When a is a value, the solution of the equation - A (2-x) = a-8x is x = - 1 / 2 Another equation: 2 (1 + 2Y) - 3 (10-3y) = 6

When a is a value, the solution of the equation - A (2-x) = a-8x is x = - 1 / 2 Another equation: 2 (1 + 2Y) - 3 (10-3y) = 6

① Substituting x = - 1 / 2 into the equation
-a(2+1/2)=a-8×（-1/2）
-3a/2=a+4
-5a=8
a=-8/5
②2(1+2y)-3(10-3y)=6
2+4y-30+9y=6
13y=34
y=34/13

X × 26 + 10 = 119, how to solve this equation,

X×26＋10＝119
26x=119-10
26x=109
x=109/26

[(x + y) square - (X-Y) square] divided by 2XY

Original formula = [(x ^ 2 + 2XY + y ^ 2) - (x ^ 2-2xy + y ^ 2)] / 2XY
=4xy/2xy
=2

How to solve x ^ 3-6x ^ 2 + 11x-6 = 0?
How to solve this cubic equation of one variable?

x-6x+11x-6=0 (x-1)(x-2)(x-3)=0 x1=1 x2=2 x3=3

It is known that the solution set of inequality x + 5 / 2 - 1 > MX + 2 / 2 is X

x＋﹙3／2﹚＞﹙mx＋2﹚／2
﹙2x＋3﹚／2＞﹙mx＋2﹚／2
2x﹢3﹣mx﹣2＞0
﹙2﹣m﹚x﹤﹣1
When m ＜ 2, X ＜ - 1 / (2-m), that is - 1 / 2-m = 1 / 2, M = 4

Use a calculator to find the approximate value of the cube root of the following numbers (accurate 0.01)
(1)1.21 (2)580 000 (3)-978 (4)-73/89

(1)1.21
The cube root is 1.66
(2)580 000
The cube root is 83.40
(3)-978
The cube root is -9.93
(4)-73/89
The cube root is 0.94

How to solve quadratic equation of one variable
For example, the square of X + 2x = 35

X²+2X=35
X²+2X+1=35+1
（X+1）²=36
X+1=±6
X=±6—1
∴X1=6—1=5
X2=—6—1=—7
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On the local boundedness of functions
If f is continuous at a point, then f is bounded in a neighborhood of the point
What does this neighborhood mean? Is it just the neighborhood of the point? Or is it a specific neighborhood? Do you have any requirements for this neighborhood?

A field is a small enough area, no matter how small
It is expressed in mathematical language
If f (x) is continuous at x0, for any small positive real number e (generally expressed by the Greek letter small sigma), f (x) is bounded within (x0-e, x0 + e)
Here is the proof:
Because the definition of function f (x) continuous at x0 is: for any positive real number a, there exists a positive real number e, when | x-x0|

If a = 2, then a + B - AB + B = () a.b.c.1 D.2
If anyone knows the answer, please come back immediately,

Choose B

The quadratic equation of one variable with the opposite number of the two roots of equation 2x2 + 3x = 1 = 0 as the root is

2x²+3x+1=0
Then X1 + x2 = - 3 / 2
x1x2=1/2
Now the two are - x1, - x2
Then (- x1) + (- x2) = - (x1 + x2) = 3 / 2
(-x1)(-x2)=x1x2=1/2
So it's X & # 178; - 3x / 2 + 1 / 2 = 0
That is, 2x & # 178; - 3x + 1 = 0