The first big question in a math test has 11 small questions, of which 1 6 is algebra problem 20

The total score is 3*6+5*2=28

The score obtained is not less than half of the total score of this major question, that is, it is greater than or equal to 14 correct answers to x algebra questions and y geometry questions.

x+y=63x+2y>=14

x and y are integers.

0=0<=y<=5

x》=2 So there are several cases.

xy respectively.

Answer 2 algebra questions correctly and 4 geometry questions: c6(2) c5(4)3 3 c6(3) c5(3).

4 2 c5(4)×c5(2)

5 1 c6(5)×c5(1)

6 0 c6(6)×c5(0)

It adds up to a total of 456

There are a total of 6 answer situations, as long as you find the number that meets the requirements.

Set: 1 6 questions answered correctly x questions, 7 11 answered 6-x questions correctly.

The big question has 6*3+5*2=28

3x+2*（6-x)>=28/2

3x+12-2x>=14

x+12>=14

x>=2

And because x<=6

So 2==6

And then because x is a positive integer.

x can be 2, 3, 4, 5, 6

So there are 5 situations in which you can answer the question.

The total score of this question is 6 * 3 + 5 * 2 = 28 points, that is, the score he gets is greater than or equal to 14 points, and there are five kinds of answers:

Type 1: Correct answers to six algebra questions, score: 18;

Type 2: Correct answers to five algebra questions and one geometry question, score: 17;

Type 3: 4 algebra questions and 2 geometry questions correctly, score: 16;

Fourth: three algebra questions and three geometry questions correctly, score: 15;

Type 5: 2 algebra questions and 4 geometry questions correctly, score: 14;

Total score: 6*3+5*2=28

Scored" = 14

Set: 1 6 questions answered correctly x questions, 7 11 answered 6-x questions correctly.

3x+2*（6-x)>=14

3x+12-2x>=14

x+12>=14

x>=2

So, x So, there are 5 cases.

6*3+5*2=28 (total score)

x+y=6 (6 questions correct).

3x+2y=>14 (not less than half of the total score of the question).

144=3*3*2*2*2*2*2, so b can only be 9 or 3 or 1, if b=3, then a 2=48, a is not an integer, if b=9, then a 2=16, a=4, can only be divisible by 4, if b=1, then a=12, can be divisible by 3, 4, 6, 12, so the answer is d

Note a = 6 under the root number

Finding the derivative yields x*x'+y*y'+z*z'=x'+y'+z'=0, if x'=0, then y*y'+z*z'=y'+z'=0, we get y=z, so (x,y,z) = (-2 a, 1 a, 1 a) or (2 a, -1 a, -1 a).

By symmetry, you may wish to set |x|>=|y|>=|z|, so m = x 2, and the maximum value of m can only be obtained at the critical point and endpoint of x. by the top x', at the critical point of x, m=(2 a) 2=2 3.

And at the endpoint, |x|=|y|, if x=y, you will get |z|=2/a > x|= 1 a, contradicting the assumptions.

So x = -y, at this time (x,y,z)=(1 root number2,-1 root number2,0) or (-1 root number2,1 root number2,0), at this time m=1 2

Therefore, the minimum value of m is 1 2.

When the three terms are equal, m is the smallest, which is 1 3

When x=y=z, the value of m is the smallest. i.e. m is smallest when x 2 = y 2 = z 2 = 1 3.

If 1, x, and y are both positive numbers, then the answer is 3

2, x, y are both negative, then the answer is -1

3, x, y is one positive and one negative, then, the answer is -1

Combined, the answer is 3 or -1

a^2+a-b^2-b=0

a+b)(a-b)+(a-b)=0

a-b)(a+b+1)=0

a=b or a+b=1

A=B, B a+a is lead B=1+1=2

a+b=1 is good, a,b is the root of the equation x 2+x-1=0.

a+b=-1

ab=-1b wheel width a+a b

b^2+a^2)/ab

a+b)^2-2ab]/ab

b a+a b has a value of -3 or 2

A 2 + A-1 = 0 and B 2 + B-1 = 0 have the same structure.

So, there are two scenarios:

1) A and b are roughly equal to each other, then the positive one is b a+a b=1+1=22) a and b are the two unequal roots of the equation x 2+x-1=0.

According to Vedder's theorem, a+b=-1, ab=-1

So b a+a b=(a 2+b 2) ab=[(a+b) 2-2ab] ab=(a+b) 2 ab-2=(-1) 2 (-1)-2=-1-2=-3

So, the values of b a+a b are 2 and -3

The parabola y=x +4x+k intersects the axis at a[- 4-k)-2,0] and b[(4-k)-2,0], and the vertex is c(-2,c).

So c = k-4, the vertex is c (-2, k-4), the distance of ab = 2 (4-k) and then the k value can be found according to the equilateral condition.

4(4-k)=(k-4)^2+^2=(k-4)^2+(4-k),(k-4)^2+3(k-4)=0,(k-4)(k-1)=0

k = 1 or k = 4

When k=4, the ABC three points coincide and are no longer triangles. But strictly speaking, it is also possible.

y=x²+4x+k=x²+4x+4+k-4=(x+2)²+k-4)

So the coordinates of vertex c are (-2,k-4).

As you can see, K-4<0, K<4

Let the coordinates of A and B be (x1,0), (x2,0), x1<0,x2<0

x1+x2=-4，x1x2=k

abc is an equilateral triangle, ab=ac=bc=(k-4) sin60 .1)

ab=|x1-x2|=√(x1-x2)^2=√(x1^2-2x1x2+x2^2)

(x1^2+2x1x2+x2^2-4x1x2)

[(x1+x2)^2-4x1x2]

(16-4k)

By (1), absin60=k-4, (absin60) 2=(k-4) 2

16-4k)*3/4=(k-4)^2，3(4-k)=(k-4)^2，k-4=-3k=1

a^2/(a^4-a^2+1)

The numerator and denominator are divided by a 2 at the same time, and we get:

1/[a^2-1+(1/a)^2]

1/[(a+1/a)^2-2-1]

1/[(a+1/a)^2-3]

a+1 a=3 substitution, get:

a+1 a=3 is squared.

a 2 + a (-2) + 2 = 9.

a^2+a^(-2)=7

The original formula = a 2 (a 4 - a 2 + 1) numerator denominator divided by a 2 = 1 (a 2 + a (-2) -1 ) = 1 6