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1) Fold the triangle ACD along AD so that C falls on AB and is denoted as E angle B + angle EDB = angle AED (1).

It can be seen by the nature of the fold.

Angle AED = Angle C (2).

Angle c = 2 angle b (3).

Syntagion (1), (2), (3) is available.

Angle b = angle edb

Hence ed=eb

It can be seen by the nature of the fold.

cd=edac=ae

Hence cd=eb

Easily see ab=ae+eb

By an equal amount of substitution there is.

ab=ac+cd

2) From the proof process of 1), it is found that the condition of c=90 has no effect on the validity of the conclusion of ab=ac+cd, so it can still be guessed that the proof process of ab=ac+cd is the same as that of 1).

3) Fold the triangle ACD along the AD so that C falls on the E angle B + the angle ACB = the angle CAF

2 angle b = angle acb

Therefore angle caf = 3 angle b

AD is the angular CAF bisector.

Therefore the angle daf = angle b

Apparently Angle B + Angle ADB = Angle DAF

Therefore the angle adb = angle b

It can be seen by the nature of the fold.

Angular edb = 2 angular adb

Therefore the angle edb = the angle b

So ed=eb

It can also be seen from the nature of the folding.

cd=edac=ae

Hence cd=eb

Obviously eb=ae+ab

There is an equivalent substitution.

cd=ac+ab

What grade is this question? The third year of junior high school?

Don't you see x taking the opposite number?

y=-kx+k

According to your assumptions.

y=x-1

10.The 2 square roots of a number are opposite to each other, i.e. x-6=-(3x-2)=-3x+2 to get x=2

Then the square root of 2 is 2

11.Original = [( 3+2)( 3-2)] to the power of 2013 The answer is -1

10、x-6=-3x+2

4x=8x=2 square root plant 2

11. Original formula = [(factory 3+2)(factory 3-2)] 2013=[-1] 2013=-1

Each valid intersection produces 2 moves, and there are a total of 10 valid intersections (the first vertical line has two, the second vertical line has two, the third vertical line has two, the fourth vertical line has two, the fifth vertical line has two (the bottom intersection does not count: you can't go back), and the sixth vertical line has no (you can't go back),) so there are 2 10=1024 moves.

<>Refine the steps for you, and you will see! Let's re-examine the question yourself and see if the so-called valid intersection is correct.

Since there is none, please post the answer to the final question, or the teacher will give you feedback after explaining, thank you.

5184 species.

The way back doesn't count:

We calculate from left to right, remembering that no matter how you go up and down, as long as you reach a foothold on the right, even if it is a path, there is only one foothold on the far right.

There are five small segments of distance from left to right, then we multiply by 5 times.

In the first paragraph, there are 3 and 2 footholds; 3*2=6 types.

The second section has 3 and 2 footholds; 3*2=6 types.

The third section has 2 and 2 footholds; 2*2=4 types.

The fourth section has 4 and 3 footholds; 4*3=12 species.

There are 3 in the fifth paragraph. 1 foothold (i.e. b); 3*1=3 types.

So it's 6*6*4*12*3=5184 methods.

It is also possible to set the polynomial to t.

（5） （3x-2）²=3x-2

3x-2=0 or 3x-2=1

x=2 3 or x=1

6）（x-2）²=4（x+3）²

x-2=2(x+3) or x-2 =-2(x+3)x=-5 or x=-4 3

7）【（3x+1）+2】²=0

3x+1=-2

x=-18）3x²-7x+3=0

b -4ac=(-7) -4 3 3=13x=(-b root) 2a=(7 root13) 6

The first problem is y1=-y2, solve the equation by yourself.

Question 2: If the sum of two angles is a right angle (90°+180°K, k z), then call these two angles "complementary angles", that is, add 90 degrees to the two angles, 90-a-2 9*(180-a)=1, and solve this number by yourself.

Multiply 24 by each number in parentheses to get (4-2)-4 5, then calculate the numbers in parentheses and subtract them to get the answer 6 5.

(1) The center of the circle is at (,0), the radius is , and the equation for the circle is (

Substituting x=0 gives y 2= y=2

Or use the diameter corresponding to the angle ACB as a right angle, which can be done using the Pythagorean theorem.

Let oc=s ac 2= oa 2+oc 2=1+s 2 bc 2=ob 2+oc 2=16+s 2 ab 2=25

ac ^2+bc^2=ab^2 17+2s^2=25 s=2

The coordinates of point C are (0,2), and the parabola of three points through ABC is found in this way, let it be y=a(x+1)(x-4) and substitute the coordinates of point C to obtain 2=-4a a=-1 2

The parabola is y=(-1 2)(x+1)(x-4)=(-1 2)x 2+(3 2)x+2

2) cd=ad

Let od=d then ad 2=oa 2+od 2=1+d 2 cd=2-d

So 1+d 2=(2-d) 2=4-4d+d 2 gives d=3 4

The coordinates of point d are (0,3,4).

The line where ad is located is y=(3 4)(x+1) substituting the circular equation (

The coordinates of point e ( are solved, and the length of AC is calculated as the root number (1 2 + 2 2) = the root number 5

The length of the CE is the root number [root number 5

So chord ac = chord ce so arc ac = arc ce

3) ae = root number [(root number 16 = 4.)

So om=2 m point is (-2,0).