Middle school math, which master will help me!

Do more questions, practice makes perfect.

There is a big problem in the compilation of mathematics textbooks in China.

First of all, I am also a junior high school student, but my grades in elementary school were very poor, and I didn't study at all, but after I went to junior high school, my math scores have improved a lot, so I will summarize it for you:

1. Cultivate good self-confidence, and when you don't do a question, you always imply that you can do it.

Second, don't stick to the question when doing the question, just know the idea, cultivate your thinking, and the idea will be greatly improved.

3. As a contemporary, it's not stupid or smart, now the simple questions I see are like teachers can only do in the eyes of others in our class.

Fourth, junior high school geometry, there are many problems in graphics, and primary school focuses on calculation, so it is very simple to learn a little bit of primary school things, but junior high school is different, you need to hone your basic graphics, as long as you understand the basics, (junior high school mathematics is nothing, except for super disgusting questions).

You can put two, three, four into one point.

For one of my things (again a figure no one will do, I will not either, but the number he gives is a single digit, I think about the single digit I can't do it I have lived in vain for so many years, the result...The teacher is also just in my forehead. I was right, but the number was calculated for me by the teacher, (the teacher didn't remember it at the time, before I reminded me.)

There is no bad learning, n only whether you are willing to learn,

First, let's calculate how many integers in 1 6 can be represented as f(x).

For 1<=x<2, [2x]=2, [4x]=4, [6x]=6, hence f(x)=12.

For 2<=x<3, [2x]=4, [4x]=8, [6x]=12, so f(x)=24.

For 3<=x<4, [2x]=6, [4x]=12, [6x]=18, so f(x)=36.

For 4<=x<5, [2x]=8, [4x]=16, [6x]=24, so f(x)=48.

For 5<=x<6, [2x]=10, [4x]=20, [6x]=30, so f(x)=60.

For 6<=x<7, [2x]=12, [4x]=24, [6x]=36, so f(x)=72.

So, in the range of 1 6, there are 6 integers that can be expressed in the form of f(x).

Following Hui Qi, we extended this law to the whole front vertical as a range 1 2004.

Noting that f(x+1 2) f(x)+6, we can see that for any integer n, if n can be expressed as f(x), then n+6, n+12, n+18....It can also be expressed in the form of f(x).

So we just need to calculate how many numbers in 1 6 can be expressed as f(x), and then in the range of 1 2004, every 6 numbers will appear.

In 1 2004, 2004 6 = 334, so in the range of 1 2004, there are 334 integers that can be expressed as the form f(x).

Thus, 1 2004 in 334 integers can be expressed in the form of [2x]+[4x]+[6x].

Hope subject, thanks!

7.(x+y) -2x-2y+1=(x+y-1) =0, so, x+y=1, so (x+y) to the power of nine hundred and ninth = 1

6. The original formula = (a-1) 2+(2b-1) 2+1 1, so there is a minimum value of 1, which is obtained when a=1 and b=1 2. The number of squares is always not less than 0.

5.The original formula = 2(x-y) +3(y-2) +1 1, so there is a minimum value of 1, which is obtained when x=y=2. The number of squares is always not less than 0.

3.(x+2) +x-3) =2x 2-2x+13=13, so 2x(x-1)=0, so x=0 or 1, substituting the desired formula to get 6

I'll only have these few questions, and the rest of you can see if it's wrong, and if you have the answer, tell me okay, I like to do this kind of question.

(1) Is there a mistake in the question? If it's a +1 a.

a-1/a）²=a²+1/a²-2=1

a-1/a=1

2) Isn't it infinity?

3) The question is (x+2). Words.

x+2）-（x-3）]²x+2)²+x-3)²-2*(x+2)*(x-3)

25=13-2*(x+2)*(x-3)

x+2)*(3-x)=6

4) Don't you dare to make a mistake again, b sweat... If 2 is changed to a ...

Let's eat, and if all of this is right, give it a point

Changed: 3If (x+2) +x-3) = 13, then (x+2) (3-x) = 6

x²+4x+4+x²-6x+9=13

2x²-2x+13=13

x²-x=0

x(x-1)=0

x=0 or x=1 substituting 6

5.Find the maximum value of the polynomial 2x -4xy+5y -12y+13. Minimum?

