Tìm GTNN của biểu thức sau: a) A= x^2-2x+y^2+4y+8 b) B= x^2-4x+y^2-8y+6 c) C= x^-4xy+5y^2+10x-22y+28
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![](https://rs.olm.vn/images/avt/0.png?1311)
\(A=5-x^2+2x-4y^2-4y=-\left(x^2-2x+1\right)-\left(4y^2+4y+1\right)+7\\ =-\left(x-1\right)^2-\left(2y+1\right)^2+7\le7\)
đẳng thức xảy ra khi \(\left\{{}\begin{matrix}x-1=0\\2y+1=0\end{matrix}\right.\Rightarrow\)\(\left\{{}\begin{matrix}x=1\\y=-0,5\end{matrix}\right.\)
vậy MAX A=7 tại \(\left\{{}\begin{matrix}x=1\\y=-0,5\end{matrix}\right.\)
\(D=\left(x-1\right)\left(x+2\right)\left(x+3\right)\left(x+6\right)\\ D=\left(x^2+5x-6\right)\left(x^2+5x+6\right)\)
đặt: \(t=x^2+5x\) khi đó:
\(D=\left(t-6\right)\left(t+6\right)\\ D=t^2-36\ge-36\)
đẳng thức xảy ra khi :
\(t=0\\ \Leftrightarrow x^2+5x=0\\ x\left(x+5\right)=0\Rightarrow\left[{}\begin{matrix}x=0\\x=-5\end{matrix}\right.\)
vậy MAX D=-36 tại x=0 hoặc x=-5
![](https://rs.olm.vn/images/avt/0.png?1311)
a/ Ta có:
\(A=x^2-6x+11\)
\(A=x\cdot x-3x-3x+3\cdot3+2\)
\(A=x\left(x-3\right)-3\left(x-3\right)+2\)
\(A=\left(x-3\right)\left(x-3\right)+2\)
\(A=\left(x-3\right)^2+2\)
Vì \(\left(x-3\right)^2\ge0\)
Nên GTNN của \(\left(x-3\right)^2\)là 0
=> \(A_{min}=0+2=2\)
mình chỉ biết a. thôi
a) ta có : \(A=x^2-6x+11\)
\(A=x.x-3x-3x+3.3+2\)
\(A=x\left(x-3\right)-3\left(x-3\right)+2\)
\(A=\left(x-3\right)\left(x-3\right)+2\)
\(A=\left(x-3\right)^2+2\)
vì \(\left(x-3\right)^2\ge0\)
nên GTNN của \(\left(x-3\right)^2\)là \(0\)
\(\Rightarrow\)\(A_{min}\)\(=0+2=2\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a) \(A=x^2+6x+11\)
\(A=x^2+6x+9+2\)
\(A=\left(x+3\right)^2+2\)
Có: \(\left(x+3\right)^2\ge0\Rightarrow\left(x+3\right)^2+2\ge2\)
Dấu = xảy ra khi: \(\left(x+3\right)^2=0\Rightarrow x+3=0\Rightarrow x=-3\)
Vậy: \(Min_A=2\) tại \(x=-3\)
b) \(B=4x-x^2+1\)
\(B=-x^2+4x-4+5\)
\(B=-\left(x-2\right)^2+5\)
\(B=5-\left(x-2\right)^2\)
Có: \(\left(x-2\right)^2\ge0\)
\(\Rightarrow5-\left(x-2\right)^2\le5\)
Dấu = xảy ra khi: \(\left(x-2\right)^2=0\Rightarrow x-2=0\Rightarrow x=2\)
Vậy: \(Max_B=5\) tại \(x=2\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a,<=> x2-4x+22+y2-8y+42-14
<=> (x2-2x2+22)+(y2-2x4+42)-14
<=> (x-2)2+(y-4)2-14
Vì (x-2)2+(y-4)2>= 0
=> F >= -14 => MIn F = -14 <=> x=2, y=4
b, <=> (x2+52+(2y)2-4xy+10x-20y) +(y2-2y+1)+2
<=> (x+5-2y )2+(y-1)2+2
Vì (x+5-2y) 2+(y-1)2 >= 0
=> G >= 2 => Min =2 <=> y=1, x= -3
\(F=x^2-4x+y^2-8y+6\)
\(F=\left(x^2-2.2x+2^2\right)+\left(y^2-2.4.y+4^2\right)-14\)
\(F=\left(x-2\right)^2+\left(y-4\right)^2-14\)
Ta có: \(\left(x-2\right)^2\ge0\forall x\)
\(\left(y-4\right)^2\ge0\forall x\)
\(\Rightarrow\left(x-2\right)^2+\left(y-4\right)^2-14\ge-14\forall x\)
