It is known that f (x) is a function defined on R which is not always zero. For any x, y ∈ R, f (x · y) = XF (y) + YF (x) holds. & nbsp; sequence {an} satisfies an = f (2n) (n ∈ n *), and A1 = 2. Then the general term formula of sequence an=______ ．

It is known that f (x) is a function defined on R which is not always zero. For any x, y ∈ R, f (x · y) = XF (y) + YF (x) holds. & nbsp; sequence {an} satisfies an = f (2n) (n ∈ n *), and A1 = 2. Then the general term formula of sequence an=______ ．

Since an = f (2n) then an + 1 = f (2n + 1) and A1 = 2 = f (2) ∵ for any x, y ∈ R, there is f (x · y) = XF (y) + YF (x) ∵ such that x = 2n, y = 2 then f (2n + 1) = 2NF (2) + 2F (2n) ∵ an + 1 = 2An + 2 × 2n ∵ an + 12n + 1 − an2n = 1 ∵ the sequence {an2n} takes A12 = 1 as the first common term

The extremum of F (x, y) = the square of X + the square of Y + XY under the condition of X + 2Y = 4

According to x + 2Y = 4. So x = 4-2y. Bring in the original equation. So f (x, y) = (4-2y) ^ 2 + y ^ 2 + (4-2y) y = 3Y ^ 2-12y + 16 = 3 (Y-2) ^ 2 + 4. So when y = 2, it is the minimum, at this time, x = 0

XY / x + y = 1, ZY / Z + y = 2, ZX / Z + x = 3, find the value of X, y, Z,

They are defined as Formula 1, 2, 3 respectively. From Formula 1 we get x = Y / (Y-1) and from formula 2 we get z = 2Y / (Y-2). Taking these two into formula 3 we get y = 12 / 7, x = 12 / 5 and z = - 12

Z / x + X / y + Y / z = 1

Solve x (x + y + Z) = 4-yz, y (x + y + Z) = 9-xz, Z (x + y + Z) = 25 XY equations

（1） X (x + y + Z) = 4-yz. = = = = > X & sup2; + (y + Z) x + YZ = 4. = = = > (x + y) (x + Z) = 4 ①. Similarly, the following two equations can be transformed to (x + y) (y + Z) = 9, ② (x + Z) (y + Z) = 25. ③ by multiplying the three equations, we can get (x + y) (y + Z) (x + Z) = ± 30. ④. (2) when (x + y) (y + Z) (x + Z) = 30, By dividing the three equations, we get y + Z = 15 / 2. X + Z = 10 / 3. X + y = 6 / 5. By adding the three equations, we get x + y + Z = 361 / 60. Then we get x = - 89 / 60. Y = 161 / 60. Z = 289 / 60. (3) when (x + y) (y + Z) (Z + x) = - 30, we can get the same solution

To solve the equations {XY / x + y = 12 / 7, YZ / y + Z = 6 / 5, XZ / x + Z = 4 / 3}

xy\X+Y=12\7
1/y+1/x=7/12 (1)
YZ\Y+Z=6\5
1/z+1/y=5/6 (2)
XZ\X+Z=4\3
1/z+1/x=3/4 (3)
From (1) - (2)
1/x-1/z=-1/4 (4)
From (3) + (4)
1/z+1/x+1/x-1/z=3/4-1/4
2/x=1/2
x=4
y=3
z=2

To solve the equations XY / (x + y) = 6 / 5, ① YZ / (x + Z) = 12 / 7, ② XZ / (x + Z) = 4 / 3, (7) - (5) x = 2, right

(7) (5) get 1 / x = 13 / 12-7 / 12 = 1 / 2
x=2

Finding dy =? From xy = arctan (Y / x)?
My calculation method is as follows:
First of all, the two sides are derived separately,
We get [1 - (1 / x ^ 2 + y ^ 2)] y + [x + (x / x ^ 2 + y ^ 2)] dy / DX = 0
Then we get [(x ^ y + y ^ 3-y) / (x ^ 2 + y ^ 2)] + [(x ^ 3 + XY ^ 2 + x) / (x ^ 2 + y ^ 2)] dy / DX = 0
Finally, I get dy = [(Y-X ^ Y-Y ^ 3) / (x + XY ^ 2 + x ^ 3)] DX
But the answer given by the teacher is dy = [(y + x ^ y + y ^ 3) / (x-xy ^ 2-x ^ 3)] DX
So, I want to ask if there's any problem with this, or am I wrong?

Xy = arctan (Y / x), the differential on both sides is YDX + XDY = 1 / [1 + (Y / x) ^ 2] * (XDY YDX) / x ^ 2 = (XDY YDX) / (x ^ 2 + y ^ 2), х XDY [1-1 / (x ^ 2 + y ^ 2)] = - YDX [1 + 1 / (x ^ 2 + y ^ 2)], х dy = - Y (x ^ 2 + y ^ 2 + 1) DX / [x (x ^ 2 + y ^ 2-1)] = [(y + x ^ y + y ^ 3) / (x-xy ^ 2-x ^ 3)] DX

Find the partial derivative of Z = (Y / x) ^ X / Y (Y / X to the power of Y / x) in (1,2) to X

Z = (Y / x) ^ (x / y) is the X / y power of Y / X. take the logarithm lnz = (x / y) ln (Y / x) lnz = (1 / y) * [XLN (Y / x)] on both sides and take y as the constant AZ / ax * 1 / z = (1 / y) * [ln (Y / x) + X * (y * - 1 / X & sup2;) / (Y / x)] AZ / AX = (1 / y) * [ln (Y / x) - (x * y / X & sup2;) * (x / y)] * zaz / AX = (1 / y)

How to find the (Y / z) power partial derivative of X

If we take the partial derivative of X, the answer is Y / Z-1 power of X