Mathematics in the first year of junior high school, high scores, fast, fast!!!

1. The equation can be shifted left and right to obtain (6k 1) x = 6k 21, that is, k≠1 6;

2. When the square of 9-(4m+3) takes the maximum value m= 3 4, then the equation is 3 5 = 3x 3, that is, x = 5 3;

3、m/a－m/(a＋6)

The solution process is as follows:

Suppose a doll is produced every day, and a total of m dolls are born, then x=m a day can be completed, and now 6 more dolls are produced every day, and a total of m dolls are born, then y=m (a+6) days can be completed, then, the number of days in advance of producing m dolls is y-x=m a m (a 6)

1. K is not equal to 1 6, and K is equal to 1 6, then the denominator is 0.

4m+3) cannot be taken as the maximum, and when m is negative infinity, the square is also infinity.

3、m/a -m/(a+6)

1.If 6kx-21=6k+x is a one-dimensional equation with respect to x, then k is not equal to (x).

2.When the square of 9-(4m+3) is taken as the maximum. Then the solution of the equation 4m-5=3x-3 is (5 3).

3.A toy factory originally planned to produce A dolls per day, but due to the increase in market demand, it is now producing 6 more dolls per day. Then the number of days in advance for the production of m dolls is (m a m a + 6).

1. Sort it out, calculate (6k 1)x=6k 21, then 6k 1≠0, that is, k≠1 6;

2. At this time, m= 3 4, then the equation is 3 5=3x 3, that is, 3x= 5, x= 5 3;

3、m/a－m/(a＋6)=[6m]/[a(a＋6)]

1.One in six is eliminated if it is equal to x.

2.m = minus three-quarters and the solution is minus five-thirds.

3. （m/a）—m/（6+a）

1.If a≠0 and p are positive integers, then the following is wrong: (c).

Calculations: 2 (x^m-1·y^n+1)^3=-x^(3m-3)·y^(3n+3)

3. (a^3)^2·(-a^2)^3=a^6·(-a^6)=-a^12

4. (2m^2)^3÷(-2m^2)^-1=-8m^6·(-2m^2)=16m^8

5, 2x·x^4+（x^2)^3=3x^5

6. 8·4^n÷2^n-1=2^(n+2)

7.The National Stadium "Bird's Nest" has a construction area of 10,000 square meters, and the 10,000 square meters are expressed in science and technology (rounded to retain 2 significant figures).

9 If x = 2 m + 1 and y = 3 + 4 m, the algebraic expression of x is used to express y

x=2^m+1

y=3+4^m=3+2^2m=3+(2^m)^2=3+[2^(m+1)/2]^2=3+(x/2)^2=3+x^2/2

10。Simplification evaluation: a 3· (-b 3) 2+(-1 2ab 2) 3, where a = 1 4 and b = 4

a^3·(-b^3)^2+(-1/2ab^2)^3

a^3*b^6-1/8 a^3b^6

7/8 a^3b^6

11.The thickness of a sheet of A4 copy paper is meters, and how many of these sheets of paper must be stacked together to make their total thickness 1m (the result is expressed in scientific notation, retaining 2 significant figures).

These questions should not be difficult, are you taking the exam?

The last one is multiplied by 10 to the fifth power.

Yes, yes, but the symbols you hit are incomprehensible.

It's very simple, if the sum of two non-negative numbers is zero, there is only one case, that is, both numbers are zero, so x=4 y=1, so the answer to the question is 2

If the time of the encounter is x hours, then the distance traveled by A is 10x km, and the distance traveled by B is 5x km ab and the distance between the two places is 5x + 10x = 15x km.

10x-5x=20

5x=20x=415x=15 4=60 km.

ab The two places are 60 kilometers apart.

Set the time to t and the distance to s.

s=10t+5t

10t-5t=20

t=4, s=60

A total of x hours were walked when they met.

10x - 5x = 20

x = 4 so a and b are apart: (10+5) * 4 = 60 km.

Let the single digit be x and the ten digits be y.

x+y=11

10y+x+63=10x+y

The solution gives x=9 y=2 and the original number is 29

Let the digits on the original two tens of digits be x, 10x+(11-x)+63=10(11-x)+x, x=2

Original two-digit = 29

2x+5=3x-4

Solve the equation: x=9

The number of apples is 2x+5=2 9+5=23.

Because it is parallel, it is x=30 degrees, 3y=180-30=50 degrees, and y=50. Emptiness.

So the absolute value of 30-50 refers to the chain of 20