X + y + Z = 2, XY + YZ + XZ = 1. Find x ^ 2Y ^ 2Z ^ 2

X + y + Z = 2, XY + YZ + XZ = 1. Find x ^ 2Y ^ 2Z ^ 2

(x+y+z)^2=x^2+y^2+z^2+2xy+2xz+2yz=4
so x^2+y^2+z^2=4-2=2

Given XYZ = 1, x + y + Z = 2, X2 + Y2 + Z2 = 16? Find the value of [1 △ (XY + 2Z) + 1 △ (YZ + 2x) + 1 △ (ZX + 2Y)]

Therefore, 1 / (XY + 2Z) = 1 / (XY + 4-2x-2y) = 1 / (X-2) (Y-2) (Y-2) let r = X-2, s = Y-2, t = Z-2, t = Z-2, let r = X-2, s = Y-2, t = Z-2, t = Z-2, let's get 1 / RS + 1 / St + 1 / RT = (R + S + T + T) / rst, and from the known can get (R + 2) (s + 2) (T + 2) (T + 2) = 1, R + S + T = - 4, (R + 2) ^ 2 + (s + 2) ^ 2 + (T + 2) ^ 2 + (T + 2) ^ 2 = 16, get the solution (R + S + S + T) / RST = - RST = - rst4 / 13

Given that XY (x + y) ^ - 1 = 1, YZ (y + Z) ^ - 1 = 2, XZ (Z + x) ^ - 1 = 3, try to find the value of XYZ (XY + YZ + XZ) ^ - 1

XY (x + y) ^ - 1 = 1, YZ (y + Z) ^ - 1 = 2, XZ (Z + x) ^ - 1 = 3, you take the reciprocal of left and right to get 1 / x + 1 / y = 11 / y + 1 / z = 1 / 21 / Z + 1 / x = 1 / 3, so that we can get 1 / x, 1 / y, 1 / Z, of course, we can also find x, y, Z (but not necessary), and we can see from the following

The square of 3x - "the square of 7x - (4x-3) - 2x" + 2 where x + - 5 / 2

Original formula = 3x? - (7x-4x + 3-2x?) + 2
=3x²-(3x+3-2x²)+2
=3x²-3x-3+2x²+2
=5x²-3x-1
=5*(-5/2)²-3*(-5/2)+1
=129/4

The solution equation is: (x2 + 3x-4) 2 + (2x2-7x + 6) 2 = (3x2-4x + 2) 2

Let u = x2 + 3x-4, v = 2x2-7x + 6, then u + V = 3x2-4x + 2
Then the original equation becomes U2 + V2 = (U + V) 2,
That is, U2 + V2 = U2 + 2uv + v2,
∴uv=0，
ν u = 0 or V = 0,
That is, X2 + 3x-4 = 0 or 2x2-7x + 6 = 0
The results show that X1 = - 4, X2 = 1. X3 = 3
2，x4＝2；

(Junior High School Mathematics) first remove the brackets: 3x 2 - [7x - (4x-3) - 2x?)

3x²-[7x-(4x-3)-2x²]
=3x²-7x+(4x-3)+2x²
= 3x²-7x+4x-3+2x²
=5x²-3x-3

3x-4y=4,5x+2y+3z=2,z=2x-7

3x-4y = 4 ① 5x + 2Y + 3Z = 2 ② z = 2x-7 ③ substituting ③ into ②, we can get: 5x + 2Y + 6x-21 = 211x + 2Y = 23; ④ from ① + ④ × 2, we can get: 25X = 50 x = 2; by substituting x = 2 into ①, we get: 6-4y = 4, y = 0.5; by substituting x = 2 into ③, we get: z = - 3

Given 4x + y = 5 and 3x + 2Y = 4, what is the value of X-Y

4x+y=5 A
3x+2y=4 B
A-B=X-Y=1

(3x-y)²-2x（4x-5y）-（x+2y）² First simplify and then evaluate (3x-y) 2 - 2x (4x-5y) - (x + 2Y) 2, where x = 2000, y = - 2

(3x-y)²-2x（4x-5y）-(x+2y)²
=9x²-6xy+y²-8x²+10xy-x²-4xy-4y²
=-3y²
x=2000,y=-2
Then the original formula = - 3 × (- 2) 2
=-3×4
=-12

If (3x squared y + 2XY squared) / M = - 3x + 2Y, then the monomial m is equal to ()

(3x squared y + 2XY squared) / M = - 3x + 2Y
xy(3x+2y)=m(2y-3x)
m=xy(3x+2y)/2y-3x