a,
`A=3x^2+5x+4`
`=3x^2+5x+25/12+23/12`
`=(xsqrt3+5/(2sqrt3))^2+23/12`
Vì `(xsqrt3+5/(2sqrt3))^2>=0`
`=>A>=23/12`
Dấu `=` xảy ra `<=>xsqrt3+5/(2sqrt3)=0`
`<=>xsqrt3=(-5)/(2sqrt3)`
`<=>x=-5/6`
Vậy `A_(min)=23/12<=>x=-5/6`
b,
`B=x^4-4x^2-12`
`=(x^2)^2-2·x^2·2+2^2-16`
`=(x^2-2)^2-16`
Vì `(x^2-2)^2>=0`
`=>B>=-16`
Dấu `=` xảy ra `<=>x^2-2=0<=>x^2=2<=>x=±sqrt2`
Vậy `B_(min)=-16<=>x=±sqrt2`