If the square of a + A + 1 = 6, then (2-A) (3 + a) - 2=______

If the square of a + A + 1 = 6, then (2-A) (3 + a) - 2=______

(2-a)(3+a)-2
=-a²-a+6-2
=-a²-a+4
=-(a²+a+1)+5
=-6+5
=-1

The square of × (3 × - 2) - 6 × is 0

3x'2-2x-6x'2=0
-3x'2-2x=0
x（-3x-2）=0
x1=0,x2=-2\3

Draw a geometric figure so that it can represent (M + n) (M + 3n) = m ^ 2 + 4Mn + 3N ^ 2

The formulas on the left and right sides are the same thing. If you simplify everything, you can't solve it

If the tangent of the circle C: x + y = 5 is made through the point m (2,1), then the tangent equation is

It's very simple. First point m (2,1) is right on the circle, right
Then the center of the circle is at the origin o (0,0), so the tangent of point m should be perpendicular to OM
Om slope is 1 / 2, so tangent slope is - 2, so tangent is y = - 2x + 5

Given x, y ∈ R, M = x2 + Y2 + 1, n = x + y + XY, then the relation between M and N is ()
A. M ≥ NB. M ≤ NC. M = nd

M-N = x2 + Y2 + 1 - (x + y + XY) = 12 [(x2 + y2-2xy) + (x2-2x + 1) + (y2-2y + 1)] = 12 [(X-Y) 2 + (x-1) 2 + (Y-1) 2] ≥ 0

In the pyramid p-abcd, PD is perpendicular to the plane ABCD, PD = DC = BC = 1, ab = 2, AB is parallel to DC, and the angle BCD is 90 degrees

PD is perpendicular to ABCD; BC belongs to ABCD = > BC is perpendicular to PD
BCD = 90 = > BC perpendicular to DC
BC is perpendicular to DC; BC is perpendicular to PD; PD intersects DC with d = > BC is perpendicular to PDC
PC belongs to PDC = > PC is perpendicular to BC

It is known that for any angle θ, y = sin θ ^ 2-2msin θ - 2m-1 is always less than 0. Try to find the value range of real number M

The original formula = sin θ ^ 2-2msin θ + m ^ 2-m ^ 2-2m-1 = (sin θ - M) ^ 2 - (M + 1) ^ 2 = (sin θ - M + m + 1) (sin θ - m-m-1) because the absolute value of sin θ is less than or equal to 1, so (sin θ - M + m + 1) is greater than or equal to 0, that is, sin θ - M-1 = sin θ - 2m-11-sin θ, sin θ belongs to [- 1,1], 1-sin θ belongs to [0,2

Plane equation of passing point m (1,0,0) and straight line: X-1 / 2 = y + / 3 = Z

The equation passing through point m and X-1 / 2 = y + 3 = Z must pass through the point x = 1
That is to say, it must have passed the x-axis,
That is to say, the over equation of x-axis and y-axis or the over equation of x-axis and z-axis are plane equations
Therefore, the plane equation is X-2 / 1 = Z or X-2 / 1 = y + 3

N is a positive integer. If the power-1 of 2 is prime, it is proved that n must be prime

As shown in the figure

[question] in the cube abcd-a'b'c'd ', P is the midpoint of a'B', and Q is the midpoint of b'c '. Are AP and CQ straight lines out of plane
Such as the title
In the cube abcd-a'b'c'd ', P is the midpoint of a'B', and Q is the midpoint of b'c '. Is AP and CQ a straight line out of plane

Connect a1c1, AC and PQ
Easy to know a1c1 parallel AC, a1c1 parallel PQ
So AC parallel PQ
So AC and PQ must be on a plane in space
So the two lines AP and CQ in this plane can not be out of plane lines