`A=2x^2-3x+5`
`A=2(x^2-3/2x+5/2)`
`A=2(x^2-2.x.(3)/4+(3/4)^2+31/16)`
`A=2(x-3/4)^2+31/8>=31/8`
Dấu `=` xảy ra
`<=>x-3/4=0`
`<=>x=3/4`
Vậy `minA=31/8` khi `x=3/4`
`b)B=x+3x^2-1`
`B=3(x^2+1/2x-1/3)`
`B=3(x^2+2.x.(1)/4+(1/4)^2-19/48)`
`B=3(x+1/4)^2-19/16>=-19/16`
Dấu `=` xảy ra
`<=>x+1/4=0`
`<=>x=-1/4`
Vậy `minB=-19/26` khi `x=-1/4`
`c)C=-2x^2+5x-7`
`C=-2(x^2-5/2x+7/2)`
`C=-2(x^2-2.x.(5)/4+(5/4)^2+31/16)`
`C=-2(x-5/4)^2-31/8<=-31/8`
Dấu `=` xảy ra
`<=>x-5/4=0`
`<=>x=5/4`
Vậy `maxC=-31/8` khi `x=5/4`
`d)D=-4x^2-6x+1`
`D=-(4x^2+6x-1)`
`D=-[(2x)^2+2.2x.(3)/2+(3/2)^2-13/4]`
`D=-(2x+3/2)^2+13/4<=13/4`
Dấu `=` xảy ra
`<=>2x+3/2=0`
`<=>2x=-3/2`
`<=>x=-3/4`
Vậy `maxD=13/4` khi `x=-3/4`
