`B = (x + 1)^2 + (3x - 4)^2`
`= x^2 + 2x + 1^2 + (3x)^2 - 2. 3x. 4 + 4^2`
`= x^2 + 2x + 1 + 9x^2 - 24x + 16`
`= 10x^2 - 22x + 17`
`= (sqrt{10})^2. x^2 - 2. sqrt{10}. 11/(sqrt{10}). x + 121/10 + 49/10`
`= (sqrt{10}x)^2 - 2 sqrt{10}x. 11/sqrt{10} + (11/sqrt{10})^2 + 4,9`
`= (sqrt{10}x - 11/sqrt{10})^2 + 4,9`
Vì `(sqrt{10}x - 11/sqrt{10})^2 >= 0 forall x in mathbb R`
`=> (sqrt{10}x - 11/sqrt{10})^2 + 4,9 >= 4,9 forall x in mathbb R`
`=> B >= 4,9`
Dấu "`=`" xảy ra `<=> sqrt{10}x - 11/sqrt{10} = 0`
`<=> sqrt{10}x = 11/sqrt{10}`
`<=> (sqrt{10}x)^2 = (11/sqrt{10})^2`
`<=> 10x^2 = 121/10`
`<=> x^2 = 121/10 : 10 = 121/100`
`<=> x = +-11/10`
Vậy `M i n B = 4,9 <=> x = +-11/10`