A few math problems in the second year of junior high school! It has to be fast

The first question upstairs is wrong, a is the biggest.

1 The circumference of the triangle is 12, and the three sides a, b, and c have the following relationship: a=b+1, b=c+1, then the lengths of a, b, and c are [5 4 3] respectively

3 Rotate the regular triangle abc around the point c at a certain angle a b c in a clockwise direction. If the angle BCA = 150 degrees, then the angle ACB = [90 degrees].

5.In the Cartesian coordinate system, the point p(0,0) is the center of the circle, the x-axis intersection coordinates of the circle with 5 are the radius, and the coordinates of the intersection point with the y-axis are [(0,5)(0,-5)].

If the 6 point p(m+03,m+1) is on the x-axis, then the coordinate of the point p is [(2,0)].

If the coordinate of point 7 A is (minus 7,0) and the distance between him and b on the x-axis is 1/2, then the coordinate of point b is [(-3,0) or (-4,0)].

8 Walk 200 meters from a library to the south, and then 100 meters to the west, to the gymnasium, if the coordinate location of the gymnasium in this case is (30, 40), then the location of the library should be [(130, 240)].

9 A certain class held a philatelic exhibition with 24 more stamps than the average of 3 stamps per person, and 26 more stamps than the average of 4 stamps per person. ] students, a total of exhibits [? stamps.

b=4 c=3

The following ones are simpler, so you can do it slowly!

2/ π/4+kπ/2(k=n)

3 30 degrees.

4 y-axis.

9 The title is not clearly written.

Method 1: Connect to the AP

ab=ac, e, f are the midpoints of ac and ab, respectively, ae=af

pe⊥ac,pf⊥ab

aep=∠afp=90°

ap=ap，ae=af,∠aep=∠afp∴△aep≌△afp

PE=PF method 2 connects EF, AB=AC, E, and F are the midpoints of AC and AB, respectively, AF=AE

AEF is an isosceles triangle with AFE= AEF PE AC and PF AB

afb=90°=∠afe+∠pfe,∠aec=90°=∠aef+∠pef

pef=∠pfe

PEF is an isosceles triangle.

pe=pf

Attestation: Connect to the AP

ab=ac, pe, and pf are the perpendicular bisectors of ab and ac, respectively, ae=af

aep=∠afp=90°，ap=ap

aep≌△afp

pe=pf

This problem is too simple, connect AP, use HL to prove that the triangle APE and the triangle APF are congruent, if you have to use "the distance from the point on the perpendicular bisector of the line segment to the two breakpoints of the line segment is equal", then connect PB and PC, then the application theorem has PB PA PC, prove that the triangle APB and the triangle APC congruence (SSS), and then obtain PE PF by the congruent triangle corresponding to the high equality.

(1) Solution: Set the selling price to be X yuan, according to what is known.

x-2500)[4*(2900-x) 50+8]=5000 to obtain x=2750

2) It also doesn't make sense when x=-1.

When x<0, the proposition is false.

3) Equipped with triangles ABC and A'b'c'where ab=a'b' bc=b'c'

The midline cd on ab is equal to the midline c'd' on a'b' because: ab=a'b' bc=b'c' cd=c’d’

So: triangle ACD and A'c'd'Congruent.

So: angle b is equal to angle b'

So: triangles abc and a'b'c'Congruent.

4) Incorrect, when, a, b are less than 0, or a<0, b>0 and a<-b, the conclusion is incorrect.

(1) Solution: Set the selling price to be X yuan, according to what is known.

x-2500)[4*(2900-x) 50+8]=5000 to get x. Let's figure it out for myself.

2) It also doesn't make sense when x=-1.

When x=1 4, the proposition is false.

3) First draw two congruent triangles (because they are originally congruent), because there are two sides equal, and then you have to use the property of the middle line to find the congruence of a small triangle that is divided, so as to obtain an angle between the two sides, you can prove that these two trigrams are congruent.

4) The conclusion is incorrect, and it is not valid when a and b are negative numbers.

(1) Treating x +y as an unknown number can be solved to get x +y = 4 (2) original formula = (a + b) (c + d) = 1997 Because 1997 is a prime number, so a + b = 1 or 1997, c + d = 1997 or 1, then a + b + c + d = 1998

3）2001^3-2x2001^2-1999=2001^2x(2001-2)-1999=1999x(2001^2-1)

So: original = 1999 2002

(1) Let x +y = a a*(a-1) = 12 (a-1 2) = 49 4

a-1/2=±7/2

a=1/2±7/2

a = 4 or -3

So x +y = 4 or x + y = -3

Since x +y 0 so x +y = 4

(1)x²+y²=4

3) Defactoring = 2001 (2001-2)-1999 2001 (2001+1)-2002=1999(2001-1) 2002(2001-1)=1999-2002

The second title is 1998

The first question can be given as an unknown number x +y. The final answer is 3 or 4

1.Let x +y = t, the original formula can be reduced to t -t-12 = 0

The solution yields t=4 or t=-3

The following is thinking about it for a while.

1x-2y

The cube of the b.

1.-a-b

The square — 4y squared.

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