Two math problems with the help of the experts . You can answer any of these questions.

1.The upper and lower bottoms of the trapezoid are 4 and 7 respectively, and if one waist length is 5, the value range of the other waist length x is

Answer: 34 (1 2) < = x < = 4

Solution: abcd, ab parallel cd, ab=4, dc=7, ad=5, bc=x, assuming that the waist of length 5 and the upper bottom form an angle of a, do ae, bf perpendicular dc at e, f points.

ab = ef = 4

ae = bf = 5sina

de = 5cosa

cf = cd - de - ef = cd - de - ab = 7 - 5cosa - 4 = 3 - 5cosa

bc = x = [bf^2 + cf^2]^(1/2) = [(5sina)^2 + 3 - 5cosa)^2]^(1/2) = (34 - 30cosa)^(1/2)

Because it is trapezoidal, de>0 and cf >0

5cosa >= 0, 3 - 5cosa>=0

So 0 <=cosa<=3 5

34-0)^(1/2) <= x <= (34 - 30*3/5)^(1/2)

34^(1/2) <= x <= 4

2.If the maximum distance from p to o in the plane where o is located is a, and the minimum distance is b(a>b), then the radius of this circle is Trouble to explain it briefly).

Both the maximum and minimum distances can be clearly seen in the straight line through the point p and the center of the circle.

If p is inside a circle, then a+b = 2r, r = (a+b) 2

If p is outside the circle, then a-b = 2r, r = (a-b) 2

1) Consider the ultimate limit state.

If the waist length of 5 is a right angle, the other waist can be calculated as 34 under the root number, and if the other waist is a right angle, the other waist is 4

So the value of x is in between.

2) The radius of this circle is (a-b) 2

Both the maximum distance and the line at the minimum distance are past the center of the circle.

Question 1 2 Question 2 When p is outside the circle r=(a-b) 2 When p is inside the circle r=(a+b) 2

When p is on a circle r=b 2

Flip the waist over to form a triangle with 5 and 3 on both sides.

The sum of the two sides is greater than the third side, and the difference between the two sides is less than the third side, and the result is 22(a-b)/2

Both the maximum distance and the line at the minimum distance are past the center of the circle.

Question: Its range should be greater than 2 and less than 8

The solution to this problem is ultimately applied to the theorem of the trilateral relationship of triangles, which is greater than 5 (7 4) and less than 5 (7 4).

Question 2: (a-b) 2

Question 2: You can first set the distance d between the point p and the center of the circle, then the maximum distance between p and the circle is d+r=a, and the minimum distance is d—r=b, and the system of equations is solved to obtain r=(a-b) 2

1,[4,34] 2,(a b) 2 Just draw a picture.

1: If the shilling a angle is a right angle, then m can be found to be 17

m=17, it is a right angle, and 0m<17, it is an acute angle.

m>17, it is an obtuse angle.

2: Let the side length be a, eb=b, ec'=a-b, then ac'=a-1, sc=a 3

pc’=（2/3）a-1

The square of BC' + the square of be = ec'of the squared.

The triangle s'pc is similar to the triangle c'be.

The result is that by subtracting b to obtain a = 2 to the third power (-one-third power) and adding 1 exactly ac' = a-1, the result is 2 to the third power.

1. Matching method: add 1 to the left and right at the same time to get: (x+1) 2=6, that is: x+1= root number 6, x = root number 6-1, or x=- root number 6-1

2. The same stare is the formula: (x+2) 2+(y+2) 2=-5, no solution!

3. Grouping Lu Zeqing formula: original formula = (m 2-2mn + n 2) + (m 2 + 4m + 4) = 0

m-n) 2+(m+2) 2=0, solution: m=n=-2

m^2*n+m*n^2=-8-8=-16

4. Analysis: Because the question contains x, it must be eliminated!

Mystic: Know: x is arbitrary! ] may wish to make x=-400, as can be seen from the title: a=0, b=-1, c=1

Direct early grip substitution:

Original formula = 0 + 1 + 1 - 0-0 - 0 = 2 [If the last term of the algebraic formula is -bc, then the answer is: 3].

5. Solution: Add the three formulas to get it.

a^2+b^2+c^2=2b-2c-6a=-11

(a^-6a+9)+(b^2+2b+1)+(c^2-2c+1)-9-1-1=-11

(a-3) 2+(b+1) 2+(c-1) 2=0 because the square of a number is non-negative!

a=3，b=-1，c=1

a+b+c=3

1) Formula, air sensitive (x 2 + 2x + 1) -1 = 5x + 1) 2 = 6 (x + 1) = positive and negative root number 3 x = positive and negative root number 3-12) formula, (x +4x+4) + (y 2 + 4y + 4) -8 = -13x+2) 2 + (y + 2) 2 = -5

Isn't the topic a bit of a problem with the cherry branches?

3）(m^2-mn+n^2)+(m^2+4m+4)=0m-n)^2+(m+2)^2=0

m=m m=-2

The square of m·n + the square of m·n = (m+n)mn=-4*4=-164)a-b=1

c-b=2c-a=1

The square of a + the square of b + the square of c—ab—ac—ba=(a-b) 2+c 2+ab-ac-bc

a-b)^2+a(b-c)+c(c-b)(a-b)^2+(b-c)(a-c)

5) Add the three formulas together to get.

a^2+b^2+c^2=2b-2c-6a=-11a^-6a+9)+(b^2+2b+1)+(c^2-2c+1)-9-1-1=-11

a-3)^2+(b+1)^2+(c-1)^2=0a=3 b=-1 c=1

a+b+c=3

x*x+2x=5,=5+1,(x+1)(x+1)=6,x+1=positive and negative root number 6, then x=positive and negative root number 6-1.

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For reference, Lu Liquid Bucket, early grinding, please bury the banquet and smile.

Let the brine with a concentration of 20% have x kg, then the salt content kg (x+10) =

x = 30

Suppose y kg of salt is added to change the concentration to 25%.

y = 5+1/3

2.If the apprentice processes x parts per hour, then the master processes 5 3x parts per hour, and there are a total of y parts in this batch.

y-240)/(x+5/3*x)*5/3*x = 3/8*y -40

y=440

You are in such a hurry, then I will calculate the answer directly, it is rare to write the process.

The answer to the first question is 2kg

The second question has 440 answers.

1. Because 1+x+x 2+x 3+......x 1999 = 3, multiplied by x on both sides, there is x+x 2+x 3+......x 1999 + x 2000 = 3x and a-x-x 2-x 3 -......x^2000=a-(x+x^2+x^3+……x^2000)=a-3x=0。

2. Finishing a(1 b+1 c+1 a) + b(1 a+1 b+1 c) + c(1 a+1 b+1 c) = 0

Since a+b+c is not 0, 1 a+1 b+1 c=0.

1.-2a

Process: 1+x+x 2+x 3+......x^1999=3x+x^2+x^3+……x 2000 = 3 x = a process: a(1 b + 1 c + 1 a) + b (1 a + 1 b + 1 c) + c (1 a + 1 b + 1 c) = 0

Because a+b+c is not 0

So the equation is 0