The monomials - 4B ^ 2 · y, - B ^ 2 · x ^ 2, a ^ 3,1 / 2axy ^ 2,5x ^ 2,3a ^ 3 · x, bxy are classified according to different standards. ^ denotes power, · denotes x sign

The monomials - 4B ^ 2 · y, - B ^ 2 · x ^ 2, a ^ 3,1 / 2axy ^ 2,5x ^ 2,3a ^ 3 · x, bxy are classified according to different standards. ^ denotes power, · denotes x sign

Quadratic monomial: 5x ^ 2
Cubic monomial: - 4B ^ 2 * y, a ^ 3, bxy
Quartic monomial: - B ^ 2 * x ^ 2, 1 / 2axy ^ 2, 3A ^ 3 * x

Classify the following monomials: 3A to the third power X, bxy, 5x to the square, - 4B to the square y, a to the third power, and - B to the square X,
The square of 1 / 2axy

（1）
Monomials with the letter A: the third power X of 3a, the third power of a, the square of 1 / 2axy
Monomials without a: bxy, the square of 5x, - 4B, y, - B, the square of X
（2）
The one letter monomial: the square of 5x, the third power of a
Monomials with multiple letters: the third power X of 3a, bxy, the square of - 4B, y, the square of - B, the square of X, the square of 1 / 2axy

Factorization is very simple... Messy and simple... Unfortunately, I can't
Can m (M + 2) - n (n + 2) be further decomposed

=m^2-n^2+2(m-n)
=(m+n)(m-n)+2(m-n)
=(m-n)(m+n+2)

The tangent equation of a circle whose center is at the origin and radius is r passing through a point P (x0, Y0) on the circle is x0x + y0y = R ^ 2. Why? How?

The slope of the line OP is Y0 / x0
Because the tangent is perpendicular to Op
So the slope of the tangent is - x0 / Y0
So the tangent can be set as y = - x * x0 / Y0 + B
And because the tangent passes through the point P (x0, Y0)
arcsinx-x
b=y0 + x0*x0/y0
That is, x0x + y0y = R ^ 2

X1 and X2 are the two real roots of the equation x plus the square of x minus 7 = 0. Solve the equation and find the value of the cubic power of X1 minus 8 and the square of x2 plus 54

If X1 and X2 are two of the equations X & sup2; + X-7 = 0, then X1 + x2 = - 1, X1 * x2 = - 7, we can see that (x1) & sup2; + (x2) & sup2; = 15 reduces (x1) & sup3; - 8 (x2) & sup2; + 54 = (x1) & sup2; * X1 - (x2) & sup2; - 7 (x2) & sup2; + 54 = (x1) & sup2; * (- 1-x2) - (x2) & sup2; - 7

In the pyramid p-abcd, ABCD is a square with side length of 3, the plane of PA ⊥ ABCD, the midpoint of PC is e, if the cosine of dihedral angle b-ae-d is - 1 / 3
In the pyramid p-abcd, ABCD is a square with side length of 3, PA ⊥ plane ABCD, the midpoint of PC is e, if the cosine value of dihedral angle b-ae-d is - 1 / 3. (1) find the length of PA; (2) find the distance from C to plane Abe

The point E is the middle point of PC, and ⊥ de ⊥ PC. according to the three perpendicular theorem, de ⊥ Pb. ⊥ de ⊥ Pb, EF ⊥ Pb prove that EDB of PA | plane connects AC, BD of PA | plane connects o, and EO ⊥ square ABCD ⊥ o is the middle point of AC

If a + B = 23 π, the minimum and maximum of cos2a + cos2b are ()
A. 1-32，32B. 12，32C. 1−32，1+32D. 12，1+22

A + B = 120 degrees, so A-B ∈ [- 120 degrees, 120 degrees], y = cos2a + cos2b = 1 + cos2a2 + 1 + cos2b2  1 + 12 (cos2a + cos2b) = 1 + cos (a + b) + cos (a-b) = 1 + cos120 degrees + cos (a-b) = 12 + cos (a-b). Because cos120 degrees ≤ cos (a-b) ≤ cos0 degrees, that is - 12 ≤ cos (a-b) ≤ 1,  12 ≤ cos2a + cos2b ≤ 32, B is selected

There are two points a (- 1,0), B (1,0) on the plane, and point P is on the circumference (x-3) 2 + (y-4) 2 = 4. Find the coordinates of point P when ap2 + bp2 is the minimum

According to the meaning of the title, if we make the symmetric point Q of point P about the origin, then the quadrilateral paqb is a parallelogram. According to the properties of parallelogram, there are ap2 + bp2 = 12 (4op2 + AB2), that is, when OP is the minimum, ap2 + bp2 takes the minimum, opmin = 5-2 = 3, PX = 3 × 35 = 95, py = 3 × 45 = 125, P (95125)

Given that the m-th power of a is equal to 2 and the n-th power of a is equal to 3, find the value of M + N + 2 of A

The second power of 6A

In the cube abcd-a1b1c1d1 with edge length a, e, F and H are the midpoint of edge BB1, CC1 and dd1 respectively
In the cube abcd-a1b1c1d1 with edge length a, e, F and H are the midpoint of edge BB1, CC1 and dd1 respectively. (1) to prove that BH is parallel to plane a1efd1; (2) to find the positive value of the angle between AF and aefd Pay attention, ask the second question! The answer is 4 / 15 and the root sign is 5! It's sin. What's the difference between sin and cos? If the question requires cos, it's cos! Thank you very much!

Go through point a and make Ao ⊥ a1e intersect a1e with O,
Because the edge length is a, △ aa1o ∽ a1eb1,
So Ao = 2 √ 5A / 5, AF = 3A / 2,
Because a1d1 ⊥ surface abb1a1,
So a1d1 ⊥ Ao, so Ao ⊥ surface a1efd1,
So sin =AO/AF=4√5a/15,