The number of common points of equations (2x + y) (x + Y-3) = 0 and (4x + 2Y + 1) (2x-y + 1) = 0 is? Why

The number of common points of equations (2x + y) (x + Y-3) = 0 and (4x + 2Y + 1) (2x-y + 1) = 0 is? Why

(2x + y) (x + Y-3) = 02x + y = 0, x + Y-3 = 0 (4x + 2Y + 1) (2x-y + 1) = 0, 4x + 2Y + 1 = 02x-y + 1 = 02x + y = 0 and 4x + 2Y + 1 = 0 are parallel, there is no intersection point, there is an intersection point with 2x-y + 1 = 0, there is an intersection point with 4x + 2Y + 1 = 0, and there is an intersection point with 2x-y + 1 = 0, so there are three intersections

Vector addition and subtraction: NQ + QP + mn-mp =?
How can I work out two answers?
(1)NQ+QP=PN...+MN=PM.-MP=2PM
(2) NQ + QP = PN mn-mp = PN. The sum is 0
What's wrong with (1)?
Wrong number on it
(1)NQ+QP=NP...+MN=PM....-MP=2PM

If P is a point in rectangle ABCD, PA = 2, Pb = 3, PC = 4, what is the length of PD?

After P, EF ∥ AB was transferred to AD and BC to e and f respectively, and then GH ∥ ad was transferred to AB and CD to G and h respectively
It is easy to prove that peag, pgbf, pfch and PhD are all rectangular
∴AG＝EP＝DH、BG＝PF＝CH、AE＝PG＝BF、DE＝PH＝CF.
In addition, EAG = FBG = EDH = FCH = 90 °
Let Ag = a, BG = B, AE = C, de = D
According to the Pythagorean theorem, there are:
a^2＋c^2＝PA^2＝4、b^2＋c^2＝PB^2＝9、b^2＋d^2＝PC^2＝16、PD^2＝a^2＋d^2.
Add a ^ 2 + C ^ 2 = 4 and B ^ 2 + D ^ 2 = 16 to get: (a ^ 2 + D ^ 2) + (b ^ 2 + C ^ 2) = 20,
Substituting B ^ 2 + C ^ 2 = 9 into (a ^ 2 + D ^ 2) + (b ^ 2 + C ^ 2) = 20, we get: (a ^ 2 + D ^ 2) + 9 = 20,
∴a^2＋d^2＝11,∴PD^2＝a^2＋d^2＝11,∴PD＝√11.

Given a = {1, 2, x2-5x + 9}, B = {3, X2 + ax + a}, if a = {1, 2, 3}, 2 ∈ B, find the value of real number a

∵ a = {1, 2, 3}, 2 ∈ B, ∵ x2-5x + 9 = 3, X2 + ax + a = 2, the solution is: a = - 23 or a = - 74, so the value of real number a is: a = - 23 or a = - 74

There is a point P, AP / Pb = 2 on the line ab of length a. the points a and B at both ends of the line slide on the X and Y axes respectively. The trajectory equation of point P is obtained
Detailed process, please

Let a (x1,0), B (0, Y2), P (x, y)
Easy to know: X1 ^ 2 + Y2 ^ 2 = a ^ 2
Because AP / Pb = 2, X1 = 2x, y2 = 3 * y / 2
Substituting the above equation, the trajectory equation is: 16x ^ 2 + 9y ^ 2 = 4A ^ 2
Happy New Year!

4 times the square of a-4ab + B to factorize the polynomial

Just square it completely
Original formula = (2a-b) square

In the straight parallelepiped abcd-a1b1c1d1, ab = 5, ad = 3, Aa1 = 4, angle DAB = 60 °, then the diagonal AC1 =?
Thank you. I figured it out to be number 43

Can you describe the hexahedron clearly? If the Aa1 vertical plane ABCD, then AC1 is 4 times root 2 (root 32)

1 + 3 = 4 = 2, 1 + 3 + 5 = 9 = 3, 1 + 3 + 5 + 7 = 16 = 4. According to this rule, the value of 1 + 3 + 5 +... + 2005 + 2007
What is the sum of 2.1 + 3 + + 5 + 7 + (2n-1) + (2n + 1)?

1+3+5+...+2005+2007=（2006÷2＋1）²=1004²
2.1+3++5+7+（2n-1）+（2n+1）=(2n÷2＋1)²=(n＋1)²
The rule is that the sum of the preceding numbers is equal to the square of the preceding numbers

Let a = {x | x2 + 4x = 0, X ∈ r} and B = {x | x2 + 2 (a + 1) x + A2-1 = 0}. If B is a subset of a, the range of real number a is obtained

∵ a = {x | x2 + 4x = 0, X ∈ r}, ∵ a = {0, - 4} ∵ B = {x | x2 + 2 (a + 1) x + A2-1 = 0}, and B ⊆ a, so ① when B = Φ, △ = 4 (a + 1) 2-4 (A2-1) ＜ 0, that is, a ＜ - 1, satisfies B ⊆ a; ② when B ≠ Φ, when a = - 1, B = {0}, satisfies B ⊆ a; when a ＞ - 1, x = 0, - 4 is the equation x2 + 2 (a + 1) x +

x4-5X2+6=0
How to calculate that the fourth power of x minus the second power of 5x plus 6 equals 0?

If we take the square of X as a whole, if we set it as m, then the formula becomes the square of M minus 5m plus 6. By factoring, we can get that M is equal to 2 or 3, so the square of X is equal to 2 or 3, so we can get X