If a and B are opposite to each other, C and D are reciprocal to each other, and the absolute value of M is 1, find the value of 2014 (a + b) - CD + 2014m

If a and B are opposite to each other, C and D are reciprocal to each other, and the absolute value of M is 1, find the value of 2014 (a + b) - CD + 2014m

a+b=0,cd=1,m=±1
So 2014 (a + b) - CD + 2014m
=0-1±2014
=-2015 or 2013
If you don't understand, I wish you a happy study!

It is known that a and B are opposite to each other, CD is reciprocal to each other, and the square of X is equal to 4. Try to find the square of X - (a + B + CD) x + (a + b) 2013 + (- CD) 2014
It is known that a and B are opposite to each other, CD is reciprocal to each other, and the square of X is equal to 4
Try to find the square of X - (a + B + CD) x + (a + b) 2013 + (- CD) 2014

∵ a, B are opposite to each other, CD is reciprocal to each other, and the square of X is equal to 4
∴a+b=0 cd=1 x=±2
The original formula = 4 - (0 + 1) x + 0 ^ 2013 + (- 1) ^ 2014
=4-x+0+1
=5-x
When x = 2, the original formula is 3;
When x = - 2, the original formula is 7

If a, B are opposite numbers and C, D are reciprocal, then the value of cubic-3 (- CD) of (a + b) is zero=

3

Let Sn be the sum of the first n terms of the arithmetic sequence an, if S3 / S6 = 1 / 3, then S6 / S12 is equal to

The arithmetic sequence S3, s6-s3, s9-s6 and s12-s9 are also arithmetic sequences
S3/S6=1/3,S6=3S3,S6-S3=2S3
S9-S6=3S3,S9=6S3
S12-S9=4S3,S12=10S3
So S6 / S12 = 3 / 10

When a car runs at a speed of 72km / s, the acceleration obtained after braking is 4m / S2 (square). After 8s, the displacement is?

72km/h=20m/s
The acceleration is 4m / s and 178; after braking for 8s, the speed will decrease by 4 * 8 = 32m / s, while the initial speed is only 20m / s, so the car stops before 8s and the final speed is 0
According to 2aX = vt & # 178; - vo & # 178;
2×4×x=20²
x=50m

Xiaoming and Xiaogang are in the 100 meter race. When Xiaogang runs 90 meters, Xiaoming is 25 meters away from the finish line. When Xiaogang reaches the finish line, how far is Xiaoming from the finish line
How many meters?

Let Xiaoming be x meters away from the end
90：（100-25）=（100-90）：（25-x)
90:75=10:(25-x)
6:5=10:(25-x)
6×25-6x=50
6x=100
x=50/3
Xiaoming is 50 / 3 meters away from the end

If equation 2 (k2-2) x2 + k2y2 + k2-k-6 = 0 represents ellipse, then the value range of K is ()
A. (−∞，−2)∪(2，+∞)B. (−2，−2)∪(2，3)C. （-2，3 ）D. (−2，−2)∪(2，2)∪(2，3)

Equation 2 (k2-2) x2 + k2y2 + k2-k-6 = 0 is transformed into x26 + K − K22 (K2 − 2) + y26 + K − k2k2 = 1. ∵ equation 2 (k2-2) x2 + k2y2 + k2-k-6 = 0 represents ellipse, ∵ 6 + K − K22 (K2 − 2) > 06 + K − K22 (K2 − 2) ≠ 6 + K − k2k26 + K − k2k2 > 0, the solution is - 2 < K − 2, and 2 < K < 3, and K ≠ 2. So D

Given that the corresponding points of rational numbers a and B on the number axis are on both sides of the origin, and | a | = | B |, then the power of (a + b) is 2011=_____

Given that the corresponding points of rational numbers a and B on the number axis are on both sides of the origin, and | a | = | B |, then the power of (a + b) is 2011=__ 0___

The total weight of the two barrels of oil a and B is 350 kg. After a quarter of the oil is poured out of barrel a, a third of the oil is poured out of barrel B. at this time, the remaining oil in barrel a is two-thirds more than that in barrel B. how many kg of the original two barrels of oil are there?

Let B be the original x kg and a 350-x kg
（350－X）×（1－1/4）＝X×（1－1/3）×（1＋2/7）
（350－X）×3/4＝X×2/3×9/7
(350-x) × 3 / 4 = x × 6 / 7 both sides multiply by 28
21×（350－X）＝24X
45X＝7350
X = 490 / 3 B barrel has 490 / 3 kg
350-x = 350-490 / 3 = 560 / 3kg

A difficult problem in the calculation of linear equation with one variable

1.7(2x-1)-3(4x-1)=4(3x+2)-1
2.(5y+1)+ (1-y)= (9y+1)+ (1-3y)
3.[ (- 2)-4 ]=x+2
4.20%+(1-20%)(320-x)=320×40%
5.2(x-2)+2=x+1