2 math problems for the first year of junior high school! 10 points reward, good answer to chase poi

1。Solution: a 0 丨a丨= -a -丨a丨= ab 0 丨b丨= -b

c 0 丨c丨= c

Original = a-b-c

2.Except|1-1/2|is greater than 0 and each item in the following absolute value is less than 0, and the original formula can be reduced to:

1 minus 1/2 plus 1/2 minus 1/3 plus 1/3 minus 1/4 +....Add 1/999 minus 1/1000

So the middle term can be completely canceled out, and the remaining 1 minus 1/1000 is equal to 999/1000, which is 999 1000

The middle one is canceled out, hehe.

1) a<0 b<0 c>0

a|+|b|-|c|=a-b-c

2) From the second absolute value to the last absolute value, the result is a negative value, remove the absolute value and add a negative sign in front of it1-1/2|+|1/3 - 1/2|+|1/4 - 1/3 |+…1 in 1000 - 1 in 999|= 1-1/2 - 1/3 + 1/2 - 1/4 + 1/3 +....-1 in 1000 + 1 in 999 = 1 in 1 - 1 in 1000 = 999 in 1000

Question 1.

Let the numbers be z

10+x=?

100x+200+10y+z=？

When you tell her to count. Subtract 200. The hundredth digit is x, the tens digit is y, and the single digit is z.

ax+b）*(x+2)=x^2-4

ax^2+2ax+bx+2b=x^2-4

then a=1 2a=-b b=-2

then the desired is 1

What is the first unit? Got this problem?

1. You can look at it the other way around. The single digit of the final result must be the third number, and the second number above the tenth. Remove the number of 10 and divide the rest by 5, subtract 4, and divide by 2 to get the first number.

2、（ax+b）*（x+2）=ax^2 +（2a+b）x +2b=x^2-4

So b=-2 a=1 gives a b=1

The first and three numbers are a, b, and c

a*2+4)*5+b]*10+c=(a+2)*100+b*10+c

Then the three numbers a, b, and c can be seen at a glance.

2. (ax+b)*(x+2)=x 2-4

a=1 b=-2

a^b=1

1. Reverse thinking; Because step 5 is multiplied by 10, we know that the third number must be the single digit of the final number; Subtract the third number from the final number and divide it by 10, then the remaining single digit is the second number (because the number obtained in steps 1 to 2 is an even number multiplied by 5 and then more than ten digits), subtract the second number from this number and divide it by 5, and then subtract 4 by 2 to get the first number.

2, ax 2+(2a+b)x+2b=x 2-4 is true, then a=1; 2b=14 (i.e., b=-2); 2a+b=0 constant establishment; Get a=1, b=-2

1.Assuming that these three numbers are a, b, and c, then the question is derived:

a*2+4）*5+b]*10+c

10a+20+b]*10+c

100a+200+10b+c

100a+10b+c+200

Mr. Wang subtracts 200 from your result, and the hundred, ten, and single digits of a three-digit number you get are the three numbers you want.

1. The cost in transportation is that it costs M yuan for every 1km of each ton of cargo, so it costs ABM yuan to run AKM cargo BT.

Total cost = ABM + 25 2 = 3650 yuan.

2. The value of the quadratic polynomial 5x2+mx-nx2+6x-3x2-9x+8 of x has nothing to do with the value of x, so the coefficient of x is 0, 5x2+mx-nx2+6x-3x2-9x+8 is sorted out: (2-n) x 2+(m-3)x+8=0, n=2, m=3

The value of mn(m+n)3+100(m2+n2)-6n3+mn is 1348

3. The value of the polynomial 9x7 [to the seventh power]-8x5 [to the fifth power]+7x3-6x+m is 2, x is all odd power, and the odd power of the number that is opposite to each other is also opposite.

At x=2008, the value of the polynomial 9x7 [to the seventh power]-8x5 [to the fifth power]+7x3-6x+m is 2, i.e., 9x7 [to the seventh power]-8x5 [to the fifth power]+7x3-6x=2-m

So when x=-2008, the polynomial 9x7 [to the seventh power]-8x5 [to the fifth power]+7x3-6x=m-2

That is, when x = -2008, the value of the polynomial 9x7 [to the seventh power]-8x5 [to the fifth power]+7x3-6x-m is 2

The answer to 1 (1) is a*b*m.

2) The answer is: 25 * 2 + 300 * 12 * 1 = 36502 I can't understand the formula you write, x and * are not easy to distinguish, and the positions of numbers and letters are not regular.

3. If you take the value of x, you subtract both equations and cancel them out, and you get 24094

Question 1. The denominator in each parenthesis is proposed first, and the formula becomes.

Then the numbers in parentheses are sub-changed by the Gaussian summation formula.

Simplify it even further.

It can be found that the first number and denominator of the numerator are the same, and the formula is reduced to .

Put the denominator 2 out, and the formula is.

Use the Gaussian summation formula again, and get.

The result is 9 2

Question 2. Here is the idea of using Gaussian summation, ie.

A column of numbers in order, the sum of the first number and the last number = the sum of the second number and the penultimate second number = ....= the sum of the nth number and the nth number from the reciprocal.

So the answer to this question is:

I tried to explain it clearly, but the steps to solve the problem will probably be difficult to understand, if there is still something unclear, you can continue to ask me.

1.Replace 1 2 + (1 3 + 2 3) + (1 4 + 2 4 + 3 4) + ....1/10+2/10+…+9 10) into nine groups.

The sum of the 12 natural numbers 12 to 23 is (12+23)+(13+22)+....17+18）=6×35=210

Divided into six groups, the sum of each group is exactly 35

1. Gaussian's method is: (first term + last term) Number of terms 2 (Note: It can only be used for the addition of several consecutive natural numbers, as if it is a natural number).

For example, 1+2+ 3+4+5+6+7+8+9=(1+9) 9 2=45

The number of terms is 1 to 9, and there are 9 numbers in total.

So 1 2 + (1 3 + 2 3) + (1 4 + 2 4 + 3 4) + ....1/10+2/10+…+9/10）

Ask for it separately.

…Note: The formula described above can be used below).

To sum up, so 1 2 + (1 3 + 2 3) + (1 4 + 2 4 + 3 4) + ....1/10+2/10+…+9 10) = The natural number I mentioned above doesn't seem to be correct, because here several successive and regularly increasing numbers can be added together, and you can also use the "Gaussian method")

Second, if the numbers 1 to 8 are the same as the landlord's question, write them that way.

It is clearly visible that the front and back numbers, and the middle two numbers ......

So ditto: 12 + 23 = 13 + 22 = 14 + 21 = 15 + 20 = 16 + 19 = 17 + 18

,2,1 ,Complement 4No, slightly 5Feasible 6The first four are triangular squares and six.