Hurry, hurry, math problems, detailed processes, good plus points

It is known that the absolute value of ab minus 2 and the absolute value of b minus 1 are opposite to each other.

then ab-2=0, ab=2

b-1=0，b=1

So a=2, b=1

No, b=1, (a+1)(b-1) one, the denominator is meaningless.

1/1 AB plus (A+1) (B-1) 1/1 plus (A+2) (B+2) 1/1 ......A+2008) (B+2008).

Belch... Question, right, how can absolute value and absolute value be inverse??

The absolute values are both non-negative, and the two absolute values are opposite to each other, indicating that they are both 0, so ab-2=0, b-1=0....So b=1 a=2, the minus sign in the last formula is plus, right? That is, 1 (1*2)+1 (2*3)+

1/(2009*2010)..Do it with item elimination.

Equivalent to (1 1-1 2) + (1 2-1 3) + (1 3-1 4)...1 2009-1 2010), all of them are gone, leaving 1-1 2010.

The answer is 2009 2010

Because the absolute value is definitely greater than or equal to 0

Another reason: "the absolute value of ab minus 2 and the absolute value of b minus 1 are opposite to each other" then ab-2=b-1=0

Get b = 1 a = 2

One of the questions (A+1) (B-1) is written incorrectly, and if it should be B+1, then it will be 1 1 2 +1 2 2 3 +1/2009×2010=1-1/2+1/2-1/3+..1/2008-1/2009+1/2009-1/2010=1-1/2010=2009/2010

The fifth term in the sequence is a5=(-1) (5+1) (15-2).

The 100th term in the sequence is a100=(-1) (100+1) (300-2).

The time origin is the time when B departs.

Take a as the distance origin.

Then the relationship between the distance traveled by A and the time is:

sA = 6 * 10 * 60 + 6 * t

Then the relationship between the distance traveled by B and the time is:

s B = 15*t

Problem 1 is: When two people meet, sA = s B.

That is, 6*10*60+6*t=15*t

Solve for t=400 seconds.

Problem 2 is: s = s A - s B = 6 * 10 * 60 + 6 * t - 15 * t = 3600-9 ts = 3600-9 t

Problem 3: Substituting s=900 into the algebraic formula of problem 2 s=3600-9t, that is, 900=3600-9t

Solve for t = 300 seconds.

This answer is absolutely correct)

1. The distance traveled by A is 6*(60+t), and the distance traveled by B is B=15t

When s A = s B, catch up. That is, 6 * (60 + t) = 15t, and the solution is t = 40 after the departure of two B, the distance s s of A and B to get together

When t 40, s = s A-s B = 6*(60 + t)-15t = 360-9t, when t > 40, s = s B - s A = 15 t-6 * (60 + t) = 9 t - 360, when t = 40, s = 0

3. 9t-360=900, then t=140

A walks first, and when B starts walking, A has already walked 6m s*600s=3600m t=distance difference speed difference.

400s at this time B walked TS with v=15m s

s=15tIf you want s to be 900 meters, then 15t=900

So t=60s

6m*10*60=3600m, the distance difference at the beginning (6m*10*60) divided by (15-10)*60 (distance difference divided by speed difference) The first question is solved.

s=t *(15-10)*60 Condition(t 12) s=(t-12)*(15-10)*60 Condition(t 12) The second question is solved.

In the first case, s=t*(15-10)*60 brings s equal to 900 into the second case, s=(t-12)*(15-10)*60, brings s=900 in, and finds t, and the third question is solved.

