`A=(4x-5)²+2022`
Ta có:`(4x-5)²≥0∀x`
`⇒(4x-5)²+2022≥2022∀x`
Dấu "=" xảy ra khi `4x-5=0⇔4x=5⇔x=5/4`
Vậy `A_(min)=2022` khi `x=5/4`
`B=x²-8x+7`
`=(x²-8x+16)-9`
`=(x²-2.x.4+4²)-9`
`=(x-4)²-9`
Ta có:`(x-4)²≥0∀x`
`⇒(x-4)²-9≥-9∀x`
Dấu "=" xảy ra khi `x-4=0⇔x=4`
Vậy `B_(min)=-9` khi `x=4`
`C=4x²+16x+3`
`=(4x²+16x+16)-13`
`=[(2x)²+2.2x.4+4²]-13`
`=(2x+4)²-13`
Ta có:`(2x+4)²≥0∀x`
`⇒(2x+4)²-13≥-13∀x`
Dấu "=" xảy ra khi `2x+4=0⇔2x=-4⇔x=-2`
Vậy `C_(min)=-13` khi `x=-2`
