By the equation: (1): (x-1) &# 178; = 4 (2): (2x-1) &# 179; = - 8

By the equation: (1): (x-1) &# 178; = 4 (2): (2x-1) &# 179; = - 8

（1）
（x-1）²=4
x-1=±2
x-1=2
x1=3
x-1=-2
x2=-1
（2）：（2x-1）³=-8
2x-1=-2
2x=-1
x=-1/2

-How to solve the equation of 2x square + 1 / 2x + 19 / 8 = 0~

The original equation is: 16x ^ 2 - 4x - 19 = 0, △ = 16.77
√△ = 4√77
∴x1 = (1 +√77)/8 ,x2 = (1 - √77)/8

How to solve these two equations, 5x + y = 84, 6x + 3Y = 108, depends on the process

5x+y=84
6x+3y=108
15x+3y=252
6x+3y=108
9x=144
x=16
5*16+y=84
y=84-80
y=4
x=16
y=4

The process of solving the equation is X-2 / 5x = 18 / 25

x-2/5x=18/25
3/5x=18/25
x=18/25÷3/5
x=6/5
If you don't understand, you can chase
If you have any help, please remember to adopt it. Thank you

Given that m (x, y) a (0, - 1 / 2) B (- 1,0) is collinear, the minimum value of 2 ^ x + 4 ^ y is

Vector am = (x, y + 1 / 2)
Vector BM = (x + 1, y)
But M (x, y) a (0, - 1 / 2) B (- 1,0) is collinear
So (x + 1) (y + 1 / 2) = XY
So x / 2 + y + 1 / 2 = 0
x+2y=-1
And 2 ^ x + 4 ^ y = 2 ^ x + 2 ^ (2Y) > = 2 √ (2 ^ x * 2 ^ (2Y)) = 2 √ 2 ^ (x + 2Y) = 2 √ (1 / 2) = √ 2
So the minimum value is √ 2

When Xiao Hua read a decimal point, he misread it to 6300.1
Is the question wrong?

630.01

The two trains of Party A and Party B depart from ab at the same time and meet for the first time at a distance of 75 km from A. after meeting, the two trains continue to move forward and return to their destination
Return immediately, the second meeting is 55 kilometers away from B. how many kilometers is the distance between AB and B?

It was the first time that a met 75 kilometers away from a, and a walked 75 kilometers
At this time, the two cars took a total journey
The second time we met, the two cars took three whole journey
A walked 75x3 = 225km
The second meeting was 55 kilometers away from B
The distance between AB and ab is 225-55 = 170km

Well, the addition and subtraction of rational numbers in junior high school should use a simple method

0.0 do your own thing do it yourself

On the map of scale yes, the distance between a and B is 20 cm. Two trains leave from a and B at the same time. A travels 55 km per hour, B 45 km per hour. How many hours do the two trains meet?

40 × 20 = 800 (km), 800 △ 55 + 45, = 800 △ 100, = 8 (hours); answer: two cars meet in 8 hours

Given that the average of data ABC is 8, what is the average of data a + 1, a + 2, a + 3?
It's a + 1, a + 2, a + 3. It's not a + 1, B + 2, C + 3;

26-b-c