I'm sorry. I'll only be the first. Mathematical induction proves that I am not good at analytic geometry.

f(x)=lnx+2 x+1 and defines the domain as (0,+.)

f (x) = 1 x -2 (x+1) 2 =x 2+1 x*(x+1) 2 0, h(x) is an increment function on (0,+ when x 1, f(x) f(1)=1

When n=1, ln n+1)=ln2

3ln2=ln8 1, ln2 1 3 , i.e. the proposition is true when n=1

Let n=k, the proposition holds, i.e., ln(k+1) 1 3 +1 5 +....1/（2k+1） ．

When n=k+1, ln(n+1)=ln(k+2)=ln(k+1)+ln(k+2) (k+1) 1 3 +1 5 +....1/（2k+1） +ln（k+2）/（k+1） ．

When x1, lnx+2 (x+1) 1, i.e. lnx (x-1) (x+1).

Let x=(k+2) (k+1) then there is ln(k+2) (k+1) 1 (2k+3) then there is ln(k+2) 1 3 +1 5 +....1 (2k+1) +1 (2k+3), i.e., the proposition is also true when n=k+1

Therefore, the inequality holds,

1) Connect the OD

oa=od∠oda=∠oad=∠ead

de⊥mn∠ade+∠ead=90°

ade+∠oda=∠ode=90°

de is the tangent line.

2)ad=3√5

If o does of ad over f, then f is the midpoint of ad.

af=3√5/2

oad=∠ead

rt△ead∽rt△dao

oa/af=ad/ae=√5

oa=3√5/2*√5=15/2

1)af=2t,be=t

feb∽△foa

bf/af=be/oa

2t-4)/2t=t/8

t=42)00t=2, f at point b, s=s(oeba)=202 in summary, s=t 2+16,0e(8-t,4), f(8,2t), a(8,0)ae 2=t 2+16

af^2=4t^2

ef^2=t^2+(2t-4)^2

If ae=af,t 2+16=4t 2 t=4 3 if ae=ef,t 2+16=t 2+(2t-4) 2 t=4 if af=ef,4t 2=t 2+(2t-4) 2 t=4(2-3).

I didn't even give points, I didn't answer.

1. (50 16-460) 16=km Hour analysis: A chases B after 16 days, and A travels 460 kilometers more than B.

(1 14-1 21)=5292(km (10+6)+80 (10-6)=25 (hours) Analysis: Go with the water when you go, and go against the water when you return. The speed of the water flow should be added to the downstream water, and the speed of the water should be subtracted from the reverse water.

4. Solution: Set the rowing speed to x kilometers per hour. The speed of the water is y kilometers and the hourly speed of the downstream is twice the speed of the countercurrent, then there is x+y=2(x-y) to sail 90 kilometers downstream, and in 9 hours, there is 9(x+y)=90 to solve the system of equations, x= y=

The paddling speed is kilometre-hour, and the water velocity is kilometre-hour.

5. Solution: There are x kilograms of salt, and y kilograms of water need to be added.

x/(30-x)*100%=25%

x = 6, then the original moisture is (30-6) = 24 kg.

To be diluted into 15% brine, there is.

6/(24+y)*100%=15%

y=16 and 16 kg of water need to be added.

I don't quite understand whether to ask how much water to add or how much water to add in total. I replied about how much more water to add, because there were still 24 kilograms of water in the brine. ）

I have something to do, I can't answer the rest of the questions today, if you don't need it urgently, you can ask me, and I will answer it for you tomorrow. If you need it urgently, I will give you a reference for the time being, and see if there are any other friends who can give you all the answers.

1.(50x16-460)/16=

2.Let the distance be x

x (x 21 + x 14) (x 14-x 21) = 126 gives x = 630

The last one. Let it take Y hours for B to catch up with A, and A's velocity is x

20y=xy+5

20y+xy=35

Get x=15 from the fourth to the bottom.

Let the distance be x, and meet after y hours.

x/10+x/15）*y=x

x/15)*y+30=x/2

Get x=300

You marked the question number!! Eldest brother!!

So many topics, classmates, you didn't write your holiday homework, ......

I wiped it, and I saw that my brain was dizzy, what else should I do.

