Let a = {(x, y) | y = - 4x + 6} B = {(x, y) | y = 3x-8} then the intersection of a and B equals Choose to raise A. {(3, - 1)} B {(4, - 2)} C {(2, - 1)} D {(2, - 2)} hope the process of solving the problem can be written

Let a = {(x, y) | y = - 4x + 6} B = {(x, y) | y = 3x-8} then the intersection of a and B equals Choose to raise A. {(3, - 1)} B {(4, - 2)} C {(2, - 1)} D {(2, - 2)} hope the process of solving the problem can be written

y=-4x+6
y=3x-8
By solving this equation, we can get x = 2, y = - 2, and choose D

If a, B, C are rational numbers, and a

You have said the answer upstairs, but you need to know the method. Otherwise, what should you do when you encounter this kind of problem? Knowing how to do it is a very simple mathematical problem
Mark - 4, - 1,0,2,3 on the number axis. A is on the left side of - 4, C is between - 1 and 0, B is between 2 and 3. It's obvious that there are three segments (don't try to find out where a, B and C are, just know that they are in three segments). Then - A, - B, - 3, - 2, - 1, C can be marked on the number axis (- A is in the symmetrical range of a and origin, that is, the right side of 4, and so on). You can see that from left to right, it's more and more obvious, The whole process is less than 20 seconds, very labor-saving
The method that Ji said is also very good

2X ^ 2 + 3x + 5 is greater than or equal to 0

2(x²+3x/2+9/16)-9/8+5>=0
2(x+3/4)²+31/8>=0
Certainly
So x is any real number

Whose derivative is one in X

That is to ask
∫1/xdx
=lnx +C

Derivation of implicit function XY ^ 2-e ^ (XY) + 2 = 0
The question is how to derive e ^ (XY)?

Derivation on both sides
y^2+x*2y*y'-e^(xy)*(xy)'=0
y^2+2xyy'-e^(xy)*(y+xy')=0
The solution is y '= (Ye ^ (XY) - y ^ 2) / (2x + Xe ^ (XY))

How to do derivative of implicit function y * e ^ XY?

If we take the derivative of X, we have to get the following result
y'*e^xy+y*e^xy*(xy)'
=y'*e^xy+y*e^xy*(y+xy')

X ＾ 2 y ^ 2 = e ^ XY implicit function derivation

X (y ^ 2) - e ^ XY + 2 = 0  (y ^ 2 + 2XY & # 39; y) - e ^ XY (y + XY & # 39;) = 0  (2XY - Xe ^ XY) y & # 39; = (Ye ^ XY - y ^ 2)  (2XY - Xe ^ XY)  (Ye ^ XY - y ^ 2)  / (2XY - Xe ^ XY)

Can logarithmic derivation be used for implicit function derivation?

Yes
such as
y^x = x^y
xlny = ylnx
d(xlny) = d(ylnx)
lnydx + (x/y)dy = lnxdy + (y/x)dx
[lnx - (x/y)]dy = [lny - (y/x)]dx
dy/dx = [lny - (y/x)]/[lnx - (x/y)]

What is the derivative of 1 minus the square of X

1. The derivative of 1-x & # 178; is (1-x & # 178;) '= - 2x
2. (1-x) &# 178; = 1-2x + X & # 178;, then its derivative is (1-2x + X & # 178;) '= 2x-2

What is the derivative of y = x ^ 1 / 2?
Is it equal to 2 √ X / 1? Why?

y'=1/2x^2
Y = x ^ n derivation = NX ^ (n-1)