\(F=-14\Leftrightarrow\hept{\begin{cases}\left(x-2\right)^2=0\\\left(y-4\right)^2=0\end{cases}\Leftrightarrow}\hept{\begin{cases}x=2\\y=4\end{cases}}\)
Vậy \(F_{min}=-14\Leftrightarrow\hept{\begin{cases}x=2\\y=4\end{cases}}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(D=x^2+20y^2+8xy-4y+2009\)
\(\Leftrightarrow D=x^2+16y^2+4y^2+8xy-4y+1+2008\)
\(\Leftrightarrow D=\left(x^2+8xy+16y^2\right)+\left(4y^2-4y+1\right)+2008\)
\(\Leftrightarrow D=\left[x^2+2.x.4y+\left(4y\right)^2\right]+\left[\left(2y\right)^2-2.2y.1+1^2\right]+2008\)
\(\Leftrightarrow D=\left(x+4y\right)^2+\left(2y-1\right)^2+2008\)
Vậy GTNN của \(D=2008\) khi \(\left\{{}\begin{matrix}x+4y=0\\2y-1=0\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x+4.\left(0,5\right)=0\\y=0,5\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x=-2\\y=0,5\end{matrix}\right.\)
a) \(C=x^2-4xy+5y^2+10x-22y+28\)
\(\Leftrightarrow C=x^2-4xy+4y^2+y^2+10x-20y-2y+1+25+2\)
\(\Leftrightarrow C=\left(x^2-4xy+4y^2\right)+\left(10x-20y\right)+\left(y^2-2y+1\right)+2+25\)
\(\Leftrightarrow C=\left(x-2y\right)^2+10\left(x-2y\right)+\left(y-1\right)^2+2+25\)
\(\Leftrightarrow C=\left[\left(x-2y\right)^2+10\left(x-2y\right)+25\right]+\left(y-1\right)^2+2\)
\(\Leftrightarrow C=\left(x-2y+5\right)^2+\left(y-1\right)^2+2\)
Vậy GTNN của \(C=2\) khi \(\left\{{}\begin{matrix}x-2y+5=0\\y-1=0\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x-2.1+5=0\\y=1\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x=-3\\y=1\end{matrix}\right.\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a) Ta có A = x2 - 2x - 1 = (x2 - 2x + 1) - 2 = (x - 1)2 - 2 \(\ge\) -2
Dấu "=" xảy ra <=> x - 1 = 0 => x = 1
Vậy Min A = -2 <=> x = 1
b) Ta có B = 4x2 + 4x + 8 = (4x2 + 4x + 1) + 7 = (2x + 1)2 + 7 \(\ge\)7
Dấu |"=" xảy ra <=> 2x + 1 = 0 => x = -1/2
Vậy Min B = 7 <=> x = -1/2
c) Ta có C = 3x - x2 + 2
= -(x2 - 3x - 2)
= -(x2 - 3x + 9/4 - 9/4 - 2)
= -[(x - 3/2)2 - 17/4)
= -(x - 3/2)2 + 17/4 \(\le\frac{17}{4}\)
Dấu "=" xảy ra <=> x - 3/2 = 0 => x = 3/2
Vậy Max C = 17/4 <=> x = 3/2
d) Ta có D = -x2 - 5x = -(x2 + 5x) = -(x2 + 5x + 25/4 - 25/4) = -(x + 5/2)2 + 25/4 \(\ge\frac{25}{4}\)
Dấu "=" xảy ra <=> x + 5/2 = 0 => x = -5/2
Vậy Max D = 25/4 <=> x = -5/2
e) Ta có E = x2 - 4xy + 5y2 + 10x - 22y + 28
= (x2 - 4xy + 4y2) + 10x - 20y + y2 - 2y + 28
= (x - 2y)2 + 10(x - 2y) + 25 + (y2 - 2y + 1) + 2
= (x - 2y + 5) + (y - 1)2 + 2 \(\ge\)2
Dấu "=" xảy ra <=> \(\hept{\begin{cases}x-2y+5=0\\y-1=0\end{cases}}\Rightarrow\hept{\begin{cases}x=-3\\y=1\end{cases}}\)
Vậy Min E = 2 <=> x = -3 ; y = 1
a: \(A=x^2-2x+1+y^2+4y+4+3\)
\(=\left(x-1\right)^2+\left(y+2\right)^2+3>=3\)
Dấu '=' xảy ra khi x=1 và y=-2
b: \(B=x^2-4x+4+y^2-8y+16-14\)
\(=\left(x-2\right)^2+\left(y-4\right)^2-14>=-14\)
Dấu '=' xảy ra khi x=2 và y=4