A math problem in the third year of junior high school urgent!

On the diagram question. It is known that bad= adc, abc= bac> bad. Therefore, you can take the point E on the AD line segment so that AEC=ABC, since the two angles are facing the same line segment AC, so ABEC four points are concircular (this theorem should be known, right?).

Therefore ACB= AEB> ADB, that is.

It's a very simple question.

Make the circumscribed circle of ABC, cross AD to E, and connect BE

Because aeb= adb + dbe (the outer angles of the triangle are equal to the sum of the non-adjacent inner angles) aeb adb

Because acb = aeb (the circumferential angle of the same arc is equal), then acb adb

Take g as the center of the circle and cg as the radius to make a circle, then the d point is outside the circle, because the circumferential angle is greater than the outer angle of the circle, so acb > adb

From the title, it can be seen that the three points of a, b, and c can determine a circle O, and the straight line L and the circle O are tangent to the point C, connecting any point D and A on L, and Ad must intersect with the circle O at a point N, by the circumferential angle theorem, the angle ACB = the angle Anb, but the angle Anb is the outer angle of the triangle BND, so the angle Anb is greater than the inner angle ADB, so the angle ACB is greater than the angle ADB

It's really difficult, and even though I'm a junior high school student, I can't do it.

Analysis: (1) Use the sum of the inner and outer angles of the polygon.

The sum of the inner angles of the polygon = 180 degrees (n-2), and the sum of the outer angles of the polygon is always 360 degrees.

2) Find the inner angles, then know the outer angles, and then find the number of sides of a regular polygon surrounded by multiple regular pentagons through the outer angles. That is, if you know how many outer corners there are, you will know how many edges there are in the polygon you require, and then you will know how many regular pentagons there are in a circle.

Solution: 1. The sum of the internal angles of the regular pentagon = 180 degrees (5-2) = 540 degrees, and one internal angle = 540 5 = 108 degrees.

2. An inner angle of the inner polygon sought: 360 degrees 108 degrees 2 = 144 degrees An outer angle of the polygon sought: 180 degrees 144 degrees = 36 degrees 3.

360 degrees 36 degrees = 10 (strips).

So a total of 10 regular pentagons are needed, and 7 are needed.

Find the center of the circle (-1..)-4) to the point (3.)-1) is outside the circle if the distance is greater than five.

equal to five on a circle.

Less than five within a circle.

The circular equation (x+1) +y+4) =5 substitutes(3,-1) into the circular equation and the points (3,-1) are on the circle.

Draw a circle on a circle according to the coordinates, and you can judge it by finding the points (3, -1).

Because a+b=-2

ab=-5 assumes a=0 then the equation x(squared)+2x - 5=0 is not true, so the root of x(squared)+2x - 5=0 is equal to 0

So there is a(squared)+2a+ab=a(a+2+b)=0

Solution: a, b are the two real roots of the equation: x +2x - 5=0 a +2a=5

and ab (i.e. x1x2) = c a = (-5) 1 = -5 a +2a + ab = 5-5 = 0

Because ab = -5, a (squared) + 2a + ab = a (squared) + 2a - 5

and x (squared) + 2x - 5 = 0 so a (square) + 2a + ab = a (squared) + 2a - 5 = 0

a 2 (squared) + 2a - 5 = 0 (because a is the root), ab = - 5 (the relationship between the root and the coefficient): so a 2 + 2a + ab = 5-5 = 0

(1) -3x square + 5x + 1

3(x^2-5x/3)+1

3(x 2-5x 3+(5 6) 2-(5 6) 2)+1=-3(x-5 6) 25 6) 2+37 12When x=5 6 -3x square +5x+1 maximum value 37 12(2) Find the minimum value of 2x square -7x+2.

2(x^2-7x/2)+2

2(x^2-7x/2+(7/4)^2-(7/4)^2)+2=2(x-7/4)^2-49/8+2

2(x-7/4)^2-33/8

When x = 7 4, the minimum value of 2x square - 7x + 2 - 33 8

1, -3x square + 5x + 1 = -3 (x 2-5 3) + 1 = -3 (x - 5 6) 2 + 61 36

So: the maximum value of -3x square + 5x+1 is x square - 7x+2=2(x-7 4) 2-33 16 So: The minimum value of 2x square - 7x+2 is: -33 16