B = $-x^{2}$ – 2$y^{2}$ – 2xy -2x +2y + 2013 Tìm gtln 07/12/2021 Bởi Katherine B = $-x^{2}$ – 2$y^{2}$ – 2xy -2x +2y + 2013 Tìm gtln
Đáp án: `B=-x^2-2y^2-2xy-2x+2y+2013` `=-(x^2+2y^2+2xy+2x-2y-2013) ` `=-(x^2+2xy+y^2+2x+2y+y^2-4y-2013)` `=-[(x+y)^2+2(x+y)+1+y^2-4y+4-2018]` `=-[(x+y+1)^2+(y-2)^2-2018]` Ta có: $\left\{\begin{matrix}(x+y+1)^2≥0& \\(y-2)^2≥0& \end{matrix}\right.$ `-> (x+y+1)^2+(y-2)^2>=0` `-> (x+y+1)^2+(y-2)^2-2018>=-2018` `-> -[(x+y+1)^2+(y-2)^2-2018]<=2018` Dấu “=” xảy ra `<=>` $\left\{\begin{matrix}(x+y+1)^2=0& \\(y-2)^2=0& \end{matrix}\right.$ `->` $\left\{\begin{matrix}x+y+1=0& \\y-2=0& \end{matrix}\right.$ `->` $\left\{\begin{matrix}x+y=-1& \\y=2& \end{matrix}\right.$ `->` $\left\{\begin{matrix}x=-3& \\y=2& \end{matrix}\right.$ Vậy `B_(max)=2018 <=>`$\left\{\begin{matrix}x=-3& \\y=2& \end{matrix}\right.$ Bình luận
Đáp án: $GTLN_B=2018↔\begin{cases}y=2\\x=-3\\\end{cases}$ Giải thích các bước giải: $B=-x^2-2y^2-2xy-2x+2y+2013$ $-B=x^2+2y^2+2xy+2x-2y-2013$ $-B=x^2+2xy+y^2+2x+2y+y^2-4y-2013$ $-B=(x+y)^2+2(x+y)+1+y^2-4y+4-2018$ $-B=(x+y+1)^2+(y-2)^2-2018$ ta có $\begin{cases}(x+y+1)^2 \geq 0\\(y-2)^2 \geq 0\\\end{cases}$ $↔(x+y+1)^2+(y-2)^2 \geq 0$ $↔(x+y+1)^2+(y-2)^2-2018 \geq -2018$ $↔-B \geq -2018$ $↔B \leq 2018$ dấu = xảy ra khi $\begin{cases}x+y+1=0\\y-2=0\\\end{cases}$ $↔\begin{cases}y=2\\x=-1-y=-3\\\end{cases}$ vậy $GTLN_B=2018↔\begin{cases}y=2\\x=-3\\\end{cases}$ CHÚC BẠN HỌC TỐT Bình luận
Đáp án:
`B=-x^2-2y^2-2xy-2x+2y+2013`
`=-(x^2+2y^2+2xy+2x-2y-2013) `
`=-(x^2+2xy+y^2+2x+2y+y^2-4y-2013)`
`=-[(x+y)^2+2(x+y)+1+y^2-4y+4-2018]`
`=-[(x+y+1)^2+(y-2)^2-2018]`
Ta có: $\left\{\begin{matrix}(x+y+1)^2≥0& \\(y-2)^2≥0& \end{matrix}\right.$
`-> (x+y+1)^2+(y-2)^2>=0`
`-> (x+y+1)^2+(y-2)^2-2018>=-2018`
`-> -[(x+y+1)^2+(y-2)^2-2018]<=2018`
Dấu “=” xảy ra `<=>` $\left\{\begin{matrix}(x+y+1)^2=0& \\(y-2)^2=0& \end{matrix}\right.$
`->` $\left\{\begin{matrix}x+y+1=0& \\y-2=0& \end{matrix}\right.$
`->` $\left\{\begin{matrix}x+y=-1& \\y=2& \end{matrix}\right.$
`->` $\left\{\begin{matrix}x=-3& \\y=2& \end{matrix}\right.$
Vậy `B_(max)=2018 <=>`$\left\{\begin{matrix}x=-3& \\y=2& \end{matrix}\right.$
Đáp án:
$GTLN_B=2018↔\begin{cases}y=2\\x=-3\\\end{cases}$
Giải thích các bước giải:
$B=-x^2-2y^2-2xy-2x+2y+2013$
$-B=x^2+2y^2+2xy+2x-2y-2013$
$-B=x^2+2xy+y^2+2x+2y+y^2-4y-2013$
$-B=(x+y)^2+2(x+y)+1+y^2-4y+4-2018$
$-B=(x+y+1)^2+(y-2)^2-2018$
ta có $\begin{cases}(x+y+1)^2 \geq 0\\(y-2)^2 \geq 0\\\end{cases}$
$↔(x+y+1)^2+(y-2)^2 \geq 0$
$↔(x+y+1)^2+(y-2)^2-2018 \geq -2018$
$↔-B \geq -2018$
$↔B \leq 2018$
dấu = xảy ra khi $\begin{cases}x+y+1=0\\y-2=0\\\end{cases}$
$↔\begin{cases}y=2\\x=-1-y=-3\\\end{cases}$
vậy $GTLN_B=2018↔\begin{cases}y=2\\x=-3\\\end{cases}$
CHÚC BẠN HỌC TỐT