Tìm GTNN:
a) P=2x2+5y2+4xy+8x-4y+15
b) C=2x2+4y2+4xy-3x-1
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27 tháng 9 2021
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 \(x=y=-\dfrac{1}{2}\)
27 tháng 9 2021
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) \(E=9x^2+2y^2+6xy-6x-8y+10=\left[9x^2+6x\left(y-1\right)+\left(y-1\right)^2\right]+\left(y^2-6x+9\right)=\left(3x+y-1\right)^2+\left(y-3\right)^2\ge0\)
\(minE=0\Leftrightarrow\) \(\left\{{}\begin{matrix}x=-\dfrac{2}{3}\\y=3\end{matrix}\right.\)
K
3
H9
HT.Phong (9A5)
CTVHS
4 tháng 8 2023
a) \(M=x^2-3x+10\)
\(M=x^2-2\cdot\dfrac{3}{2}\cdot x+\dfrac{9}{4}+\dfrac{31}{4}\)
\(M=\left(x^2-2\cdot\dfrac{3}{2}\cdot x+\dfrac{9}{4}\right)+\dfrac{31}{4}\)
\(M=\left(x-\dfrac{3}{2}\right)^2+\dfrac{31}{4}\)
Mà: \(\left(x-\dfrac{3}{2}\right)^2\ge0\) nên: \(M=\left(x-\dfrac{3}{2}\right)^2+\dfrac{31}{4}\ge\dfrac{31}{4}\)
Dấu "=" xảy ra
\(\left(x-\dfrac{3}{2}\right)^2+\dfrac{31}{4}=\dfrac{31}{4}\Leftrightarrow\left(x-\dfrac{3}{2}\right)^2=0\)
\(\Leftrightarrow x-\dfrac{3}{2}=0\Leftrightarrow x=\dfrac{3}{2}\)
Vậy: \(M_{min}=\dfrac{31}{4}\) với \(x=\dfrac{3}{2}\)
b) \(N=2x^2+5y^2+4xy+8x-4y-100\)
\(N=x^2+x^2+4y^2+y^2+4xy+8x-4y-120+16+4\)
\(N=\left(x^2+4xy+4y^2\right)+\left(x^2+8x+16\right)+\left(y^2-4y+4\right)-120\)
\(N=\left(x+2y\right)^2+\left(x+4\right)^2+\left(y-2\right)^2-120\)
Mà:
\(\left\{{}\begin{matrix}\left(x+2y\right)^2\ge0\\\left(x+4\right)^2\ge0\\\left(y-2\right)^2\ge0\end{matrix}\right.\) nên \(N=\left(x+2y\right)^2+\left(x+4\right)^2+\left(y-2\right)^2-120\ge120\)
Dấu "=" xảy ra:
\(\left\{{}\begin{matrix}\left(x+2y\right)^2=0\\\left(x+4\right)^2=0\\\left(y-2\right)^2=0\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}-4+2y=0\\x=-4\\y=2\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}y=2\\x=-4\\y=2\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=-4\\y=2\end{matrix}\right.\)
Vậy: \(N_{min}=120\) khi \(\left\{{}\begin{matrix}x=-4\\y=2\end{matrix}\right.\)
19 tháng 12 2020
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
19 tháng 12 2020
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
NV
Nguyễn Việt Lâm
Giáo viên
20 tháng 8 2021
\(A=\left(x^2+4y^2+1-4xy+2x-4y\right)+\left(x^2-4x+4\right)-3\)
\(A=\left(x-2y+1\right)^2+\left(x-2\right)^2-3\ge-3\)
Dấu "=" xảy ra khi \(\left(x;y\right)=\left(2;\dfrac{3}{2}\right)\)
MN
1
AH
Akai Haruma
Giáo viên
30 tháng 10 2023
Lời giải:
$A=(x^2+4y^2+4xy)+x^2+5-8x-12y$
