Using fraction to calculate (x / 2Y) ^ 2 * y / 2x - X / y ^ 2 - X / y ^ 2 △ 2Y ^ 2 / X （x/2y)^2*y/2x -x/y^2 -x/y^2÷2y^2/x x+1/x*(2x/x+1)^2-(1/x-1 - 1/x+)

Using fraction to calculate (x / 2Y) ^ 2 * y / 2x - X / y ^ 2 - X / y ^ 2 △ 2Y ^ 2 / X （x/2y)^2*y/2x -x/y^2 -x/y^2÷2y^2/x x+1/x*(2x/x+1)^2-(1/x-1 - 1/x+)

（x/2y)^2*y/2x -x/y^2 -x/y^2÷2y^2/x
=（x^2/4y^2)*y/2 -x/y^2 -x/y^2*x/2y^2
=x/8y -x/y^2 -x^2/2y^4

Calculation: (2x-y) 2-4 (X-Y) (x + 2Y)

The original formula = 4x2-4xy + y2-4 (x2 + xy-2y2) = 4x2-4xy + y2-4x2-4xy + 8y2 = - 8x + 9y2

Even if the sphere x ^ 2 + y ^ 2 + Z ^ 2 = 1 is the volume of the part sandwiched by the plane z = 0 and z = 1

Z = 0, is the xoy plane, z = 1, the plane is tangent to the top of the ball, so the volume calculated is the volume of the hemisphere. The rectangular coordinates are transformed into spherical coordinates x = rsin φ cos θ, y = rsin φ sin θ, z = RCOs φ, f (R, θ, φ) = (rsin φ cos θ) ^ 2 + (rsin φ sin θ) ^ 2 + (RCOs φ) ^ 2 = R ^ 2 (sin φ) ^ 2 + R ^ 2 (COS φ) ^ 2

The rectangle ABCD, ab = 1, BC = 2 are known. The △ abd is folded along the straight line of the diagonal BD of the rectangle
A. There is a position where the line AC is perpendicular to the line BD. B. there is a position where the line AB is perpendicular to the line CD. C. There is a position where the line ad is perpendicular to the line BC. D. for any position, three pairs of lines "AC and BD", "AB and CD", "ad and BC" are not perpendicular

As shown in the figure, AE ⊥ BD, CF ⊥ BD, according to the title, ab = 1, BC = 2, AE = CF = 63, be = EF = FD = 33, a, if there is a certain position, making the straight line AC perpendicular to the straight line BD, then ⊥ BD ⊥ AE, ⊥ BD ⊥ plane AEC, thus BD ⊥ EC, which is in contradiction with the known, exclude a; B, if there is a certain position, making the straight line AB perpendicular to the straight line BD

In the parallelogram ABCD, the following formula: vector 1AD = AB + BD 2ad = AC + CD 3aD + AB = AC 4AB + BC = AC 5ad = AB + BC + CD 6ad = DC + ca
It's not true

6 is wrong ad = AC + CD

In the parallelogram ABCD, vector AB = (1,0), vector AC = (2,3), then vector ad. vector BD equals a.4 B. - 4 c.9 D. - 9

Vector AB = (1,0) = vector DC, vector AC = (2,3)
Vector ad = vector AC - vector DC = (2,3) - (1,0) = (1,3)
Vector BD = vector ad - vector AB = (1,3) - (1,0) = (0,3)
therefore
Vector ad · vector BD = (1,3) · (0,3) = 1 × 0 + 3 × 3 = 9
Choose C

In the parallelogram ABCD, the vector AB = vector a, the vector ad = vector B, then the vector AC + the vector Ba is equal to what is the best detailed answer, oh, tomorrow night

Since a quadrilateral is a parallelogram, then vector AC = vector AB + vector ad = vector a + vector B, and vector Ba = - vector AB = - A, so vector AC + vector Ba = (a + b) - a = B;
Or: AC + Ba = (AB + AD) + Ba = (AB + AD) - AB = ad = B

In the parallelogram ABCD, point O is the intersection of two diagonals. Let vectors AB = A and BC = B, and try to use vectors a and B to represent vectors OA and ob

(1) Let o be the midpoint of AC,
∵ vector AB + vector BC = vector AC = 2, vector Ao,
2 vector Ao = a + B,
Vector Ao = 1 / 2 (a + b)
The vector OA = - 1 / 2 (a + b)
(2) From vector Ao + vector ob = vector ab
Ψ vector ob = vector ab - vector Ao
=a-1/2（a+b）
=1/2（a-b）

In the parallelogram ABCD, O is the diagonal intersection, and Ba and BC are used to represent Co
This is a vector problem! Sorry, the sign of that vector won't! But please,

∵CA=CB+BA,CO=1/2CA
∴CO=1/2(CB+BA)

As shown in the figure, in the quadrilateral ABCD, ∠ a = ∠ C = 90 °, be bisection ∠ ABC, DF bisection ∠ ADC, judge whether be and DF are parallel

The proof of dfdfdfproof: the proof of dfdfproof: the result of the proof of dfproof: a \ \\\\\\\ \\\\\#; the \\\\\\\\\\\\\\\\\\\\\\\\\\\\\\= CFD