Đáp án+Giải thích các bước giải:
`C=x^2-3x+10`
`C=x^2-2.x.\frac{3}{2}+\frac{9}{4}+\frac{31}{4}`
`C=(x-\frac{3}{2})^2+\frac{31}{4}`
`(x-\frac{3}{2})^2≥0∀x∈R`
`⇔(x-\frac{3}{2})^2+\frac{31}{4}≥\frac{31}{4}`
Dấu `"="` xảy ra khi
`x-\frac{3}{2}=0`
`⇔x=\frac{3}{2}`
Vậy `C_{min}=\frac{31}{4}` khi `x=\frac{3}{2}`
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`D=2x^2-10x+22`
`D=2x^2-10x+\frac{25}{2}+\frac{19}{2}`
`D=2(x^2-5x+\frac{25}{4})+\frac{19}{2}`
`D=2(x^2-2.x.\frac{5}{2}+\frac{25}{4})+\frac{19}{2}`
`D=2(x-\frac{5}{2})^2+\frac{19}{2}`
`(x-\frac{5}{2})^2≥0∀x∈R`
`⇔2(x-\frac{5}{2})^2≥0`
`⇔2(x-\frac{5}{2})^2+\frac{19}{2}≥\frac{19}{2}`
Dấu `"="` xảy ra khi
`2(x-\frac{5}{2})^2=0`
`⇔(x-\frac{5}{2})^2=0`
`⇔x-\frac{5}{2}=0`
`⇔x=\frac{5}{2}`
Vậy `D_{min}=\frac{19}{2}` khi `x=\frac{5}{2}`
