Mathematics topics in the first year of junior high school. Urgent, solve. Urgent. The first year of

The questions are very simple (-72)-(37)-(22)-17 (-16)-(12)-24-(-18) Solution: Original=-72+37+22-17 Solution: Original=-16+12-24+18 =(-72+22)+(37-17) =(-16-24)+(12+18) = -50+20 = -40+30 = -30 = -10 23-(-76)-36-(-105) (32)-(27)-(72)-87 Solution:

Original formula = 23 + 76-36 + 105 Solution: Original formula = -32 + 27 + 72-87 = (23 + 105) + (76 + 36) = (72-32) + (27-87) = 128 + (-40) = 40 + (60) = 88 = -20 The question is very simple, how can it not be? Make use of the law of subtraction:

Subtracting a number is equal to adding the opposite of that number! Steps: 1 Convert subtraction to addition 2 Use the commutative and associative properties of addition to make it easy 3 Use the addition law to do it.

Trouble, thanks!

1.Original formula = -72 + 37 + 22-17 = -89 + 59 = -30 2Original formula = -16 + 12-24 + 18 = -40 + 30 = -10 3

Original formula = 23 + 76-36 + 105 = 204-36 = 168 4Original = -32 + 27 + 72-87 = -119 + 99 = -20 I believe you can do it.

Let's merge the same denominator first.

3 and 2/5 + 5 and 3/5 = 3 + 5 + (2 5 + 3 5) = 9-2 and 7-1 and 1/8 = -2-1 - (7 8 + 1 8) = -4-3 and 5/12 + 5 and 5/12 = -3 + 5 + (-5 12 + 5 12) = 2

Finally, add the three numbers together to get 7

Make it into a fake score first, then pass the score, and do the math yourself.

2015 to 44 is a total of 1972 numbers, which is an even number, so the multiplication of even numbers and negative numbers is a positive number.

It can be turned into 2014, 2015x2013, 2014x2012, 2013... 44/45x43/44＝43/2015

1 Let the BCD be x, 2x, 3x x+2x+3x+(180-2x)=360 x=45, and the ABCD be 90, 45, 90, 145 2 respectively, 3x+y+2z=28, multiply the two sides by 4, and then subtract 5x-3y+z=7 to get 7x+7y+7z=105, so x+y+z=35 3 p-2m+3n=0 multiplied by 2 and subtract 2p-3m+4n=0 to get m:n=2 and then substitute to get p:n=1 So p:

m:n=1:2:

1 4 Let x,2x,3x x+2x+3x+(3x-9)=360 x=41 180-(3x-9)=66 5 Let A, B, and C be xyz x 2+y 8+z 6= x 4+y 4+z 2=8 x 4+y 8*5+z 3= x=6 y=8 z=9 So A, B, and C are 6kg, 8kg, and 9kg respectively

Solution: 1: 4000 + (x-1) (1-25) 4000 4000x (1-20).

2: Set up the purchase of x computers.

1) If two malls charge the same amount, then:

6000+(1-25%) 6000(x-1)=(1-20%) 6000x, solution: x=5, When buying 5 computers, the two shopping malls charge the same (2 If you buy more favorably in shopping mall A, then:

6000+(1-25%) 6000(x-1) (1-20%) 6000x, solution: x 5, when the number of computers purchased is greater than 5, the purchase of shopping mall A is more favorable; ）

3) If you go to shopping mall B to buy more discounts, then:

6000+(1-25%) 6000(x-1) (1-20%) 6000x, solution: x 5, when the number of computers purchased is less than 5, the purchase of shopping mall B is more favorable; Hope.