Đáp án:
MinA `= -2` khi `x = y = -1/2`
Giải thích các bước giải:
` A = 5x^2 - 2xy + 4x + y^2 – 1`
`=4x^2 + 4x + 1 + x^2 - 2xy + y^2 - 2`
`=(2x + 1)^2 + (x - y)^2 - 2`
`(2x + 1)^2 + (x - y)^2 >= 0 `
`⇔ (2x + 1)^2 + (x - y)^2 - 2 >= - 2`
hay `A >= - 2`
Dấu bằng xảy ra khi
`2x + 1 = 0 ⇔ x = -1/2 ⇒ y = -1/2`
Vậy MinA `= -2` khi `x = y = -1/2`
$5x^{2}-2xy+4x+y^2-1$
$=(4x^{2}+4x+1)+(x^2-2xy+y^2)-2$
$=(2x+1)^{2}+(x-y)^2-2$
Ta có: $(2x+1)^{2}$ $\geq0$ , $(x-y)^{2}$ $\geq0$
-> $(2x+1)^{2}+(x-y)^2$ $\geq0$
-> $(2x+1)^{2}+(x-y)^2-2$ $\geq-2$
Dấu bằng xảy ra khi $\begin{cases} 2x+1=0\\\\x-y=0 \end{cases}$
⇔ $\begin{cases} x=-\dfrac{1}{2}\\\\x=y \end{cases}$
⇔ $x^{}=y=\dfrac{1}{2}$ để A min = -2
Chúc bạn học tốt !!!!!
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