`A(1;-1);B(4;-3);C(5;5)`
`M∈∆:y=2x-1`
`=>M(x;2x-1)`
Ta có:
`MA^2=(x-1)^2+(2x-1+1)^2=x^2-2x+1+4x^2=5x^2-2x+1`
`MB^2=(x-4)^2+(2x-1+3)^2`
`=x^2-8x+16+4x^2+8x+4=5x^2+20`
`MC^2=(x-5)^2+(2x-1-5)^2=x^2-10x+25+4x^2-24x+36=5x^2-34x+61`
`=>MA^2+MB^2+MC^2`
`=15x^2-36x+82`
`=15(x^2-2x. 6/ 5+{36}/{25})+{302}/5`
`=15(x-6/ 5)^2+{302}/5≥{302}/5 ∀x`
(Vì `(x-6/ 5)^2≥0 ∀x`)
Dấu "=" xảy ra khi `x-6/ 5=0<=>x=6/ 5`
`=>y=2x-1=2. 6/5-1=7/5`
`=>M(6/5;7/5)`
Vậy `MA^2+MB^2+MC^2` có $GTNN$ bằng `{302}/5` khi `M(6/5;7/5)`