The answer to this question is C

First of all, the total investment in the second quarter of 2005 was 100 million yuan, and in 2004 it was 5,730.8 billion yuan; Total investment in the third quarter of 2005 was compared to that in 2004. (It shouldn't be difficult to calculate this, mainly because of being careful and serious; And it is necessary to understand the meaning of the year-on-year comparison: it is compared with the same period last year; Then there is the comparison of "growth"; Asked what happened in the second quarter.

These are the details).

Secondly, it is necessary to calculate how much more investment in the second quarter and the third quarter than the same period, that is, the second quarter of 05 was more than 04 million yuan, and the third quarter of 05 was more than 04 million yuan.

Then calculate how many percent of the year-on-year growth in the second quarter and the third quarter, the year-on-year growth in the second quarter is divided by equals, and the year-on-year growth in the third quarter is divided by the equals, both quarters are more than 05 years than 04 years, so they should be the extra divided by 04 years of investment).

And then by comparison, it is obvious that the second quarter is more than the third quarter, how many percent more: that is, 3 percentage points higher.

To verify the answer of B, if it is three percent higher, then the reverse deduction of the past two quarters should be: , this value obviously does not mean that it has been calculated.

So the answer is c

It shouldn't be b, it seems to find the percentage of the total in the two quarters separately, and then subtract it, and the percentage should be the percentage.

12+29 13+29 25+29 10=29x(12+13+25+10)= 1740,4,10,10 Four numbers, use four operations to form an equation, so that the result is equal to 24, and the equation is.

3. The sum of the two numbers A and B is 112, the quotient of the number A divided by the number B is 6, and the two numbers of A and B are respectively A = (96) and B = (16).

x 6 + x = 112 x = 96

4. Enclose the wire with a length of 120 cm into a rectangle, so that the length is 12 cm more than the width, length = (36) cm, width = (24) cm.

2(x+12)+2x=120, find x=24

5. As shown in the figure, it is known that the area of rectangular ABD is 24 square centimeters, the area of triangular ABE is 5 square centimeters, the area of triangular AFD is 6 square centimeters, and F is the midpoint of DC, so the area of triangular AEF is ( ) square centimeters.

There is no diagram, I can't answer it, the main thing is that I don't know where point e is.

6. The box contains all kinds of ballpoint pens, of which the red ones occupy, and then put 8 red ballpoint pens in the box, and then the red ballpoint pens account for the total, then there were originally red ballpoint pens ( ).

There are omissions in the question and cannot be answered.

5 is divisible by 72, and the quotient of this number divided by 72 is ( ).

Since it can be rounded, it means that the last digit can be integered by 2, 0, 2, 4, 6, 8 are all possible. Third, think about it.

8. A and B ride bicycles at the same time from opposite sides of the two places, A travels 16 kilometers per hour, B travels 14 kilometers per hour, and the two meet at the midpoint 2 kilometers, then the distance between the two places is ( ) kilometers.

9. If the delegation to a conference is between 150,200 people, if there is one less person for every 4 people in the room, when the house is divided; 1 person per 6 people per room; There are more than 6 people in every 7 people in one room, and there are 167 delegates in total.

4x6x7=168

10. If there is a 56 decimeter long piece of wood, if it takes 30 minutes to sawn into a short timber of 8 decimeters per section, then it takes ( ) minutes to sawn a short timber with a length of 7 decimeters per section.

30 <56 (8-1)>=30 8=minutes segment.

56 7) +1> = minutes.

2. Answer questions.

1. For a project, A will be completed in 16 days and B alone in 12 days; If A does the rest of the work separately and it takes 14 days to complete the whole work, how many days does A do it take first?

No diagram, no answer.

3. A and B depart from two places 30 kilometers apart at the same time; Walking in opposite directions, A walks 3 kilometers per hour and B travels 2 kilometers per hour. There is also a dog that starts at the same time, in the same place and in the same phase with A, running 7 kilometers per hour, and the dog turns back and runs to A after touching B. The dog goes back and forth between the two, and asks how many kilometers did the dog run when A and B met?