Solve two math problems, two math problems, find the solution, want the process?

1. By -2 x 1 there are: 0 x+2, x-1 0So.

y=1-x-2|x|+x+2=3-2|x|It can be seen that when the absolute value of x is larger, the smaller y is, and the smaller the absolute value of x, the greater y is.

When x=0|x|Minimum, when x=-2|x|Max, so.

y max = 3 - 2 * 0 = 3

y min=3-2*|-2|=-1

Y max + y min = 2

2. Substituting C=A+B into B+C=D, obtaining D=B+C=A+2B, and then substituting the values of C and D into C+D=A, obtaining A=-3B, and then there.

c=-2b,d=-b

So a+b+c+d=-5b, where b is a positive integer, so when it is the smallest, a+b+c+d is maximum.

That is, b = 1, and a + b + c + d is the maximum value of -5

1.-2<=x<=1, y=-(x-1)-2|x|+(x+2)=3-2|x| ;

ymax=y(0)=3;ymin=y(-2)=-1;So ymax+ymin=2;

c, a integer, so b is a positive integer. Add the left and right sides of the three equations to get 2b+c=0;

Substituting the second equation d=-b; thus c=-2b; a=-3b;

Then a+b+c+d= is a positive integer, so the maximum value is =5

1. Draw a diagram with the primary method (easy to understand), and then follow the formal procedure to find the inflection point and concave and convex interval.

Note: Please fill in the calculation process by yourself.

2. There are two knowledge points: variable substitution and logarithm. It is also a regular technique for finding the limit, so please digest it.

For reference, please smile.

I only know that the second question is 1, and I forgot how to write the first question.

1.It is obtained from the cosine theorem.

a^2=b^2+c^2-2bc*cosa

b^2+c^2+bc

b+c)^2-bc;

substitution a=7, b+c=8; Get.

49=64-bc;i.e. bc=15;

recombined b+c=8; i.e. c=8-b.

b(8-b)=15;

b^2-8b+15=0;

b-3)(b-5)=0;

b=3 or 5;

c=5 or 3;

2.(b+c) 4=(c+a) 5=(a+b) 6=k, so b+c=4k

c+a=5k

a+b=6k

Add 2 (a+b+c) = 15k

a+b+c=

So a=, b=, c=

So a maximum.

cosa = (b 2 + c 2-a 2) 2bc = (so maximum internal angle = 120 degrees.

1.Cosine theorem, a 2 = b 2 + c 2-2bc cosa49 = b 2 + c 2 + bc

49=(b+c)^2-bc

bc=15 and b+c=8....So b = 3 or 5

2.(b+c) 4=(c+a) 5=(a+b) 6=k, so b+c=4k

c+a=5k

a+b=6k

Add 2 (a+b+c) = 15k

a+b+c=

So a=, b=, c=

So a maximum.

cosa = (b 2 + c 2-a 2) 2bc = (so maximum internal angle = 120 degrees.

1。According to b + c = 8, and the cosine theorem a square = b square + c square - 2a * b * cosa. Solving.

2。The same is the use of the cosine theorem, and the proportional equation can be homogenized.

a^2=b^2+c^2-2*b*c*cosa =(b+c)^2-2bc(cosa+1)

49=64-2bc

bc=15 and b+c=8....

The solution is b=3 or b=5

In the second question, let b+c=4k, there is a+b+c=, and substitute (b+c):(c+a):(a+b)=4:5:6

Get a=, b=, c=

cosa=（b^2+c^2-a^2）/2bc=（25+9-49）/（2*15）=-1/2

Therefore a=120

Trigonometric operations involving inverse trigonometric functions, I hope you can understand it.

To be continued. Finding the derivative, finding the extreme value (monotonic interval), and finding the maximum value on the closed interval are routine problems. Pay attention to the process and sketches

For reference, please smile.

1. Set the number of students enrolled in '06 as x and '07 as y.

x y=8 7 3y-2x=1500 solution: x=2400 y=2100

According to the title, the number of children enrolled in primary schools is decreasing year by year, so the number of children in 2008 exceeded 2,400. Hole difference.

1.Let the center of the circle be (a,b) and the radius be r, then the circle equation is (x-a) 2+(y-b) 2=r 2

Intersect the x-axis y=0 to get x 2-2ax + a 2+.0;So x1+x2=2a

Intersect the y-axis x=0 and get y2-2bx+b2+.0;So y1+y2=2b

then the sum of the intercepts is x1+x2+y1+y2 = 2b+2a = 2

In addition, a-4) 2+(b-2) 2=(a+1) 2+(b-3) 2 gives 5a-b-5=0

The combination of the two formulas gives a=1, b=0 radius r=root number 13

So the trajectory of the circle is (x-1) 2+y 2=13

2.Let point c be (x0,y0) to get point d (2 x0 -1,2 y0).

AC and OD are the midline, so the intersection point is the center of gravity, and the intersection coordinates are one-third of the sum of the coordinates of the three vertices, so the coordinates of the intersection point p are (x,y)=(2 x0 -1,2 y0) 3

x0,y0)=(3x+1) 2 ,3y 2) to get the equation of the trajectory of p.

3x+1)^2+9y^2=4