`a,A=x^2+x+1`
`=x^2+x+\frac{1}{4}+\frac{3}{4}`
`=(x+\frac{1}{2})^2+\frac{3}{4}`
`Do (x+\frac{1}{2})^2\geq0`
`=>(x+\frac{1}{2})^2+\frac{3}{4}\geq\frac{3}{4}`
Dấu bằng xảy ra khi` x+\frac{1}{2}=0`
`=> x=\frac{-1}{2}`
Vậy` A_(min)=\frac{3}{4} khi x=\frac{-1}{2}`
`b,B=x^2-x+1`
`=x^2-x+\frac{1}{4}+\frac{3}{4}`
`=(x-\frac{1}{2})^2+\frac{3}{4}`
`Do (x-\frac{1}{2})^2\geq0`
`=>(x-\frac{1}{2})+\frac{3}{4}\geq\frac{3}{4}`
Dấu bằng xảy ra khi` x-\frac{1}{2}=0`
`=> x=\frac{1}{2}`
Vậy` B_(min)=\frac{3}{4} khi x=\frac{1}{2}`
`c,C=x^2-3x+5`
`=x^2-3x+\frac{9}{4}+11/4`
`=(x-3/2)^2+11/4`
`(x-3/2)^2\geq0`
`=>(x-3/2)^2+11/4\geq11/4`
Dấu bằng xảy ra khi `x=3/2`
Vậy `C_(min)+11/4 khi x=3/2`
`d,D=x^2-x-5,75`
`=x^2-x+\frac{1}{4}-6`
`=(x-\frac{1}{2})^2+-6`
`Do (x-\frac{1}{2}^2\geq0`
`=>(x-\frac{1}{2}-6\geq\-6`
Dấu bằng xảy ra khi` x-\frac{1}{2}=0`
`=> x=\frac{1}{2}`
Vậy` D_(min)=6 khi x=\frac{1}{2}`
