Math problems in the first year of junior high school, kneel and ask for detailed solutions!! Each q

1.Move all the right sides of the equation to the left side of the equation and get the sum of the two squares, and the sum of the squares adds to zero, then both squares are zero, so we get a=c b=d so it's a parallelogram.

2.Place the equation on the right side of the equation then the square of m = a -6m = 24 b = 9 so a = 16 m = -4 b = 9

3.The mode is m+b, the median is n+b, and the mean is x+b5If the number of rooms in the dormitory is x the number of students is y, then the formula is.

4x+19=y

y-6(x-1) is between 0 and 6, but is not equal to 0 and 6, resulting in x=11 y=63

x=12 y=67

x=10 y=59

Question 1. Move the item to move the right to the left.

Then you can come up with a perfectly squared formula.

Wow Ka Ka. I'll do it all.

1. Mathematics in riddles.

a) Number puzzles:

1) The Rising Sun 9

2) Mrs. Mo into 2

3) With white, it becomes black 7

4) People are as big as it, but the sky is not as big as him 1

5) I don't have the surname of Wang, I must have me when I open my mouth 5 "Five" is like "King", five + mouth = I (me).

6) Take off the poor hat and remove the poor root 8

7) One minus one is not zero (each with a number) 3

2) Serial riddles: A hundred years of life is short in the blink of an eye (play a mathematical term) Geometry.

c) Fractional Enigma:

7 2 (playing an idiom) No three, no four.... .3＜7/2＜4

Ku 6 (typed) 30 "mouths" divided by 6 equals five "mouths": I.

x Fen = 4 (play a seasoning name) mustard: [according to the structural analysis of the word, the following is wrong, the "eye" is horizontally buckled: four, combined with the seasoning, combined with the button: mustard] 2, calculated.

43. It is known that the sum of the angle AOC and the angle AOB is 180°, om and on are the bisector of the angle AOC and the angle AOB respectively, and the angle MO=30°, please draw a graph and calculate the degree of the angle AOB.

Let aoc=x, aob=180-x aom=x 2 aoc=(180-x) 2

mon=∠aoc-∠aom=(180-x)/2-x/2=30° x=60 ∠aob=180-x =120

Note: aob- aoc= boc).

4. Simplify, and then evaluate.

Known (a+b-4) 2+|abc-2|=0, find the value of 2[abc+(-3a)]-3(2b-abc).

Answer: (a+b-4) 2+|abc-2|=0 gets:

a+b=4abc=2

2[abc+(-3a)]-3(2b-abc)

2[2-3a]-3(2b-2)

4-6a-6b+6

10-6(a+b)

5. Fill in the blanks. 1. Please write a one-dimensional equation with x=1 2 as the root---3x+

2. On the world map, --- is used to represent cities, and the screen is composed of ---. A small house is used to represent the city.

1） 9（2） 2

ii) Geometry.

c) 7 2 No three, no four.... .Bitter 6 : I x Fen mustard. 2. Calculation.

43. Let aoc=x, aob=180-x aom=x 2 aoc=(180-x) 2

mon=∠aoc-∠aom=(180-x)/2-x/2=30° x=60 ∠aob=180-x =120

Note: aob- aoc= boc).

4. Simplify, and then evaluate.

Known (a+b-4) 2+|abc-2|=0, find the value of 2[abc+(-3a)]-3(2b-abc).

Answer: (a+b-4) 2+|abc-2|=0 gets:

a+b=4abc=2

2[abc+(-3a)]-3(2b-abc)=2[2-3a]-3(2b-2)

4-6a-6b+6

10-6(a+b)

5. Fill in the blanks. A 3x+

1. One: (1) Nine (2) Two (3) Seven (4) One (5) Five (6) Eight (7) Three Two: Geometry.

Three: 1, not three, not four, 2, I 3, mustard.

2. It is 4° or 120°

4. Cause(a+b-4) 2+|abc-2|=0, and (a+b-4) 2 0, |abc-2|≥0

abc=2, a+b=4So 2 [abc + (-3a)] -3 (2b-abc) = 2 (2-3a) -3 (2b-2) = 10-6 (a + b) = -14

5、x-1/2=1

A small house is used to represent the city.

Let the whole process be x

x+20) 2 15+(x-20) 2 10=x The solution yields x=10

Before the plus sign is the time taken for speed 15, after the plus sign is the time taken for speed 10, and the sum is the time of the whole process divided by the average speed after the equal sign is the time taken for the whole process.

The distance between the two ends is calculated using the sum difference formula.

