Give me a few math problems in junior high school and ask me about one math problem in junior high s

Your teacher told you to do it.

The top vertex value of the nth triangle is 3n-2, the lower left is 3n-1, and the lower right is 3n.

Because 2010 = 3 670

So 2010 is at the lower right vertex of the 670th triangle.

Hope it works.

2010 3=670 divisible, so at the bottom right of 670 triangles (multiples of 3).

670 triangles at the lower right vertex.

Make a small circle with a diameter bc (bc=2).

Through B, C respectively make a small circle tangent, Jiaotong University Garden in A, D. Connect to AD.

ad=2、ab=2、cd=2

The area of a square ABCD: 2*2=4

1.In the above problem-solving process, the wrong step is [ ], because there are positive and negative values after the square is opened. Please write the correct answer process.

Solution: Shift the term to get 4(2x - 1) = 25(x+1).

Direct squared yields 2(2x - 1) = 5(x+1).

x = -7 or x= -1 3 2（1）

y² -2y - 399 = 0

y - 1)² 400 = 0

y - 1)² = 400

y - 1 = ±20

y = 21 or y = -19

2）(1/4) x² +3 = 6x

x² +12 = 24x

x² -24x + 12 = 0

x - 12)² 144 + 12 = 0x - 12 = ± 132

x = 12 ± 2√33

1. The mistake is in the second step, you have to add a plus or minus sign after removing the square symbol, that is, 2 (2x - 1) = + 5 (x + 1) 2, (1) (y-1) 2 = 399 + 1 = 400 (y-1) = plus and 20

2）1/4x2 -6x+36=-3+36（1/2x-6）2 =33

x can also be found x2 -24x=-12

x-12）2 =144-12

I'm not going to count it when it's all here.

1.The reason for the mistake in step 2 is that the squares of the two numbers are equal, and the two numbers may be equal or opposite to each other.

2.（1）y² -2y - 399 = 0y²-2y+1=399+1

y+1）²=400

y+1=+20 or -20

y=19 or y=-21

2）1/4x²+3=6x

x²-24x+12=0

x²-24x+12²=-12+12²

x-12）²=132

(1) The second step: After opening the square, there are positive and negative signs, and there are not only positive signs, but also negative signs.

2）y² -2y - 399 =(y-1)²-400=(y-1)²-20² =(y-1+20)(y-1-20)=(y+19)(y-21)=0

So y=-19 or y=21

The left and right sides of the original formula are multiplied by 4 first, that is: x +12 = 24x, and the move direction (the right one moves to the left) is x -24x + 12 = 0

x-12) -12 +12=0 i.e. (x-12) -12*11=0 It should be the next thing you can do Yes, there is a root number in the answer, remember that there is a positive or negative.

This angle bad=angleCDA=120 degrees, right?

If that's it, it's done.

ad//bc，∠bad=∠cda=120º。then CBA= BCD=60

AC divides BCD equally, so BCA=30

So cd=ab=1 2 bc ad=bc-2*ab*cos60=bc-ab=bc-1 2 bc=1 2 bc

Circumference = BC+AB+DC+AD=BC+1 2 BC+1 2 BC+1 2 BC =5 2 BC=10 BC=4 Radius=1 2 BC=2

The shadow area is the S sector AOD -S triangle ado ad=ao=od=1 2 BC=radius.

So S sector aod = 1 6 S circle = 1 6 * 2 * 2 = 2 3

S triangle ado = 1 2 * 2 * 2 * sin60 = root number 3

The area of the shaded part of the figure = 2 3 - root number 3

Handsome guy The question is wrong, please correct it and add a question How is the angle bcd equal to 120°?

There is a problem with this question, if the BCD is less than 90 degrees, how can it be 120 degrees?

[1] AE is perpendicular to BC and DF is perpendicular to BC

Because AC bisects BCD, ADC=120, the angle BCD=ABC=60 degrees, and the angle DCA=ABC=30

Let ab=x, because bc is the diameter, so the angle bac=90 degrees, then bc=2x, and because ab=cd, then cd=ab=x

Because the angle bae = 30 degrees, so be=x 2, cf=x 2, the circumference of the quadrilateral abcd is ab+ad+cd+bc=x+(2x-x 2-x 2)+x+2x=10

The solution is x=2, so bo=co=x=2

2] Because ab=ad=cd=2, the area of the shaded part is equal to the area around ab,cd, so the area of the shaded part in the figure is (semicircle-abcd) 3=[2*2*