Ask for math questions and answers for the first grade of junior high school!!!! 5

Someone goes to a tile store to buy a regular polygonal tile with a dense floor, he can't possibly buy ( ) aRegular triangle bRegular quadrilateral c

Regular hexagon dRegular octagon 7, the following statement is correct ( ) aIf the two lines are truncated by the third line, the same side inner angle complements b

The middle line of the triangle is the straight line c. that passes past the vertex and bisects the opposite sideTwo straight lines parallel to the same line dThere is only one and only one straight line parallel to the known line at a point on the plane 8, and the distance from the point p(5, 7) to the x-axis is ( )c

5 d.7 9. The triangle suitable for a= b= c is ( )aAcute triangle b

Right triangle cObtuse triangle d10. The lengths of the two sides of an isosceles triangle are 3 and 7 respectively, so its circumference is ( ) or 17 d

None of the above is true.

Big brother, you have to pass on the topic. (I don't know which version you are) but I still advise you to do your own thing!

Since your teacher has given you this homework, you should complete it carefully!

If you can't, you can ask, but you can't be opportunistic and fool the teacher!

Untie; 1.|2x+3|=5

Categorical thinking: If 2x+3 is a positive number, then: 2x+3=52x=2, x=1. If 2x+3 is negative, then: -(2x 3)=5,-2x-3=5,2x=-8,x=-4.".

2.|x-|2x+1||=3

When x>=-1 2 2 2x+1>=0

Originalized to |-x-1|=3

Because x>=-1 2 so -x-1<0

Primitives to -x-1=-3

The solution is x=2

When x>=-1 2 2 2x+1>=0

Originalized to |3x+1|=3

Because x<-1 2 so 3x+1<0

Primitives to 3x+1=-3

The solution yields x=-4 3

5.If the equation for x ||x-2|-1|=a has three integer solutions, then the value of a is - from the known conditions that a>=0

x-2|-1=±a

x-2|=±a+1

So |x-2|=1+a or |x-2|=1-ax-2=1+a,-1-a,1-a,a-1

If 1+a=-1-a, a=-2<0, round off.

If 1+a=1-a, a=0, then -1-a=a-1, there are only two solutions, and they are rounded.

If 1+a=a-1, 1=-1, round off.

In the same way, -1-a = 1-a is not true.

If -1-a=a-1, then 1+a=1-a, be discarded.

If 1-a=a-1, a=1

At this time, x-2=2,-2,0, which is in line with the title.

So a=1 I'm sorry I didn't finish it for you, I've done my best, because I'm also in the first year of junior high school, my.

High-tech No. 1 Middle School.

1.Solve equations|2x+3|=5 (Absolute.)

2x+3=±5

2x=±5-3

x=(±5-3)÷2

x1=1 x2=-4

2.Solve equations|x-|2x+1||=3

x-|2x+1|= 3

x=±3+|2x+1|

x=±3±（2x+1）

x±（2x+1）=±3

x+2x+1=3

3x = 2x = two-thirds.

x-（2x+1）=3

x-2x-1-3=0

x=-4③x+2x+1=-3

3x=-4x=-four-thirds.

x-（2x+1）=-3

x-2x-1+3=0

x=2, so x1=two-thirds.

x2 = -4x3 = - four-thirds.

What is the value of x4=2, is the equation (x 3-1) m=1-2 3m m=1 (3x-3)???

1.solution, go absolutely worth: 2x+3=5 shifts to solve 2x+3=5 and 2x+3=-5 x=1 or -4

2.Solution, it's definitely worth it to go :x-|2x+1|=±3

Category Discussion: When x-|2x+1|=3, -|2x+1|=3-x ∵-2x+1|0 3-x 0 x 3 is definitely worth it: -2x-1=3-x x=-4 contradicts x3 as previously deduced.

So the solution doesn't exist.

