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\(-5x^2-2xy-2y^2+14x+10y-1\\ =-\left(x^2+2xy+y^2\right)-\left(4x^2-2\cdot2\cdot\dfrac{7}{2}x+\dfrac{49}{4}\right)-\left(y^2-10y+25\right)+\dfrac{55}{4}\\ =-\left(x+y\right)^2-\left(2x-\dfrac{7}{2}\right)^2-\left(y-5\right)^2+\dfrac{55}{4}\le\dfrac{55}{4}\\ Max\Leftrightarrow\left\{{}\begin{matrix}x=-y\\2x=\dfrac{7}{2}\\y=5\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-y\\x=\dfrac{7}{4}\\y=5\end{matrix}\right.\Leftrightarrow x,y\in\varnothing\)
Vậy dấu \("="\) ko xảy ra
a: Ta có: \(-x^2+3x\)
\(=-\left(x^2-3x+\dfrac{9}{4}-\dfrac{9}{4}\right)\)
\(=-\left(x-\dfrac{3}{2}\right)^2+\dfrac{9}{4}\le\dfrac{9}{4}\forall x\)
Dấu '=' xảy ra khi \(x=\dfrac{3}{2}\)
\(A=\left(2x-1\right)^2+9\ge9\\ A_{min}=9\Leftrightarrow x=\dfrac{1}{2}\\ B=2\left(x^2-2\cdot\dfrac{3}{4}x+\dfrac{9}{16}\right)+\dfrac{1}{8}=2\left(x-\dfrac{3}{4}\right)^2+\dfrac{1}{8}\ge\dfrac{1}{8}\\ B_{min}=\dfrac{1}{8}\Leftrightarrow x=\dfrac{3}{4}\\ C=\left(4x^2+4xy+y^2\right)+2\left(2x+y\right)+1+\left(y^2+4y+4\right)-4\\ C=\left[\left(2x+y\right)^2+2\left(2x+y\right)+1\right]+\left(y+2\right)^2-4\\ C=\left(2x+y+1\right)^2+\left(y+2\right)^2-4\ge-4\\ C_{min}=-4\Leftrightarrow\left\{{}\begin{matrix}2x=-1-y\\y=-2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-\dfrac{3}{2}\\y=-2\end{matrix}\right.\)
\(D=\left(3x-1-2x\right)^2=\left(x-1\right)^2\ge0\\ D_{min}=0\Leftrightarrow x=1\\ G=\left(9x^2+6xy+y^2\right)+\left(y^2+4y+4\right)+1\\ G=\left(3x+y\right)^2+\left(y+2\right)^2+1\ge1\\ G_{min}=1\Leftrightarrow\left\{{}\begin{matrix}3x=-y\\y=-2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{2}{3}\\y=-2\end{matrix}\right.\)
\(H=\left(x^2-2xy+y^2\right)+\left(x^2+2x+1\right)+\left(2y^2+4y+2\right)+2\\ H=\left(x-y\right)^2+\left(x+1\right)^2+2\left(y+1\right)^2+2\ge2\\ H_{min}=2\Leftrightarrow\left\{{}\begin{matrix}x=y\\x=-1\\y=-1\end{matrix}\right.\Leftrightarrow x=y=-1\)
Ta luôn có \(\left(x-y\right)^2+\left(y-z\right)^2+\left(z-x\right)^2\ge0\)
\(\Leftrightarrow2x^2+2y^2+2z^2-2xy-2yz-2xz\ge0\\ \Leftrightarrow x^2+y^2+z^2\ge xy+yz+xz\\ \Leftrightarrow x^2+y^2+z^2+2xy+2yz+2xz\ge3xy+3yz+3xz\\ \Leftrightarrow\left(x+y+z\right)^2\ge3\left(xy+yz+xz\right)\\ \Leftrightarrow\dfrac{3^2}{3}\ge xy+yz+xz\\ \Leftrightarrow K\le3\\ K_{max}=3\Leftrightarrow x=y=z=1\)
A= 2x^2 + y^2 - 2xy -2x+3
A= x^2-2xy + y^2 + x^2 - 2x+ 1 +2
A= (x-y)^2 + (x-1)^2 + 2
(x-y)^2> hoặc = 0 với mọi giá trị của x
(x-1)^2 > hoặc =0 với mọi giá trị của x
=> (x-y)^2 + (x-1)^2 > hoặc =0 với mọi giá trị của x
=> (x-y)^2 + (x-1)^2 + 2 > hoặc =2
=> A lớn hơn hoặc bằng 2
=> GTNN của A=2 tại x=y=1
a.
