\(a)\\A=x^2-2x+5\\=(x^2-2x+1)+4\\=(x-1)^2+4 \geq 4\)
Dấu "=" xảy ra khi : `(x-1)^2=0`
`<=> x=1`
Vậy `A_(min)=4 <=> x=1`
`b)` `B=2x^2+6x-5`
`=2(x^2+3x-5/2)`
`=2(x^2+2.x . 3/2+9/4-5/2-9/4)`
`=2[(x+3/2)^2-19/4]`
`=2(x+3/2)^2-19/2>=-19/2`
Dấu "=" xảy ra khi : `(x+3/2)^2=0`
`<=> x=-3/2`
Vậy `B_(min)=-19/2 <=> x=-3/2`
`c)` `C=(x+1)(2x-1)`
`=2x^2-x+2x-1`
`=2x^2+x-1`
`=2(x^2+1/2x-1/2)`
`=2(x^2+2.x. 1/4+1/16-1/2-1/16)`
`=2[(x+1/4)^2-9/16]`
`=2(x+1/4)^2-9/8>=-9/8`
Dấu "=" xảy ra khi : `(x+1/4)^2=0`
`<=> x=-1/4`
Vậy `C_(min)=-9/8 <=> x=-1/4`
`d)` `D=(x^2-2x)(x^2-2x+2)`
Đặt `x^2-2x=t` ( với `t>=0` )
`D=t(t+2)`
`=t^2+2t`
`=t^2+2t+1-1`
`=(t+1)^2-1>=-1`
Dấu "=" xảy ra khi : `(t+1)^2=0`
`<=> t=-1`
`<=> x^2-2x=-1`
`<=> x^2-2x+1=0`
`<=> (x-1)^2=0`
`<=> x=1`
Vậy `D_(min)=-1 <=> x=1`
