`a)A=-x^2-3x+10`
`A=-(x^2+3x-10)`
`A=-[x^2+2.x. 3/2+(3/2)^2-49/4]`
`A=-(x+3/2)^2+49/4<=49/4`
Dấu `=` xảy ra
`<=>x+3/2=0`
`<=>x=-3/2`
Vậy `maxA=49/4` khi `x=-3/2`
`b)B=-9x^2-6x-20`
`B=-(9x^2+6x+20)`
`B=-[(3x)^2+2.3x.1+1^2+19]`
`B=-(3x+1)-19<=-19`
Dấu `=` xảy ra
`<=>3x+1=0`
`<=>3x=-1`
`<=>x=-1/3`
Vậy `maxB=-19` khi `x=-1/3`
`c)C=4x^2-12x+20`
`C=(2x)^2-2.2x.3+3^2+11`
`C=(2x-3)^2+11>=11`
Dấu `=` xảy ra
`<=>2x-3=0`
`<=>2x=3`
`<=>x=3/2`
Vậy `minC=11` khi `x=3/2`
`d)D=x(x+1)`
`D=x^2+x`
`D=x^2+2.x. 1/2+(1/2)^2+3/4`
`D=(x+1/2)^2+3/4>=3/4`
Dấu `=` xảy ra
`<=>x+1/2=0`
`<=>x=-1/2`
Vậy `minD=3/4` khi `x=-1/2`
