Known x 3＝y 4＝z 5. Find XY + YZ + ZX The value of x2 + Y2 + Z2

Known x 3＝y 4＝z 5. Find XY + YZ + ZX The value of x2 + Y2 + Z2

Set X
3＝y
4＝z
5 = K (K ≠ 0), then x = 3k, y = 4K, z = 5K,
∴xy+yz+zx
x2+y2+z2＝3k•4k+4k•5k+5k•3k
(3k)2+(4k)2+(5k)2＝47k2
50k2＝47
50．

If x ^ 2Y + XY ^ 2 + y ^ 2Z + YZ ^ 2 + Z ^ 2x + ZX ^ 2 + 3xyz = K (x + y + Z) * (XY + YZ + ZX), then the value of K is

Left = x ^ 2Y + XY ^ 2 + y ^ 2Z + YZ ^ 2 + Z ^ 2x + ZX ^ 2 + 3xyz = (x ^ 2Y + XYZ + ZX ^ 2) + (y ^ 2x + XYZ + ZY ^ 2) + (Z ^ 2Y + XYZ + XZ ^ 2) = x * (XY + YZ + ZX) + y * (XY + ZX + YZ) + Z * (YZ + XY + ZX) = (x + y + Z) * (XY + YZ + ZX) so k = 1. In addition, simply expand the (x + y + Z) * (XY + YZ + ZX) on the right directly to get x

If x + y + Z = 6, XY + YZ + ZX = 11, x ^ 3 + y ^ 3 + Z ^ 3-3xyz

(x+y+z) ²= six ²
x ²+ y ²+ z ²+ 2(xy+yz+zx)=36
x ²+ y ²+ z ²= 36-22=14
x ³+ y ³+ z ³- 3xyz 
=(x ³+ 3x ² y+3xy ²+ y ³+ z ³)- (3xyz+3x ² y+3xy ²) 
=[(x+y) ³+ z ³]- 3xy(x+y+z) 
=(x+y+z)(x ²+ y ²+ 2xy-xz-yz+z ²)- 3xy(x+y+z) 
=(x+y+z)(x ²+ y ²+ z ²+ 2xy-3xy-xz-yz) 
=(x+y+z)(x ²+ y ²+ z ²- xy-yz-xz) 
=18

XY / x + y = 1 YZ / y + Z = 2 ZX / Z + x = 3 how much is x equal to

∵ XY / (x + y) = 1 ∵ x + y) / (XY) = 1 / 1 ∵ 1 / x + 1 / y = 1. (1) ∵ YZ / (y + Z) = 2 ∵ y + Z / (YZ) = 1 / 2 ∵ 1 / y + 1 / z = 1 / 2. (2) ∵ ZX / (Z + x) = 3

XYZ = 1 x + y + Z = 2 x square + y square + Z square = 16 find 1 + (YZ + 2x) 1 + (ZX + 2Y) 1

Because z = 2-x-y, 1 / (XY + 2Z) = 1 / (XY + 4-2x-2y) = 1 / (X-2) (Y-2), similarly, 1 / (YZ + 2x) = 1 / (Y-2) (Z-2); 1 / (ZX + 2Y) = 1 / (Z-2) (X-2) let r = X-2, s = Y-2, t = Z-2, the problem becomes 1 / RS + 1 / St + 1 / RT = (R + S + T) / rst, and the three known conditions in the problem become (R + 2) (s + 2) (T + 2) = 1

If the definition field of function f (x) = root sign [(a ^ 2-1) x ^ 2 + (A-1) x + 2 / A + 1] is r, find the value range of real number a When a ^ 2-1 ≠ 0, why △≤ 0!

If the definition field of function f (x) = root sign [(a ^ 2-1) x ^ 2 + (A-1) x + 2 / A + 1] is r
Then (a ^ 2-1) x ^ 2 + (A-1) x + 2 / A + 1 is always greater than or equal to 0
That is, a ^ 2-1 > 0, and △ = (A-1) ²- 4(a^2-1)*2/（a+1 ）=（a-1） ²- 8(a-1)≤0
The solution is 1 < a ≤ 9