The fourth power of a - the square of a of 3, the square of B + the fourth power of B

The fourth power of a - the square of a of 3, the square of B + the fourth power of B

The fourth power of a - the square of a of 3, the square of B + the fourth power of B
=The fourth power of a - the square of 2a, the square of B + the fourth power of B - the square of a, the square of B
=(A's square-b's Square) - A's square-b's Square
=(A's Square - B's square + AB) (A's Square - B's Square - AB)
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9m square * (a-b) to the third power + 49 * (B-A) to the third power
.

Do you decompose factors?
9m^2(a-b)^3+49*(b-a)^3
=9m^2*(a-b)^3-49*(a-b)^3
=(a-b)^3*(9m^2-49)
=(a-b)^3*(3m+7)*(3m-7)

It is known that AB is the diameter of ⊙ o, point P is a moving point on the extension line of AB, the tangent of ⊙ o is made through P, the tangent point is C, the bisector of ⊙ APC intersects AC at point D, then ⊙ CDP is equal to ()
A. 30°B. 60°C. 45°D. 50°

As shown in the figure, connect OC, ∵ OC = OA, PD bisects ∵ APC, ∵ cpd = ∵ DPA, ∵ a = ∵ ACO, ∵ PC is the tangent line of ⊙ o, ∵ OC ⊥ PC, ∵ CPO + ∵ cop = 90 °, ∵ cpd + ∵ DPA + (∵ a + ∵ ACO) = 90 ° and ∵ DPA + ∵ a = 45 ° respectively, that is ∵ CDP = 45 °. Therefore, select C

If AB = a, C is the point on AB, m and N are the midpoint of AC and BC respectively, then Mn=--------------

1/2 a

As shown in the figure, the known point a is a trisection point on the semicircle with Mn as the diameter, point B is the midpoint of an, and point P is the point on the radius on. If the radius of ⊙ o is l, then the minimum value of AP + BP is ()
A. 2B. 2C. 3D. 52

Make a symmetric point a 'of point a about Mn, connect a' B, intersect Mn at point P, then PA + Pb is the smallest, connect OA ', AA', ob, ∵ point a and a 'are symmetric about Mn, point a is a trisection point on semicircle, ∵ a' on = ∠ AON = 60 °, PA = PA ', ∵ point B is the midpoint of arc an ^, ∵ Bon = 30 °, ∵ a' O

When a takes what value, the solution of the equation (x-1) / (X-2) - (X-2) / (1 + x) = (2x + a) / (X-2) (x + 1) is negative?

Multiply (x + 1) (X-2) (x-1) (x + 1) - (X-2) ^ 2 = 2x + A
2x-5=a
x=(a+5)/2

Point a moves on the curve x.x + y.y = 1, and point B (3,0), then the trajectory equation of point P in line AB is obtained
The answer is (2x-3y) V2 + 4yv2 = 1
Who can give the calculation process

Let a (a, b)
P(x,y)
Then (a + 3) / 2 = x, a = 2x-3
(b+0)/2=y,b=2y
A on X & sup2; + Y & sup2; = 1
a²+b²=1
So (2x-3) & sup2; + 4Y & sup2; = 1

When x = 2, the value of the polynomial ax ^ 3 + BX + 4 is 8. Then when x = - 2, what is the value of the polynomial?

When x = 2, ax & sup3; + BX + 4 = 88a + 2B + 4 = 88a + 2B = 4. When x = - 2, ax & sup3; + BX + 4 = - 8a-2b + 4 = - (8

It is known that the line L is the intersection of the plane ab1d1 passing through the vertex of the cube abcd-a1b1c1d1 and the plane of the lower bottom surface ABCD
It is known that the line L is the intersection of the plane ab1d1 passing through the vertex of the cube abcd-a1b1c1d1 and the plane of the lower bottom surface ABCD

Extend CB to e so that be = CB
Abcd-ab1c1d1 is a cube, dd1 = BB1, ad = BC, ADD1 = ebb1 = 90,
∵ be = CB, ad = BC, ∵ EB = ad, dd1 = BB1, ∵ ADD1 = ebb1, ≌ ADD1 ≌ ebb1,
∴AD1＝EB1.
Abcd-ab1c1d1 is a cube, d1c1 = AB, b1c1 = BC, b1c1d1 = EBA = 90 degrees
∵ b1c1 = BC, be = CB, ∵ b1c1 = EB, d1c1 = AB, ∵ b1c1d1 = EBA,
∴△B1C1D1≌△EBA,∴B1D1＝EA.
According to AD1 = EB1, b1d1 = EA, aeb1d1 is a parallelogram, a, e, B1, D1 are coplanar, and d1b1 ‖ AE
It is obvious that AE is the intersection of plane ac1d1 and plane ABCD, where ∥ line L = AE and ∥ d1b1 ∥ line L

If the polynomial K (K-2) x cube - (K-2) x square-5 is a quadratic polynomial about X, then what is the value of K?

If the polynomial K (K-2) x & # 179; - (K-2) x & # 178; - 5 is a quadratic polynomial about X,
Then the coefficient of the cubic term must be 0, but the coefficient of the quadratic term cannot be 0
{ k(k-2)=0
k-2≠0
The solution is k = 0