Find a solution to a math problem, find a solution to a math problem

Question 1 -99 Bring 1 and 100 up and there are 98 items in the middle, just observe it, start with 2, the sum of each adjacent two items is 1 and -1, so the sum of the middle 98 terms cancels out 0, just count 1-100, and the answer is -99

Question 2 You can add the operator symbols according to the permutation and combination of 12 numbers, which is actually very simple, so try it.

1. 1. 1+2-3-4+5+6-7-8+·· 97+98-99-100=97+98-99-100=-4 from the last 4 terms, and 1+2-3-4=-4 from the first four terms, a total of 25 such regular arrays, so the final answer is -4*25

2. There are many ways to fill in 1-2+3-4+5-6-7+8-9+10-11+12=0.

As long as the negative one is equal to 39, and the positive one is equal to 39, it's fine.

The law is also that the sum of negative numbers is equal to 36 and the sum of positive numbers is also equal to 36 Explain: add these twelve numbers to 1+2+3+4+5+6+7+8+9+10+11+12 (1+13)*6 78

Then a positive number and a negative number should be equal to zero, that is, two identical numbers minus 78, 2, 39

In the first question, every four numbers in a group, and the sum is -4, a total of 25 groups, and the sum is -100

The second question, 1+2+3-4-5-6-7-8-9+10+11+12

1.Suppose that segment slip C has done n, then B has done (1+1 8) n, then A has done (1+1 6) (1+1 8) n=21 16*n, so A has 5 16 more than C.

2.Suppose the son is n years old, then the mother is 10 3*n years old, and the father 7 is 2*n years old. Then 7 2*n-10 3*n=2, the equation can be solved, n=12, so the mother is 40 years old and the father is 42 years old.

3.Suppose the mother gives the shake n yuan, then after the first day, there will be n(1-1 2) yuan, after the second day, there will be n(1-1 2) (1-1 3) yuan, and after the tenth day, there will be n(1-1 2) (1-1 3)...1-1 10) (1-1 11) yuan, so n(1-1 2) (1-1 3) .

1-1 11) = 4, so n = 44

4.I guess the landlord wrote it wrong, 2006x2008x (1 2006x2007+1 2007x2008) right? In this way, it is good to multiply directly, and get 2008 2007 + 2006 2007 = 2

5.I guess the landlord made the same mistake. Above equation = 1 2(2 1x3+2 3x5+2 5x7+2 7x9+......2/49x51)=1/2[(3-1)/1x3+(5-3)/3x5+(7-5)/5x7+(9-7)/7x9+……51-49)/49x51]=25/51

6...You can. Above = 1 2[(3-1) 1x2x3+(4-2)2x3x4+(5-3) 3x4x5+(6-4) 4x5x6+......50-48)/48x49x50]=1/2 [1/1x2 - 1/2x3 +1/2x3 -1/3x4 ……1/48x49-1/49x50]=306/1225

Method 1: 1, change the equation from the straight line to y=0, and get x=1 k, then a(1 k,0),b(0,1 (k+1)), and the area of the triangle is sk=(1 2)(1 k)[1 (k+1)].

s1=(1/2)*(1/1)*(1/2)

s2=(1/2)*(1/2)*(1/3)

s2006=(1/2)*(1/2006)*(1/2007)

Let an=s1+s2+.sn

a1=s1=(1/2)*(1/1)*(1/2)=(1/2)*[1/2]

a2=a1+s2=(1/2)*[1/2+(1/2)*(1/3)]=1/2)*[2/3]

a3=a2+s3=(1/2)*[3/4]

a4=a3+s4=(1/2)*[4/5]

a2006=a2005+s2006=(1/2)*[2006/2007)=1003/2007

s1+s2+..s2006=1003/2007

1) Since y=kx+b passes c(1,0), then y=k(x-1), saob=2.

Let the intersection point of two straight lines be d, sacd=(1 2)saob=1=(1 stool 2)*1*yd

yd=2, which is exactly point b, intersects the y-axis, so b=2, b=-k, so k=-2.

The equation for the nucleated width of the straight line is.

y=kx+b=-2x+2

According to the meaning of the sacd: (saob-sacd)=1:5

sacd=1/3

1/2)*1*yd=1/3

yd=2/3

CAD=45°, then xd=2-2 3=4 3

2/3=k(4/3-1)

k=2 y=2(x-1)=2x-2

b = -2 method two:

1、s1=(1/2)*(1/2)*1

s2=(1/2)*(1/2)*(1/3)

s2006=(1/2)*(1/2006)*(1/2007)

s1+s2+……s2006=(1/2)(1-1/2007)

2. y=kx+b (k is not equal to 0) passes through the point c(1,0).

y=k(x-1)

a(2,0)b(0,2)

saob=2

The intersection of two straight lines is d

sacd=1=（1/2）*1*yd

yd=2=b

k=-2 y=kx+b=-2x+2

2) If the area ratio of the division is 1:5, find k,b

sacd=1 3.

yd=2/3

x=4/32/3=k(4/3-1)

k=2 y=2(x-1)=2x-2

b=-2

Obtained by the formula 1, x=7-y

Bringing it into equation 2, 900 (7-y) = 1200y, that is, 6300-900y = 1200y

6300=2100y

y=3 brings y=3 into the 1 formula, and gets, x=4

The solution of the system of equations is x=4, y=3

This is a two-dimensional one-time equation, and there are generally two solutions: substitution and elimination, addition and subtraction.

