Ask the master to explain the math problem of the first year of high school and give the process of

1.From x=5-4a+a*a, we can see that x>=1, so the set m is the set of all real numbers greater than or equal to 1.

From y=4a*a+4a+2, y>=1, so the set n is the set of all real numbers greater than or equal to 1.

These two steps are based on the minimum formula of the quadratic function.

So the correct answer is a

2。From the definition of the set operation: a b=, it can be seen that the result of the operation is to take any element x from set a, and then take any element y from set b, and substitute them into the formula z=xy(x+y), and the obtained z is one of the elements of set a b.

When x=1, y=3, z=1*3(1+3)=12x=1, y=4, z=1*4(1+4)=20x=2, y=3, z=2*3(2+3)=30x=2, y=4, z=2*4(2+4)=48 The product of all these z's = 12*20*30*48=345600

1. The range of x in m is x 5, the range of y in n is y 1, m and n are both sets of numbers, n contains m, and m is a subset of n.

2. a=, b=, where the elements are combined in pairs.

Product of all elements = 3 4 4 5 5 6 6 8 = 345600

Question 1: Think of x=5-4a+a*a and y=4a*a+4a+2 as two functions. Just find their ranges.

You'll find that their ranges are all from one to positive infinity. (1 where is the closed interval) and their range is the set of m and n. Now it's time to do the questions.

Very simple. Question 2: Simpler. Bring the data in. Divide into several groups to find z. There are four groups.

1, x=1 y=3, z=12 2, x=1 y=4, z=203, x=2 y=3, z=30 4, x=2 y=4, z=48, and then multiply the four z.

There are 2011 zeros at (- 0) and there are also 2011 zeros at (0, +.

If x=0 is defined, then f(0)=0

5.Prove that f(x)=x +3x-7 monotonically increases high school direct derivative on r (if it is not a high school student, this method is forgotten) junior high school takes either two points on r, and x1f(x1)=x1 +3x1-7

f(x2)=x2³+3x2-7

f(x1)-f(x2)=(x1 -x2)+3(x1-x2)=(x1-x2)(x1 +x1x2+x2)+3(x1-x2)<0f(x) is monotonically increasing on r.

f（0）=0

The odd function is symmetrical with respect to the origin, f(x)=-f(-x)f(x)=0 -f(-x)=0

The odd function is about the origin being symmetrical and passing the origin. There are 2011 zeros on the x negative half, so there are also 2011 on the x positive half, plus the origin, a total of 2011*2+1 4023.

When x>0, f(x)=0 has 2011 solutions, so when x<0, fx is symmetrical with it, so there are also 2011 solutions, and fx is an odd function, so f(0)=0, so there are 4023 solutions in total.

172, now let mathematics be a, history be b, ** be cThink of the three as three events, the total number of people: a + b + c - (a intersection b + a intersection c + b intersection c) + a b c intersection.

Counting the 6 overlapping people and the repeat attendance of the two lectures, there are a total of 172 people.

an=nsinnπ/2

a1=1*1

a2=0*2

a3=-1*3

a4=0*4

s4=-2 s8=-4

s2013=s2012+s2013=-2012/4*2+2013=2013-1006=1007

a+b!=!a!^2+!b!^2-2!a!*!b!^cos120°=7

a+b)*a=!a+b!*!a!cosa°=2 (sina'=√3/7 cosa'Scumbag laugh = 2 7).

3) It seems that there is something wrong with this burn a=

sn=n^2+n

an=sn-sn-1=n^2+n-[(n-1)^2+n-1]=2n

an=2 Senliang void 3bn+2=2n

bn=3^(n-1)

anbn=2n*3^(n-1)

tn=2*3^0+4*3^1+6*3^2+..2n*3^(n-1)

3tn= 2*3^1+4*3^2+6*3^3+..2n*3^n

tn-3tn=-2tn=2[3^0+3^1+3^2+..3^(n-1)]-2n*3^n=3^n-1-2n*3^n

tn=[2n*3^n-3^n+1]/2

Mispronounced: d

a. indicates that the intersection a is contained in b, then a b=a is correct.

b. denotes the union of a b b, then b contains all the items of a correctly.

c, (a b) a intersection belongs to union (a b) is correct.

d. Error.

The correct answer should be cu(a, u b)=(cua) (cub), e.g. a is an aquatic animal.

b is a terrestrial animal.

u for animals. Apparently there is a C flying animal in the U in addition to A and B.

CUA is an animal other than aquatic animals, of course, it is terrestrial animal B and flying animal C.

Cu (Cua) is an animal other than B and C, which is aquatic animal A. So cu(cua)=a

A and B are all aquatic A and terrestrial animal B

Cu(A and B) are animals other than aquatic and terrestrial animals, and the flying animals Ccua are terrestrial and flying animals, as mentioned above.

cub is an aquatic and flying animal, and the definition book of the supplement has made it very clear.

CUA) intersection (cub) is the common part of the above two animal groups, of course, it is the flying animal C!

Personally, I think it is better to connect the straight line and the circle b equation to determine whether they intersect, if they intersect, the smallest circle must be in the circle b, if not, forget it, or write the answers to both cases.

Intersect, the radius of the circle c r = half of the distance from the farthest point of the straight line to the straight line of the circle b, the determination of the center of the circle is more troublesome, first set to a, b

a, b must be on a straight line perpendicular to the line l and passing through the center of the circle b) so that the answer can be found.

Disjoint, the radius of the circle c r = half of the distance from the nearest point of the line to the line of the circle b, the other calculations of the same intersection.

Let the center of the circle c be (a,b) and the radius r.

The circle c is tangent to the straight line, then there is:

The distance from the center of the circle c to the straight line is the radius r of the circle, and there is:

r^2=|a+b-2|^2/2...1) There is a problem with the equation of the circle b, and the radius cannot be found?