Mathematics problems in the first year of junior high school... There is an addition

1) x=1 3, analysis: first eliminate 1 9, that is, both sides of the equation are multiplied by 9 at the same time, and 9-8 is used to ...... 1(And so on, it can be easily calculated).

2) The known original polynomial is (-4ab+2bc-1)-(2ab-3bc+4), i.e., -6ab+5bc-5

Therefore, the original answer is: (-6ab+5bc-5)-(2ab-3bc+4)=-8ab+8bc-9

3) According to the title, if the purchase does not exceed 200 yuan, the payment will not exceed 200 yuan; If the purchase exceeds $200 but does not exceed $500, the payment will exceed $180 but not exceed $450; If the purchase is more than $500, the payment will be more than $450. The person used 150 yuan and 405 yuan for two purchases respectively, so the person could not get a discount for the first purchase, and the second purchase should enjoy a 10% discount, and the actual value of the items for the two purchases was 150 yuan and 450 yuan respectively.

He saved 45 yuan in this event.

The brackets in the first question are wrong.

2. （2ab-3bc+4)=-4ab+2bc-1

The solution is ()=-6ab+5bc+5, and then -6ab=5bc+5-(2ab-3ab+4)=-8ab+8bc+1

1。Because the distance from the point on the bisector of the angle to both sides of the angle is equal, the height of the adb is equal to the height of the bec, i.e., the ratio of 2:3 to the base.

And because ec=6-ae, i.e., at this time, ad=6-ae is equal to 2:3. Because the time is t, ad is equal to 1t and ae is equal to 2t.

That is, 1t: (6-2t) = 2:3.

The solution is t = 12/7.

2.In the same way, to make adb congruent with bec, we have to make add equal to ec, that is, so that 1t is equal to (6-2t), and the solution is: t=2

3x+2a=x-7

2x=-7-2a

x=(-7-2a)/2=2

So a=3x-11=x+7

2x = 18x = 9 The original solution of the equation is 9

To adopt!!

3x+2a=x-7

Substituting x=2 gives a=-11 2

Substituting a=-11 2 into the original formula, we get x=9

At the beginning, when it is -7, x=2, at this time, the original equation of x=2 generations, that is, 3*2+2a=2-7, is solved to obtain 2a=-11, and then substituted into 3x+2a=x+7 to solve x=9

Bring x=2 into equation 3x+2a=x-7 to get the value of a, and then bring the value of a into equation 3x+2a=x+7 to solve the equation to get x=?

By substituting x=2 into the equation by 3x+2a=x-7, x=2, the solution is a=

Since the equation is actually 3x+2a=x+7

Therefore, substituting a= into the equation gives x=5

1, (1), 20 + x = 2 * (16 + 15-x) x = 14 (2), 6 are drawn, and 9 are left.

20+x=2*(16+9-x) x=10 Not feasible for reason x>9

2,3*(6+x)=16+18-x x=4

1. (1) According to the question, there are x people in the library in the classroom and y people in the laboratory, then there are x+y=15 and 20+x=(16+y)*2, and x=14 y=1 can be solved

2) It is not feasible, the same approach, and finally the solution of y is negative, then the division is not feasible.

2. It is the same as the first question, so I will not explain it in detail here.

Hello non-mainstream S wings!

Solution: Because OE divides the BOD equally

So bod=2 boe

And because aod: boe=4:1

Then aod: bod=2:1

Whereas aod+ bod=180°

So aod=120°, bod=60°

So boe=30°

Because BOC and AOD are opposite each other.

So boc=120°

coe = boc + boe 120° + 30 ° = 150° again because of bisect coe

So eof=1 2 coe=150° 2=75°.

Because aod: boe=4:1

and OE splits the BOD equally

So so and because of so and because of of the deuce coe

So

Because OE bisects the angle DOB, the angle DOE = the angle BOE is the same: the angle COF = the angle EOF

Let the angle boe=x, then the angle doe=x, and the angle aod=4x, so 4x+x+x=180

So x=30

then the angle doe = 30

Then the angle coe = 180-30 = 150

So the angle eof=1 2 angle, eoc=1, 2*150=75

Because oe deuces because of coe so eof 1 2 coe 75

I'll do the second question:

q is a prime number; m, n is an integer, and q=mn.

=>m and n must have one of them as 1

=>p=1+q

=>p=3,q=2

=> = (3 3 + 2 2) (1 2 + 2 1) = 31 3

31/33

2. Is this really a junior high school question- -I'm not sure about the last one.

It's really strong now, and the questions in the first year of junior high school are all tender and difficult

18888888888888, hello:

Because |x|9, so x 9

Because y 4, so y 2

So the coordinates of the point p that meet the conditions are: (9,2),(9,2),(9,2),(9,2).

By |x|=9 gives x=3 or -3

y = 4 gives y=2 or -2

The coordinates of all points are (3,2)(3,-2).

That's pretty much it. The knowledge points involved in this question are: the definition of the coordinates of the point, the meaning of the absolute value, and the solution of the square root.

By |x|=9 to get x=9 or x=-9

y=2 or y=-2 is obtained from y = 4

So there are four p-points that meet the condition:

p（9，2）p（9，-2）p（-9，2）p（-9，-2）

|x|9, so x 9

y 4, so y 2

The coordinates of the point p that meet the conditions are: (9,2), (9, 2), (9,2), (9, 2).

The coordinates on the x*x+y*y=13 circle are satisfied.