This question is relatively simple with the substitution elimination method.

There is also a kind of old punching method called the exchange method.

1) b (2)x-y=5,(10x+y)-(x+y)8=5 ,x=6,y=1, i.e. this two-digit number is 61

3) (800*, discounted.)

4) If the high service life is x hours, the 10-watt energy-saving lamp will spend 10*x*, and the 40-watt white woven lamp will spend 40*x*; When x = 2 hours, then both costs are the same, when x > 2, then the 10 watt energy-saving lamp costs less than the 40 watt white woven lamp, the life of the two lamps under normal filial piety conditions are more than 2 hours, all buy 10 watt energy-saving lamps excitedly.

4 Yes (0,0)(2,1)(0,1); 2,1)(2,2)(Cha Zheng 0,1) ; 1,0)(1,2)(Xuhuai Song 0,2);

1,0)(2,0)(1,2) or Two diagonal lines are used as axes of symmetry respectively.

Let y=k1 return to x-k2x2

If x=-1, y=5 and x=1, and y=1 are missing, then k1=-2k2=-3 is obtained by the above formula.

Get: y=-2 x+3x2

Method 1: 1, change the equation from the straight line to y=0, and get x=1 k, then a(1 k,0),b(0,1 (k+1)), and the area of the triangle is sk=(1 2)(1 k)[1 (k+1)].

s1=(1/2)*(1/1)*(1/2)

s2=(1/2)*(1/2)*(1/3)

s2006=(1/2)*(1/2006)*(1/2007)

Let an=s1+s2+.sn

a1=s1=(1/2)*(1/1)*(1/2)=(1/2)*[1/2]

a2=a1+s2=(1/2)*[1/2+(1/2)*(1/3)]=1/2)*[2/3]

a3=a2+s3=(1/2)*[3/4]

a4=a3+s4=(1/2)*[4/5]

a2006=a2005+s2006=(1/2)*[2006/2007)=1003/2007

s1+s2+..s2006=1003/2007

1) Since y=kx+b passes c(1,0), then y=k(x-1), saob=2.

Let the intersection point of two straight lines be d, sacd=(1 2)saob=1=(1 stool 2)*1*yd

yd=2, which is exactly point b, intersects the y-axis, so b=2, b=-k, so k=-2.

The equation for the nucleated width of the straight line is.

y=kx+b=-2x+2

According to the meaning of the sacd: (saob-sacd)=1:5

sacd=1/3

1/2)*1*yd=1/3

yd=2/3

CAD=45°, then xd=2-2 3=4 3

2/3=k(4/3-1)

k=2 y=2(x-1)=2x-2

b = -2 method two:

1、s1=(1/2)*(1/2)*1

s2=(1/2)*(1/2)*(1/3)

s2006=(1/2)*(1/2006)*(1/2007)

s1+s2+……s2006=(1/2)(1-1/2007)

2. y=kx+b (k is not equal to 0) passes through the point c(1,0).

y=k(x-1)

a(2,0)b(0,2)

saob=2

The intersection of two straight lines is d

sacd=1=（1/2）*1*yd

yd=2=b

k=-2 y=kx+b=-2x+2

2) If the area ratio of the division is 1:5, find k,b

sacd=1 3.

yd=2/3

x=4/32/3=k(4/3-1)

k=2 y=2(x-1)=2x-2

b=-2

Because the height is missing 3, then the wisdom triangle is divided into two small right triangles, and according to the Pythagorean theorem, the hypotenuse of the former search state can be found to be the root number 3*(1+3), so the area is 6 root number 3

1. Do the handwheel 48 35 (let A be 1, then B is 5 Tubi 6, C is 35 48) pure faith.

2. Let the son be x years old, the mother 10 3, and the father 7 2, (7 2) x-(10 3)=2 x=12 The son is 12 years old.

Quote from the 3rd floor: because:

b I'm really wondering how the hell am=an was launched? I can give many counterexamples, the limit case, m is the midpoint of bc, point n coincides with point c, and a=b>c

Use the method of rotation.

Rotate the ABM 90° clockwise to the ACM'

There is a basis for this, first of all, ab=ac, so ab and ac can coincide.

Because man=45°, bam+ can=45° due to rotation, so cam'=∠bam

synthetic, so angular nam'=45°

It's spinning again, am=am'

Add an=an

So amn am'n

So mn=m'n

Again, since it is a rotation, b= acm'=45° bm=cm'

So ncm'=90°

In RT NCM'Medium, cn +cm'²=m'n So cn squared plus bm squared = mn squared.

i.e. b 2 = a 2 + c 2

(a-1) +b-1) +c =2c-1(a-1) +b-1) +c-1) =0 (the sum of greater than or equal to 0 is 0, and all of them are 0).

a-1=0,b-1=0,c-1=0

a=1b=1c=1

Listen to lectures, do basic questions, practice more, and after practice, it is important to summarize and think.

And if you can't do the questions you have done, or if you do it wrong, take a good look at it, do it again, and try not to make a mistake when you do it again next time.

During the exam, we must ensure that we are 99% sure that we can do what we can do, so I don't think I will fail in the exam.