Môn Toán Lớp 8 Giups mình với ạ!!!!!!!!!!!!!!!!!!!!!!!!!!!!1
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Giải thích các bước giải:
\(x^2+2xy+6x+6y+2y^2+8=0\\ \Leftrightarrow\left(x+y\right)^2+6\left(x+y\right)+y^2=-8\) Ta có \(y^2\ge0\Leftrightarrow\left(x+y\right)^2+6\left(x+y\right)\le-8\) \(\Leftrightarrow\left(x+y\right)^2+6\left(x+y\right)+9\le1\\ \Leftrightarrow\left(x+y+3\right)^2\le1\\ \Leftrightarrow\left|x+y+3\right|\le1\\ \Leftrightarrow-1\le x+y+3\le1\\ \Leftrightarrow2017\le B\le2019\) \(B_{min}=2017\Leftrightarrow\left\{{}\begin{matrix}x+y+2021=2017\\y=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-4\\y=0\end{matrix}\right.\) \(B_{max}=2019\Leftrightarrow\left\{{}\begin{matrix}x+y+2021=2019\\y=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-2\\y=0\end{matrix}\right.\)
$\\$
`x^2+2xy+6x+6y +2y^2+8=0`
`<=>(x+y)^2 + 6 (x+y) + 9 – 1 = -y^2`
`<=> (x+y+3)^2 – 1^2=-y^2`
`<=>(x+y+3-1)(x+y+3+1)=-y^2`
`<=> (x+y+2)(x+y+4)=-y^2`
`<=> (x+y+2021 – 2019)(x+y+2021 – 2017)=-y^2`
`<=> (B-2019)(B-2017)=-y^2`
Do `y^2\ge 0 => -y^2 ≤0`
`=> (B-2019)(B-2017) ≤ 0`
TH1 :
`B-2019 ≤ 0, B-2017 ≥ 0`
`<=>B ≤ 2019, B ≥ 2017`
`<=> 2017 ≤ B ≤ 2019` (T/M)
TH2 :
`B-2019 ≥ 0, B-2017 ≤ 0`
`<=>B ≥ 2017, B ≤ 2017` (L)
Do đó : `2017 ≤ B ≤ 2019`
$\bullet$ `2017 ≤ B`
`=>min B=2017`
Dấu “`=`” xảy ra khi :
`x+y+2021=2017, y=0`
`<=> x+2021=2017, y=0`
`<=>x=-4, y=0`
Vậy `min B=2017<=>x=-4, y=0`
$\bullet$ `B ≤ 2019`
`=> max B=2019`
Dấu “`=`” xảy ra khi :
`x+y+2021=2019, y=0`
`<=>x+2021=2019,y=0`
`<=>x=-2,y=0`
Vậy `max B=2019<=>x=-2,y=0`