When x-|2x+1|=3, -|2x+1|=-3-x ∵-2x+1|0 -3-x 0 x -3 is definitely worth it::-2x-1=-3-x gives x=2 in line with the previous inference of x -3, so x=2

3.|x+4|-|x-2|=x+1

Category discussion: When x -4, it is definitely worth it: -x-4+x-2=x+1 gives x=-7

When -4 x 0, it's definitely worth going: 4+x+x-2=x+1 gives x=-1

When 0 x 2, it is definitely worth going: 4+x-2+x=x+1 gives x=-1 (which does not fit the previous inference, and is rounded).

When x 2, it's definitely worth going: 4+x-x+2=x+1 gives x=5

That's all for today, it's going to bed late, add some points, and I'll do the remaining two questions for you tomorrow.

∴x=1 x=-4

2. x-|2x+1|=±3

i.e. |2x+1|=x+3 or |2x+1|=x-32x+1=x+3,2x+1=-x-3 or 2x+1=x-3,2x+1=3-x

i.e. x=2, x=-4 3 or x=-4, x=2 3 because the absolute value must be 0

So x+3 0 or x-3 0

i.e. x 3 or x 3

So just take x=2 x=-4 3

3.When x 2 is x+4-x+2=x+1 x=5, when -4 x 2 is x+4-2+x=x+1 x=-1, when x -4 i.e. -x-4-2+x=x+1 x=-7 is troublesome, so let's stop it, I hope it's useful.

33km, if the original road length is x meters, then it will take (x 8) hours to go, and the length of the road back is 3+x, that is the time, which is (3+x) 9, because there is an extra 1 8 hours, so it is (x 8) = (3+x) 9+1 8

Suppose that the original road is s kilometers long. Then there are:

s+3)/9 -s/8=1/8

Doing the math gives us s=15

The answer is 15. Let the original road length x, then the road length when coming back is (x+3) Then the time taken by the student from the starting point to the destination is x 8 The time taken when coming back is (x+3) 9 And because it takes 1 8 hours longer to come back than to go, there is an equation x 8+1 8=(x+3) 9 The solution x is equal to 15

Let the length of the original road x, then the time x 8, and the other road is (x+3), and the time is (x+3) 9, then it can be obtained; (x+3) 9-x 8=1 8, and the solution is x=15km,

Let ab=c,ac=b,bc=a,ad=h, to prove that b+c has a cosine theorem, we can get cosa=(b +c -a) 2bc, because a is 90°, so cosa 0, there is b +c a .2)

1)+(2), B +C +2CB A +2Ah A +2Ah+H ,(B+C).

I don't like the problem of no picture the most, alas!

The first step is to buy a fountain pen A, a fountain pen B, and a C fountain pen C, to obtain two equations (1) 4A+5B+6C=60

2) 3a+4b+5c=48

Subtract the two formulas to obtain a+b+c=12, and the total number of pens purchased is 12.

The second step is to set up the purchase of a fountain pen n, and the remaining money after the purchase of a fountain pen should be between 4 yuan and 5 yuan on average. Inequalities are available.

4≤48-3n/12-n≤5

Multiply 12-n on both sides (12-n is greater than zero).

48-4n, 48-3n, 60-5n

Subtract 48-4n from the same amount

0 n 12-n

Add n again to get 2n 12

i.e. n 62100 Let the speed and length of the train be v, s800+s v=45, respectively

Answer: a, c

A 3 yuan, 4 yuan, 5 yuan.

If A is 11 branches, 33 yuan, B 1 branch, 4 yuan, C 2 branches, 10 yuan, a total of 47 yuan, if A is 7 branches, 21 yuan, B 4 branches, 16 yuan, C 2 branches, 10 yuan, a total of 47 yuan, let the length of the train be x

x/[(45-35)/2]=800/[(45-35)/2+35]x/[10/2]=800/[5+35]

x/5=800/40

x/5=20

x=100 m

The first question, as long as you calculate the maximum and minimum possibility, first of all, originally to 60, each drop one yuan, the last only to spend 47, save 13 yuan, if a pen to buy the most, other pens each, then up to a pen to buy 12, if you buy 11, then is the original price of a total of 44 yuan + 2 B a total of 10 yuan + 1 C 6 yuan = a total of 60 yuan, cheap one yuan is 11 + 2 + 1 = 13 yuan.

So choose A11.

The second question, I won't do it, thank you.