4=2x -4xy+y +4y -12y+13=(2x-y) +4(y-three-two) +4 (2x-y) 0,(y-three-two) 0 When the minimum (2x-y) =0,(y-three-thirds) =0, the original 4 has not been seen yet.

Buddy. A test paper, you really dare to put it.

1. One of the abc numbers is negative and two are positive numbers, so abc is a negative number, |abc|/abc=-1

And bc |ab|*ac/|bc|*ab/|ac=(-1)*(1)=1So(|.)abc|/abc）^2009÷（bc/|ab|*ac/|bc|*ab/|ac|）=1

2. The thickness of the fold x times is meters, and the three answers are obtained according to this:

1) 14 times (2) m (3) 27 times.

b/|b| c/|c|There must be two values in +1 and a -1 in order to =1 in the end, so one of a, b, and c is a negative |abc|is positive abc is negative so|abc|abc = -1 to the power of 2009 or -1 bc |ab|×ac/|bc|×ab|ac|It is possible to go down with bc ac ab |ab*bc×ac| -a^2*b^2*c^2/|a^2*b^2*c^2|(2 for squared) -- 1-1/1=-1

Analysis: Because a |a|+|b|/b+|c|c=1, so two of a, b, and c must be 1 and one of -1, so abc=-1, so |abc|/abc=-1，bc/|ab|*ac/|bc|*ab/|ac|The variable formula is bc*ac*ab |ab*bc*ac|=a^2*b^2*c^2/|a^2*b^2*c^2|, so bc |ab|*ac/|bc|*ab/|ac|=1, so (|abc|/abc）^2009÷（bc/|ab|*ac/|bc|*ab/|ac|）=1/1=-1。

2.(1) 10 times.

2) m. 3) 24 times.

Analysis: Let the thickness of the folded paper be s, and the number of folds is n, it is not difficult to find the law of s=, and then substitute the value to find the answer, pay attention to the conversion of units.

Note: Divide, * is multiplier, 2 is to the power, and n is to the nth power.

1) g is the intersection point of L2 and the X axis, so Y=0 is substituted into L2: Y=2X+8 to get X=-4, so the area of the quadrilateral BCME at point C, and the point M is in **? Can't find it! Are you sure the question is correct? Solution:

100 to the 80th power is 10 160.

It is written as 1 followed by 160 zeros

**There are 80 squares in total, and the natural numbers less than 100 are from 0 to 99 for a total of 100.

Each grid can have 100 ways to put it, so there are 100 80 ways in total.

In this question, does the natural number include 0? I remember that my elementary school didn't seem to include 0, if it didn't include 100 to the 80th power, including 101 to the 80th power, the reason is that there can be 100 kinds of numbers in each square, a total of 10 rows and 8 columns, that is, 80 squares, so it must be 80 power.

There are a total of 80 grids, and the natural number less than 100 is 1-99, a total of 99, so there are 88 power species of 99, 0 is no longer a natural number, it was a few decades ago, and now it has changed, so the upstairs is wrong, and the idea is right.

There are 3 people sitting at a table and 2 people not sitting, Yuhui sitting at a table with 5 people and 4 people not sitting, and 7 people sitting at a table and 6 people not sitting! A table sits 9 and 8 does not sit, then the number of people minus one is just right, a table sits 3, 5, 7, and 9, then all the number of people minus 1 should be a common multiple of 3*5*7*9, that is, the common multiple of 945, notice that 945 itself is divided by 11 and the remainder is -1 [945 -1 (mod11)], let all the people be 945n+1 (n is a positive integer), then the remainder of 945n+1 divided by 11 is -n+1+11m=0 (m is a non-negative integer)[945n+1 1-n (mod11)], so n = 11m + 1, the total number of people in the Zen let the Ega void game is 945n + 1 = 10395m + 946 (m is a non-negative integer), and the minimum number is 946 people.

11 people at a table is just right.

The description is a multiple of 11.

There were 3 people at one table and 2 people didn't sit at the table.

So in addition to 3 and 2

35, 68, 101, 134...

There are 5 people sitting at a table and 4 people not sitting), so in addition to 5, 4 (can be 59,114,169,224... Behind the same with the spring of the tree.

The last discovery was.

Do the math yourself.

It's going to be down. Just use.

A table is counted.

Multiply by 11 in turn. Remainder.

Last + the number of seats on the non-existent.

There will be a string of numbers.

Count each conditional file rollover as a change.

When it's all the same, it's it. ~！y