A few math problems, junior high school, I fought until late at night... Pleading for help...

1. Start with a1, -1....It has been looping with the numbers -1.

2004 3=668, Gu a2004=-12, according to the quadratic of (a-1) +|b-2|=0 to get a-1=0 b-2=0 to solve a=1 b=2, that is, to find (1*2) 1/1 + (2*3) 1/1......+2007*2008).

1 per 1 * 2 + 1 per ...... (2 x 3).+2007*2008).

a1＝1a2＝1/(1-a1)＝2

a3＝1/(1－a2)＝－1

a4＝1/(1－a3)＝1/2

a5＝1/(1－a4)＝2

Three numbers 1, 2, 2, -1 appear in cycles.

a1＝a4＝a7＝……

a2＝a5＝a8＝……

a3＝a6＝a9＝……

3 is divisible by 2004, so a2004 a3 12, by the power of (a-1) |b-2|0 gets a 1, b 2 so, 1 (a+k)(b+k) 1 (1+k)(2+k) 1 (1+k) 1 (2+k) 1 (a+k) 1 (b+k).

And because b a+1, so, 1 ab+1 (a+1)(b+1)+1 (a+2)(b+2)+1/(a+2006)(b+2006)

1/a－1/b＋1/(a+1)－1/(b+1)＋1/(a+2)－1/(b+2)＋…1/(a+2006)－1/(b+2006)

1/a－1/(b+2006)

a2=1/(1-1/2)=2

a3=1/(1-2)=-1

a4=1/(1+1)=1/2

a5=1/(1-1/2)=2

The result is a cyclic sequence t=3

2004 3 = 668 divisible.

So a2004=-1

2.(a-1) to the power of +|b-2|=0, resulting in a-1=0, a=1, |b-2|=0，b=21/ab+1/(a+1)(b+1)+1/(a+2)(b+2)+.1/(a+2006)(b+2006)=1/(1x2)+1/(2x3)+1/(3x4)+.1 (2007x2008)

Haha, for the first question, how many do you count to see, a1=1 2, a2=1 (1-1 2)=2, a3=1 (1-2)=-1, a4=1 (1--1)=1 2, see no, the law is out, four numbers are a cycle, so a2004=1 2

For the second question, the power of (a-1) is +|b-2|=0, we can get a=1, b=2, so the equation is 1 2+1 2*3+.1/2007*2008=1/2+1/2-1/3+..1/2006-1/2007+1/2007-1/2008=1-1/2008=2007/2008.

One question is too long and I don't want to read it

Question 2 a=1, b=2, just dissolve

a is the smallest positive integer.

The opposite of a=1b is still itself.

b = 0c is 3 greater than the largest negative integer

c=-1+3=2

2a+3c)*b

01 floor from the load.

-a 3-5 is the opposite of 2.

a/3=5-2=3

a=-9 b=-1/9

Original = -18 + 3 4 = -69/4

-a 3-5 is the opposite of 2, then a = -4

A and b are reciprocal to each other, then b = -1 4

So 2a+3 4ab=

The meaning of the topic is unclear, and there are many solutions.

a=-3 of the solution

So b=3

3*ab/2a+4=27/2

2.According to the title, a=1

b=0c=-1+3=2

2a+3c)*b

2a+3c)*0

1, d2, (1) add 8, add the same number on both sides of the equation at the same time, and the result is still the equation.

2) Multiply by 12, multiply both sides of the equation by the same number at the same time, and the result is still the equation.

3. (1) Subtract 6 from both sides to get 6-1 2x-6=3-6 merge similar terms to get -(1 2)x=-3

Multiply both sides by -2 to get (-2) -1 2)x=-3 -2 coefficients to 1, to get x=6

2) Subtract 4 from both sides to get 9y+4-4=7y-3-4 and merge similar terms to get 9y=7y-7

Subtract 7y from both sides to get 9y-7y=7y-7y-7y, and combine similar terms to get 2y=-7

Multiply both sides by 1 2 at the same time to get (1 2) 2y=(1 2) (7) coefficients to 1, get y=

1 d2 1) plus 8 both sides of the equation plus the same constant, the equation does not change2) multiply by 12 sides of the equation multiply the same (not 0) constant, the equation does not change3 : 6-1 2x=3;

Solution: Subtract 6 on both sides to get 6-1 2x-6=3-6, combine similar terms to get -1 2x=-3, multiply both sides by -2 to get (-1 2) (2)x=(-3) (2), and the coefficient is 1 to get x=6

9y+4=7y-3.

