`a) (3x-1)^2016+(2y-3)^2018=0`
Do `(3x-1)^2016>=0; (2y-3)^2018>=0` với `∀x;y`
`=>(3x-1)^2016=0; (2y-3)^2018=0`
`=> 3x-1=0; 2y-3=0`
`=> x=1/3; y=3/2`
Khi đó `M=15x^2+7xy=15.(1/3)^2+7. 1/3. 3/2`
`=15. 1/9+7/2=5/3+7/2=31/6`
Vậy `M=31/6` khi `x=1/3; y=3/2`
`b) (3x-5)^2006+(y^2-1)^2008+(x-z)^2100=0`
Do `(3x-5)^2006>=0; (y^2-1)^2008>=0; (x-z)^2100>=0` với `∀x;y;z`
`=> (3x-5)^2006=0; (y^2-1)^2008=0; (x-z)^2100=0`
`=> 3x-5=0; y^2-1=0; x-z=0`
`=> x=5/3; y^2=1; x=z`
`=> x=z=5/3; y=+-1`
Vậy `(x;y;z)=(5/3;1;5/3);(5/3;-1;5/3)`
