then a can only be the number of 1 after dividing by 5.

i.e. class [1]) You can try to give a counterexample. Letting a be class [2], class [3], and so on will not satisfy the condition.

Okay, that's it, I hope it helps.

The same class, or the remainder of an integer divided by an integer, is the same number.

The integers a and b belong to the same "class", and the remainders of the integers a and b divided by 5 are the same, so that the remainder of a-b divided by 5 is 0, and vice versa is also true, so the sufficient and necessary condition for the integers a and b to belong to the same "class" is "a-b [0]" so it is correct.

The so-called same class means that the remainder of the division by 5 is the same, that is, a class.

In a-b, such as 11-6, 12-7, 13-8, 14-9, all the remaining are multiples of 5, because the same remainder will be subtracted, we can know that a and b have the same remainder, so a and b belong to the same class.

Equal difference and d1≠0

Also equal, d2

i.e.: a(n)-a(n-1)=d1

and 1 a(n)-1 a(n-1)=d2

1 a(n)-1 a(n-1)=-d1 a(n)a(n-1)=d2 d2 remains constant only if a(n)a(n-1) is constant, at which point a(n)a(n-1)=k

And a(n)=a(n-1)+d1, so a(n-1) 2+a(n-1)d=k, a(n-1) is constant.

This is not possible due to d1≠0.

Proportional series b(1), b(2)=b(1)q, b(3)=b(1)q 2, b(4)=b(1)q 3

b(1), b(3)-4, b(4)-13 are equal differences, i.e., b(1)+b(3)-5=2b(2)-2

b(2)+b(4)-14=2b(3)-8

Substituting the above conditions.

b(1)+b(1)q^2-5=2b(1)q-2b(1)q+b(1)q^3-14=2b(1)q^2-8q=2,b(1)=3,b(2)=6,b(3)=12,b(4)=24

Solution: Let the number column be an equal difference series, the first term is a1, and the tolerance is d, then there is an-a(n-1)=d, d≠0

1/an-1/a(n-1)

a(n-1)-an]/[ana(n-1)]

d/[ana(n-1)]

For an equal variance series, -d [ana(n-1)] is a fixed value.

ana(n-1)

a1+(n-1)d][a1+(n-2)d]

a1^2+(n-2)a1d+(n-1)a1d+(n-1)(n-2)d^2

d^2n^2+(2a1d-3d^2)n-3a1d+2d^2+a1^2

For this polynomial to be a fixed value, i.e. independent of n, it is required.

d^2=0 2a1d-3d^2=0

and d≠0, so you can't find a d that satisfies the topic

2a +b = 3, get b = 3-2a

a√1+b²

a²+a²b²

a²+a²(3-2a²)

a²+3a²-2a^4

-2(a^4-2a²)

-2(a²-1)²+2

Since a is a positive number, there is a maximum value of 2 when a=1

First of all, don't call me stupid, after all, I've been reading for a long time, and no one has answered the question for so long, so I'll make do with it first.

After all, I'm a scumbag, let's say okay first, all I have now is Meng, and I'll ask the teacher tonight.

Get an accurate reply to you around 9:15.

Now I'm starting to analyze, judging from the question, if I am blinded, I will be blinded by a=1, b=1, then this question will be easy to do.

Original max = root number two.

But if you do this, it is not necessarily the maximum, so you have to continue the number of sets.

But in my experience, this topic should be root number two.

I'll ask the teacher in the evening, and I'll give you an accurate answer later!

2a^2+b^2=3

b^2=3-2a^2

Original = [a 2(1+b 2)]=a 2(1+3-2a 2)] = 2[a 2(2-a 2)]<2[a 2+(2-a 2)] 2 = 2

The original maximum is: 2

where b 2 = 3-2a 2, and bringing in the required formula gives a* root number (4-2a 2).

Take a to the root number to get 4a 2-2a 4=2(2a 2-a 4)=-2(a 2-1) 2+2

And -2(a 2-1) 2 is certainly less than or equal to 0

So to get the maximum, so there is -2(a 2-1) 2=0a, and b is a positive number.

So when a=1 and b=1, there is a maximum root number of 2

Let a=x, root number, (1+b 2)=y

Then the condition becomes 2x 2+y 2=4 Find the maximum value of xy Seeing such a change, do you think of some inequality?

First, the domain is defined as r, which satisfies symmetry.

f（1）=2+2=4

f(-1)=0+4=0

i.e. f(1) = f(-1).

f(x) is an even function! PS: This method is a big problem.

The big questions should be explained by f(x) and f(-x).

i.e. prove f(x) = f(-x).

If the landlord is from Guangdong, this question is generally not a big question in the college entrance examination.

I'll show you how.

First of all, let's analyze |x+1|+|x-3|If the absolute value of x+1 is greater than 0, then x+1 can be expelled, and if x-3 is also greater than 0, then when x is greater than 3, it is x+1+x+3

If x+1 is less than zero, and x-3 is less than zero, then open -2x+4, i.e., if x+1 is less than or equal to 0, x-3 is greater than or equal to 0, which is impossible, because x+1 is significantly larger than x-3.

Therefore, x+1 is greater than 0, x-3, and when light rain is 0, x+1+3-x is opened, which is 4

And so it works.

Drawing an image is about a trapezoidal upside down.

I also thought of another way.

Is it possible to use the formula that belongs to r, so it can be understood to bring in the value directly? I did it for a long time.

Evidence: f(-x)=log2[(1-x) (1+x)]=log2[(1+x) (1-x)] 1)=-log2[(1+x) (1-x)].

f(x) So f(x) is an odd function [not an even function!] In the defined domain (-1,1), let -11-x1>0,1-x2>0,x1-x2<0---g(x1) so g(x) increments over the defined domain, and log2(x) also increments, so the composite function f(x) increments on (-1,1).

Hope it helps.