Tìm GTNN :
x^2 + y^2 + xy + 3x + 3y + 2018
Tìm GTNN :
x^2 + y^2 + xy + 3x + 3y + 2018
Đặt \(A=x^2+y^2+xy+3x+3y+2018\)
\(4.A=4x^2+4y^2+4xy+12x+12y+8072\)
\(4.A=\left(4x^2+4xy+y^2\right)+3y^2+12x+12y+8072\)
\(4.A=\left[\left(2x+y\right)^2+2\left(2x+y\right).3+9\right]+3\left(y^2+2y+1\right)+8060\)
\(4.A=\left(2x+y+3\right)^2+3\left(y+1\right)^2+8060\)
Mà \(\left(2x+y+3\right)^2\ge0\forall x;y\)
\(\left(y+1\right)^2\ge0\forall y\)\(\Rightarrow3\left(y+1\right)^2\ge0\forall y\)
\(\Rightarrow4.A\ge8060\)
\(\Leftrightarrow A\ge2015\)
Dấu "=" xảy ra khi :
\(\hept{\begin{cases}2x+y+3=0\\y+1=0\end{cases}}\Leftrightarrow\hept{\begin{cases}x=-1\\y=-1\end{cases}}\)
Vậy ...
1) tìm GTNN của: \(x^2+xy+y^2-3x-3y+2018\)
Đặt P =\(x^2+xy+y^2-3x-3y+2018\)
= \(x^2+\left(xy-3x\right)+y^2-3y+2018\)
= \(x^2+x\left(y-3\right)+y^2-3y+2018\)
= \(x^2+2.x.\dfrac{y-3}{2}+\dfrac{\left(y-3\right)^2}{4}-\dfrac{\left(y-3\right)^2}{4}+y^2-3y+2018\)
= \(\left(x+\dfrac{y-3}{2}\right)^2+\dfrac{-y^2+6y-9+4y^2-12y}{4}+2018\)
= \(\left(x+\dfrac{y-3}{2}\right)^2+\dfrac{3y^2-6y-9}{4}+2011\)
= \(\left(x+\dfrac{y-3}{2}\right)^2+\dfrac{3}{4}\left(y^2-2y-3\right)+2018\)
\(=\left(x+\dfrac{y-3}{2}\right)^2+\dfrac{3}{4}\left(y-1\right)^2+2015\)
Với mọi x;y có \(\left(x+\dfrac{y-3}{2}\right)^2\ge0\) ; \(\dfrac{3}{4}\left(y-1\right)^2\ge0\)
\(\Rightarrow\left(x+\dfrac{y-3}{2}\right)^2+\dfrac{3}{4}\left(y-1\right)^2+2015\ge2015\) với mọi x;y
\(\Rightarrow P\ge2015\) với mọi x;y
\(P=2015\Leftrightarrow\) \(\left\{{}\begin{matrix}x+\dfrac{y-3}{2}=0\\y-1=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}2x+y-3=0\\y=1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=1\\y=1\end{matrix}\right.\)
Vậy ......
tìm gtnn của B = x^2 + y^2 +xy -3x-3y
tìm gtnn của B = x^2 + y^2 +xy -3x-3y+2015
TÌM GTNN CỦA BIỂU THỨC: A=X^2+XY+Y^2-3X-3Y+2022
cho 2 số dương x,y tm xy=1 , tìm GTNN của A= x^2+3x+y^2+3y + 9/(x^2+y^2+1)
Tìm GTNN \(G=x^2+xy+y^2-3x-3y\)
Ta có: \(G=x^2+xy+y^2-3x-3y\)
\(=\left(x^2+2xy+y^2\right)-3\left(x+y\right)-xy\)
\(=\left(x+y\right)^2-3\left(x+y\right)-xy\)
Mà \(\left(x+y\right)^2\ge0\Leftrightarrow x^2+y^2\ge2xy\)
\(\Leftrightarrow x^2+2xy+y^2\ge4xy\Leftrightarrow\left(x+y\right)^2\ge4xy\)
\(\Leftrightarrow xy\le\frac{\left(x+y\right)^2}{4}\Leftrightarrow-xy\ge-\frac{\left(x+y\right)^2}{4}\)
\(\Rightarrow G\ge\frac{\left(x+y\right)^2-3\left(x+y\right)-\left(x+y\right)^2}{4}\)
\(\Leftrightarrow G\ge\frac{3\left(x+y\right)^2}{4}-3\left(x+y\right)\)
Đến đây để cho dễ nhìn, ta đặt \(t=x+y\)
\(\Rightarrow G\ge\frac{3t^2}{4}-3t=3\left(\frac{t^2}{4}-\frac{2t}{2}+1\right)-3\ge3\left(\frac{t}{2}-1\right)^2-3\ge3\)
Dấu "=" xảy ra \(\Leftrightarrow\frac{t}{2}=1\Leftrightarrow t=2\Leftrightarrow\hept{\begin{cases}x+y=2\\x=y\end{cases}\Leftrightarrow x=y=1}\)
Vậy \(MIN_G=-3\Leftrightarrow x=y=1\)
Tìm GTNN của Q=x^2+xy+y^2-3x-3y+1999
R=2x^2+2xy+y^2-2x+2y+15
Giúp mk với
Lời giải:
$2Q=2x^2+2xy+2y^2-6x-6y+3998$
$=(x^2+2xy+y^2)+x^2+y^2-6x-6y+3998$
$=(x+y)^2-4(x+y)+(x^2-2x)+(y^2-2y)+3998$
$=(x+y)^2-4(x+y)+4+(x^2-2x+1)+(y^2-2y+1)+3992$
$=(x+y-2)^2+(x-1)^2+(y-1)^2+3992\geq 3992$
$\Rightarrow Q\geq 1996$
Vậy $Q_{\min}=1996$ khi $x+y-2=x-1=y-1=0\Leftrightarrow x=y=1$
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$R=(x^2+2xy+y^2)+x^2-2x+2y+15$
$=(x+y)^2+2(x+y)+x^2-4x+15$
$=(x+y)^2+2(x+y)+1+(x^2-4x+4)+10$
$=(x+y+1)^2+(x-2)^2+10\geq 10$
Vậy $R_{\min}=10$ khi $x+y+1=x-2=0$
$\Leftrightarrow x=2; y=-3$
Tìm GTNN:
A=2x^2+2xy+y^2-2x-2y
b=x^2+xy+y^2-3y-3x
B=x^4-2x^3+3x^2-2x+1