XY (x + y) = X3 + y3-5

XY (x + y) = X3 + y3-5

x2y+xy2=x3+y3-5
x2(y-x)+y2(x-y)=-5
(x-y)(x2-y2)=5
(x-y)2*(x+y)=5

X3 + 6xy + Y3 = X3 + Y3 + 6xy = (x + y) (x2-xy + Y2) + 6xy why X3 + Y3 + 6xy can be transformed into (x + y) (x2-xy + Y2) + 6xy? Please explain for me
Forgive me for being shallow,

It is X & sup3; + Y & sup3; = (x + y) (X & sup2; - XY + Y & sup2;) x & sup3; + Y & sup3; = x & sup3; + X & sup2; Y-X & sup2; y-xy & sup2; + XY & sup2; + y & sup3; = x & sup2; (x + y) - XY (x + y) + Y & sup2; (x + y) = (x + y) (X & sup2; - XY + Y & sup2;) this is a formula called cubic sum formula

Let x, y and Z be real numbers which are not all zero, and prove that (2yz + 2zx + XY) / (x ^ 2 + y ^ 2 + Z ^ 2) ≤ (√ 33 + 1) / 4
Let x, y and Z be real numbers which are not all zero
(2yz+2zx+xy)/(x^2+y^2+z^2)≤(√33+1)/4
It is proved that we only need to consider x ≥ 0, y ≥ 0, Z ≥ 0,
2yz+2zx+xy≤1/2xx+1/2yy+γyy+(1/γ)zz+(1/γ)zz+γxx……
Why is gamma set like this? How did you think of it?

This is mainly due to the fact that the numerator and denominator are homogeneous, and the coefficients YZ and XZ are the same, but they are not the same as XY, so we need to add a coefficient to let
1/2(x²+y²) >=xy
γx² + 1/γ z² >= 2xz
γy² + 1/γ z² >= 2yz
(1 / 2 + 1 / γ) x & # 178; + (1 / 2 + 1 / γ) y & # 178; + 2 / γ Z & # 178; > = 2yz + 2zx + XY
If (1 / 2 + 1 / γ) = 2 / γ = 1, then γ = 2
In fact, we can just get it together. Of course, we can assign coefficients to each of the three equations. Then we can get a cubic equation with three variables, and we can solve three coefficients, which are exactly the same as the coefficients

x. Y and Z are the maximum values of real numbers (XY + 2yz + 3xz) / (x ^ 2 + y ^ 2 + Z ^ 2)

Is the title copied wrong?
If it is to find the maximum value of (XY + 2yz + 2zx) / (X & # 178; + Y & # 178; + Z & # 178;), then grey is often simple!
Solution 1:
When the parameter λ is introduced, it is solved by basic inequality
2yz＝2·λy·z/λ≤λ²y²+z²/λ²
2zx＝2·λx·z/λ≤λ²x²+z²/λ²
xy≤(x²+y²)/2
∴xy+2yz+2zx
≤(λ²+1/2)x²+(λ²+1/2)+(2/λ²)z²
Let λ & # 178; + 1 / 2 = 2 / λ & # 178; → λ & # 178; = (√ 33-1) / 4
∴λ²+1/2＝(1+√33)/4
The maximum value is (1 + √ 33) / 4
Solution 2
According to embedded inequality
x²+y²+z²≥2xycosA+2yzcosB+2zxcosC
(A+B+C＝π)
Let CoSb = COSC = 2cosa, B = C, a = π - 2b
∴cosB＝2cos(π-2B)→cosB＝(-1+√33)/8.
By replacing the embedding inequality, we get the following results
The maximum value is: (1 + √ 33) / 4

x. Y and Z are three real numbers which are not all zero. Find the maximum of (XY + 2yz) / (x + y + Z)

X ^ 2 + (1 / 5) y ^ 2 > = (2 / √ 5) XY (4 / 5) y ^ 2 + Z ^ 2 > = (4 / √ 5) YZ, then x ^ 2 + y ^ 2 + Z ^ 2 > = (2 / √ 5) XY + (4 / √ 5) YZ = (2 / √ 5) (XY + 2yz) (XY + 2yz) / (x ^ 2 + y ^ 2 + Z ^ 2)

It is known that the real number XY satisfies | x | / 4 + | y | / 3

Then draw the line system of y = x-z, we can see that when (- 4,0), Z gets the maximum

If x and y are two real numbers (XY ≠ 1) and 3x & sup2; - 2005x + 2 = 0,2y & sup2; - 2005y + 3 = 0, then the value of X & sup2; / y + X / Y & sup2; and so on

Because 2Y & sup2; - 2005y + 3 = 0, so y ≠ 0, we can get 2-2005 (1 / y) + 3 (1 / y) & sup2; = 0. If 1 / y is regarded as an unknown, it is equivalent to equation 3x & sup2; - 2005x + 2 = 0. Because XY ≠ 1, X ≠ 1 / y, so x and y are two unequal roots of equation 3x & sup2; - 2005x + 2 = 0

For rational numbers, the new operation is defined as: X * y = ax + by + XY, where a and B are constants, and the common addition and multiplication operations on the right side of the equation are known as: 2 * 1 = 7, (-
For rational numbers, a new operation is defined: X * y = ax + by + XY, where a and B are constant operations. If 2 * 1 = 7, (- 3) * 3 = 3, find the value of 3 / 1 * 6

Because x * y = ax + by + XY, 2 * 1 = 7, so 2A + B + 2 = 7, you haven't finished the problem. This problem has to have another condition, so as to get another bivariate linear equation about a and B, and then form the two bivariate linear equations into bivariate linear equations. By solving the equations, you can get the values of a and B, and then solve the problem

For the rational number x, y, the new operation x # y = ax + by + XY is specified, where a and B are constant terms, 2 # 1 = 7, 3 # 4 = 17 are known, and the value of 4 # 8 is obtained

2A+1B+1*2=7
3A+4B+3*4=17
It can be solved that a =... B =
Do you need me to solve this simple quadratic equation? You should be able to

It is known that X and y are rational numbers. If a kind of operation "multiplication" is specified, its meaning is that x multiplies y = XY plus 1. Try to find the value of 2 times 4 and [1 times 4] times [- 2]

If I don't get it wrong, XY means a two digit number
1.2 times 4 = 24 + 1 = 15
2.1 times 4 = 14 + 1 = 15, so 15 times (- 2) = - 152 + 1 = - 151