If real numbers x, y and Z satisfy (x-z) square-4 (X-Y) (Y-Z) = 0, then the following equations must be true: A, x + y + Z = 0, B, x + y-2z = 0 C. Y + z-2x = 0 d, Z + x-2y = 0 try to talk about the process!

D
Let X-Y = a, Y-Z = B
From the meaning of the question, we get (a + b) ^ 2-4ab = (a-b) ^ 2 = 0
So a = B
So X-Y = y-z
So Z + x-2y = 0 chooses D

Given that real numbers x, y and Z satisfy x2 + Y2 + Z2 = 4, the maximum value of (2x-y) 2 + (2y-z) 2 + (2Z-X) 2 is ()
A. 12B. 20C. 28D. 36

When x + y + Z = 0, the maximum value of (2x-y) 2 + (2y-z) 2 + (2Z-X) 2 is 28

Given that real numbers x and y satisfy | 2x + y + 1 | ≤ | x + 2Y + 2 |, and - 1 ≤ y ≤ 1, then the maximum value of Z = 2x + y is______ ．

First, according to the constraint condition | 2x + y + 1 | ≤| x + 2Y + 2 |, - 1 ≤ y ≤ 1, we can obtain ① 2x + y + 1 ≥ 0x + 2Y + 2 ≥ 0x-y-1 ≤ 0, or ② 2x + y + 1 ≥ 0x + 2Y + 2 ≤ 0x + y + 1 ≤ 0, or ③ 2x + y + 1 ≤ 0x + 2Y + 2 & gt; 0x + y + 1 ≥ 0, or ④ 2x + y + 1 ≤ 0x + 2Y + 2 ≤ 0x-y-1 ≥ 0

It is known that a, B and C are trilateral lengths of △ ABC, and a & sup2; + B & sup2; + C & sup2; + 50 = 6A + 8b + 10C. It is proved that △ ABC is a right triangle

A & sup2; + B & sup2; + C & sup2; + 50 = 6A + 8b + 10Ca & sup2; - 6A + 9 + (B & sup2; - 8b + 16) + (C & sup2; - 10C + 25) = 0 (A-3) & sup2; + (B-4) & sup2; + (C-5) & sup2; = 0A = 3, B = 4, C = 5A & sup2; + B & sup2; = 9 + 16 = 25 = C & sup2; so △ ABC is a right triangle

If real numbers x, y and Z satisfy (x-z) square-4 (X-Y) (Y-Z) = 0, then the following equations must be true: A, x + y + Z = 0, B, x + y-2z = 0 C. Y + z-2x = 0 d, Z + x-2y = 0 try to talk about the process!