Question

tính tỗng x+y+z biết:

x\(^2\)+ y\(^2\)+ z\(^2\)+ 2x- 4y+ 6z = -14

Answer 1

x²+y²+z²+2x-4y+6z+14=0

(x+1)²+(y-2)²+(z+3)²=0

=>x+1=0=>x=-1

y-2=0=>y=2

z+3=0=>z=-3

=>x+y+z=............

Answer 2

x^2+y^2+z^2+2x-4y+6z+14=0

x^2+y^2+z^2+2x-4y+6z+1+4+9 = 0

(x+1)^2+(y-2)^2+(z+3)^2         =0

=> x+1=0 -> x = -1

=> y-2=0 -> y=2

=> z+3=0->z=-3

vậy x+y+z = -2