$12x^3y-24x^2y^2+12xy^3=12xy(x^2-2xy+y^2)=12xy(x-y)^2$
b.
$x^2-6x+xy-6y=(x^2+xy)-(6x+6y)=x(x+y)-6(x+y)=(x-6)(x+y)$
c.
$2x^2+2xy-x-y=2x(x+y)-(x+y)=(x+y)(2x-1)$
d.
$x^3-3x^2+3x-1=(x-1)^3$
e.
$3x^2-3y^2-12x-12y=(3x^2-3y^2)-(12x+12y)$
$=3(x-y)(x+y)-12(x+y)=(x+y)[3(x-y)-12]=3(x-y)(x-y-4)$
f.
$x^2-2xy-x^2+4y^2=4y^2-2xy=2y(2y-x)$
A= -x2+2x+3
=>A= -(x2-2x+3)
=>A= -(x2-2.x.1+1+3-1)
=>A=-[(x-1)2+2]
=>A= -(x+1)2-2
Vì -(x+1)2 ≤0=> A≤-2
Dấu "=" xảy ra khi
-(x+1)2=0 => x=-1
Vây A lớn nhất= -2 khi x= -1
B=x2-2x+4y2-4y+8
=> B= (x2-2x+1)+(4y2-4y+1)+6
=> B=(x-1)2+(2y+1)2+6
=> B lớn nhất=6 khi x=1 và y=-1/2
a: Ta có: \(A=x^2-2xy+5y^2+4y+51\)
\(=x^2-2xy+y^2+4y^2+4y+1+50\)
\(=\left(x-y\right)^2+\left(2y+1\right)^2+50\ge50\forall x,y\)
Dấu '=' xảy ra khi \(x=y=-\dfrac{1}{2}\)
a) \(A=x^2-2xy+5y^2+4y+51=\left(x^2-2xy+y^2\right)+\left(4y^2+4y+1\right)+50=\left(x-y\right)^2+\left(2y+1\right)^2+50\ge50\)
\(minA=50\Leftrightarrow x=y=-\dfrac{1}{2}\)
c) \(C=\dfrac{9}{-2x^2+4x-7}=\dfrac{9}{-2\left(x^2-2x+1\right)-5}=\dfrac{9}{-2\left(x-1\right)^2-5}\ge\dfrac{9}{-5}=-\dfrac{9}{5}\)
\(minC=-\dfrac{9}{5}\Leftrightarrow x=1\)
d) \(10x^2+4y^2-4xy+8x-4y+20=\left[4y^2-4y\left(x+1\right)+\left(x+1\right)^2\right]+\left(9x^2+6x+1\right)+18=\left(2y-x-1\right)^2+\left(3x+1\right)^2+18\ge18\)
\(minD=18\Leftrightarrow\) \(\left\{{}\begin{matrix}x=-\dfrac{1}{3}\\y=\dfrac{1}{3}\end{matrix}\right.\)
e) \(E=9x^2+2y^2+6xy-6x-8y+10=\left[9x^2+6x\left(y-1\right)+\left(y-1\right)^2\right]+\left(y^2-6x+9\right)=\left(3x+y-1\right)^2+\left(y-3\right)^2\ge0\)
\(minE=0\Leftrightarrow\) \(\left\{{}\begin{matrix}x=-\dfrac{2}{3}\\y=3\end{matrix}\right.\)
a ) \(x^2-x+1\)
\(\Leftrightarrow\left(x^2-2.x.\dfrac{1}{2}+\left(\dfrac{1}{2}\right)^2\right)+\dfrac{3}{4}\)
\(\Leftrightarrow\left(x-\dfrac{1}{2}\right)^2+\dfrac{3}{4}\)
Ta có : \(\left(x-\dfrac{1}{2}\right)^2\ge0\forall x\)
\(\Leftrightarrow\left(x-\dfrac{1}{2}\right)^2+\dfrac{3}{4}\ge\dfrac{3}{4}\)
Vậy GTNN là \(\dfrac{3}{4}\Leftrightarrow x=\dfrac{1}{2}.\)
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