Solution: Subtract 4+7y from both sides to get 9y+4-(4+7y)=7y-3-(4+7y)

Combining similar terms gives 2y=-7, multiplying both sides by 1 2 to get (1 2) (2)y=(1 2) (7), and the coefficient is 1, giving x=-7 2

To solve it in this way) and solve another problem:

Students eat + sleep + learn + play.

Pigs eat + sleep.

Substituting the above formula to get student pig + learn + play both sides minus play to get student-play pig + learn + play - play merge similar items to get student-play pig + learn conclusion students who can't play 1 pig in school!

Let the ** of the two merchants be x, then, after a week of price increase, the price of A is want x (1-10%), and the ** after a week of price increase is [x(1-10%)] 1+20%)

After 2 weeks, B's ** is x(1+10%)

After simplification, A's ** is B's.

Therefore, after two weeks, B's selling price is relatively high.

Let the original product be x

Then after two weeks, the selling price of store A will be x(1-10%)(1+20%)=the selling price of store B will be x(1+10%)=

Because x>0

So <

Therefore, two weeks later, the price of shop B was relatively high.

Solution: Commodity ** is x

Store A: 1 week later**=

After two weeks**=

Store B: After 2 weeks**=

Because of the "so B is higher.

Should you use a one-dimensional equation? The app is relatively large.

a-b>0 a>b

a-b=0 a=b

A-B<0 A set the original unit price of the product x yuan, two weeks later.

The selling price of A is x(1-10) (1+20), that is, yuan.

The selling price of B is x(1+10), that is.

0 so the price of store B is high.

Let ** be 1

A: 1*(1-10%)(1+20%)=

B: 1*(1+10%)=

The selling price of B is high.

Solution: Let the same ** be x, and the price of store A will be two weeks later. 2) x = The selling price of store B is 2 weeks later.

The price of shop B is high.

2. AB exclusion. Root number 6 2 = root number 6 root number 2 * root number 2

Choose C. Root number 8 = 2 times root number 2

3. (1) 3* greater than 2* (2) greater than (3) less than.

Question 1: 5 x-5 y=5 xy

5y-5x)/xy=5/xy

5y-5x=5

y-x=1y=1+x

2 questions are the same as on the 1st floor.

Question 3: x-1 x (1-1 x)=1

x-1/x÷(x-1)/x=1

x-1/(x-1)=1

x²-x-1)/x-1=1

x²-x-1=x-1

x²-2x=0

x1 = 0 (rounded), x2 = 2

Question 4: -2 x-2-x=1

2/x+x+3=0

x²+3x+2=0

x+1)(x+2)=0

x1=-1，x2=-2

Question 5: Original formula = (1+4 a-2) 1-2

1+4/a-4

4 a-36 Question: Original = 3-a 2b b a b a

3-b/2a

Corrected, I'm dazzled, sorry)

2.Question 2: Did you have a problem writing it? How can a divisor be 0?

x=-1x=-2

5.I don't know what to do.

6.I don't know what to do.

You don't know how to add, subtract, multiply and divide?

It's not that hard at all.

- -Playing soy sauce I'm the worst at math.

Let the growth rate be x

Then 200+200(1+x)+200(1+x) 2=950 divide by 200 on both sides

1+1+x+1+2x+x^2=

x^2+3x+3=

Recipe x 2 + 3 x 9 4 + 3 4 =

x+3/2)^2=

x+3/2=2

x=1/2=

That is, a growth rate of 50%.

Let the growth rate be x 2, and the total turnover in March will be 9.5 million to 2 million yuan, which will be 7.5 million yuan 750 = 200x(1+x) + 200x(1+x)x(1+x) solution x=50% The answer is c

Let the growth rate be x, then.

200+200（1+x）+200（1+x）²=9504x²+12x-7=0

2x+7）（2x-1）=0

So x=-7 2 (rounding), x=1 2=50% choose c