**The time the train is in the tunnel is 18-10=8 seconds.

Then the velocity I am 320 8 = 40 meters per second.

The light of a fixed lamp shines on the train for 10 seconds.

It can be understood that it takes ten seconds for a train to travel a distance of its length.

Then its length is 40x10 = 400 meters.

Let its length be x and its velocity be an equal relation.

320÷（18-10）=x÷10

Analysis: 1. If x kilometers are traveled at a speed of 15 kilometers per hour, then the whole journey is 2x-20 kilometers, and there is (x+x-20) + x-20) 10, which can be solved x=60, so 2x-20=100, that is, the whole journey is 100 kilometers.

2. If the length of this train is x meters, then there is.

320+x) 18=x 10, which gives x=400, i.e. the train is 40 meters long.

1.Let the full distance be y, and the distance that has been traveled is x(1) y-x=x-20

2) y [x 15+(y-x) 10]=substitution from (1) y=2x-20 to (2).

2x-20=

2x-20=

x=60y=100

2.Set the length x and the speed v

1) v=(320+x)/18

2) x=v*10

2) Substitution (1).

v=(320+10*v)/18

8v=320 v=40

x=v*10=400

1.If the whole journey is x kilometers, then the first distance is (x+20) 2, and the last distance is (x-20) 2

x+20) 2 15+(x-20) 2 10]*solution x=100

2.Let the train length x meters (320 + x) 18 * 10 = x solve x = 400

1.Suppose the distance traveled is x km, and the distance from the destination is y km y=x-20

x/15+y/10=(x+y)/

x=60, y=40

2.This train is x meters long.

320+x)/18=x/10

x=400

1.Let the first section take t1 and the second section take t2, then 15t1 = 10t2 + 20

15t1+10t2=

Then find the distance according to the time, the distance is equal to 15t1+10t22Set the train length l and the train speed v

then 18v = 320+l

10v=l

If there is xkm from the destination when the speed is changed to 10km h, the distance traveled is x+20, and the whole journey is 2x+20

The time taken for the previous distance is (x+20) 15, and the time for the next distance is x 10, resulting in a total time of (x+20) 15+x 10

And because of the average speed of the whole process.

The solution is x=40, so the whole journey is 100km

Let the travel time of 15 kilometers per hour be x, and the whole journey is y meters.

y=15x+15x-20, y/

Let the train be x meters long and the train speed y

320+x=18y, 10y=x

I don't know if it's 1991....I'm not so sure. According to the question, you can set the single digit as x and the ten digit as y.

There is a first condition that can list 2x+2 9(10-x) 2, which can be solved that x is less than 6 5According to the second condition, the number in the single digit is the same as the number in the thousand, and the number in the tenth digit is the same as the number in the hundred. That is to say, the number in the single digit and the number in the thousand digit can be set to x, and the number in the ten digit and the number in the hundred digit can be set to y.

Because it is a four-digit number, and according to the fact that x is less than 6 5, it can be seen that x cannot be zero and can only be equal to 1Then according to condition three, x+y=

The second question of this question says that the number in the single digit is the same as the number in the thousand, and the number in the tenth digit is the same as the number in the hundred.

The sum of the single digit and the ten digit is 10

So it can only be 9 and 1, thinking that 8 and 2 do not meet the condition of 1, so this four-digit number is 9191

The answer on the second floor is very good, that is, the answer is wrong, and the answer should be 1991

The digits in the single digit and the digits in the thousands, and the digits in the tens digits and the digits in the hundreds digits are swapped at the same time, and the new four-digit numbers are the same as the original four-digits.

According to the calculation, that formula is set to m. Multiply 2 by that equation, then 2m = 2 + 2 2 + 2 3 + ......2^2015

Then subtract m from the equation 2m, subtract the left from the left, and subtract the right from the right. m=2 2015-1

Let m=1 2....Ten 2 2014 (1) times 2: 2m = 2 ten 2 2 ten ....2 2014 Oct. 2 2015 (2). 2) A (1): m = 2 2015 a 1

Solution: 1 Let A repair x meters every day and B repair y meters every day, then list the following equation according to the meaning of the question:

8x+6y=390

7 (x + y) + 5 = 390 solution to x=30 y = 252If car B has traveled x hours and meets car A, then the equation can be solved

72 + 48) x + 72 * 1 4 = 340 to solve x to get 161 60 hours, and the formation minute is 161 minutes.

A drove a total of 161 + 25 = 186 minutes.

1: Set A to repair x meters every day, and B to repair Y meters every day, according to the title:

8x+6y=390

7（x+y）+5=390

Simultaneous solution of the equation is sufficient.