x. If y and Z are real numbers and X + y + Z = 4 √ (x-3) + 3 √ (y-6) = 6 √ (Z-5), then x + y + Z =? X + y + Z = 4 √ (x-3) + 2 √ (y-6) + 6 √ (Z-5) a^2+b^2+c^2+14=4a+2b+6

x. If y and Z are real numbers and X + y + Z = 4 √ (x-3) + 3 √ (y-6) = 6 √ (Z-5), then x + y + Z =? X + y + Z = 4 √ (x-3) + 2 √ (y-6) + 6 √ (Z-5) a^2+b^2+c^2+14=4a+2b+6

In the range of real number, there is x > = 3Y > = 6Z > = 5, let x = 3 + A ^ 2Y = 6 + B ^ 2Z = 5 + C ^ 2 (a, B, C > = 0), then the question is wrong, otherwise it can't be solved. It should be x + y + Z = 4 √ (x-3) + 2 √ (y-6) + 6 √ (Z-5)? X + y + Z = 4 √ (x-3) + 2 √ (y-6) + 6 √ (Z-5) can be transformed into a ^ 2 + B ^ 2 + C ^ 2 + 14 = 4A + 2B + 6C (...)