$=(x+2y)^2-6(x+2y)+x^2+5-2x$
$=(x+2y)^2-6(x+2y)+9+(x^2-2x+1)-5$
$=(x+2y-3)^2+(x-1)^2-5\geq 0+0-5=-5$
Vậy $A_{\min}=-5$. Giá trị này đạt được khi $x+2y-3=x-1=0$
$\Leftrightarrow x=1; y=1$
5 tháng 10 2021
\(A=\left(2x-1\right)^2+9\ge9\\ A_{min}=9\Leftrightarrow x=\dfrac{1}{2}\\ B=2\left(x^2-2\cdot\dfrac{3}{4}x+\dfrac{9}{16}\right)+\dfrac{1}{8}=2\left(x-\dfrac{3}{4}\right)^2+\dfrac{1}{8}\ge\dfrac{1}{8}\\ B_{min}=\dfrac{1}{8}\Leftrightarrow x=\dfrac{3}{4}\\ C=\left(4x^2+4xy+y^2\right)+2\left(2x+y\right)+1+\left(y^2+4y+4\right)-4\\ C=\left[\left(2x+y\right)^2+2\left(2x+y\right)+1\right]+\left(y+2\right)^2-4\\ C=\left(2x+y+1\right)^2+\left(y+2\right)^2-4\ge-4\\ C_{min}=-4\Leftrightarrow\left\{{}\begin{matrix}2x=-1-y\\y=-2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-\dfrac{3}{2}\\y=-2\end{matrix}\right.\)
\(D=\left(3x-1-2x\right)^2=\left(x-1\right)^2\ge0\\ D_{min}=0\Leftrightarrow x=1\\ G=\left(9x^2+6xy+y^2\right)+\left(y^2+4y+4\right)+1\\ G=\left(3x+y\right)^2+\left(y+2\right)^2+1\ge1\\ G_{min}=1\Leftrightarrow\left\{{}\begin{matrix}3x=-y\\y=-2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{2}{3}\\y=-2\end{matrix}\right.\)
5 tháng 10 2021
\(H=\left(x^2-2xy+y^2\right)+\left(x^2+2x+1\right)+\left(2y^2+4y+2\right)+2\\ H=\left(x-y\right)^2+\left(x+1\right)^2+2\left(y+1\right)^2+2\ge2\\ H_{min}=2\Leftrightarrow\left\{{}\begin{matrix}x=y\\x=-1\\y=-1\end{matrix}\right.\Leftrightarrow x=y=-1\)
Ta luôn có \(\left(x-y\right)^2+\left(y-z\right)^2+\left(z-x\right)^2\ge0\)
\(\Leftrightarrow2x^2+2y^2+2z^2-2xy-2yz-2xz\ge0\\ \Leftrightarrow x^2+y^2+z^2\ge xy+yz+xz\\ \Leftrightarrow x^2+y^2+z^2+2xy+2yz+2xz\ge3xy+3yz+3xz\\ \Leftrightarrow\left(x+y+z\right)^2\ge3\left(xy+yz+xz\right)\\ \Leftrightarrow\dfrac{3^2}{3}\ge xy+yz+xz\\ \Leftrightarrow K\le3\\ K_{max}=3\Leftrightarrow x=y=z=1\)
LM
3
22 tháng 12 2021
\(2x^2-4xy+2y^2\\ =2\left(x^2-2xy+y^2\right)\\ =2\left(x-y\right)^2\)
22 tháng 12 2021
a) 2x2-4xy+2y2
= 2x2-2xy-2xy+2y2
= 2x(x-y)-2y(x-y)
= (2x-2y)(x-y)
b) x2+4xy+4y2-9
= (x+2y)2-32
= (x+2y-3)(x+2y+3)
c) x4-x3y+x-y
= x3(x-y)+(x-y)
= (x3+1)(x-y)
NV
Nguyễn Việt Lâm
Giáo viên
12 tháng 12 2020
\(T=-2\left(x^2+y^2+1-2xy+2x-2y\right)-2y^2+8y+2004\)
\(T=-2\left(x-y+1\right)^2-2\left(y-2\right)^2+2012\le2012\)
\(T_{max}=2012\) khi \(\left\{{}\begin{matrix}x=1\\y=2\end{matrix}\right.\)
a, \(P=2x^2+5y^2+4xy+8x-4y+15\)
\(=\left(x+2y\right)^2+\left(x+4\right)^2+\left(y-2\right)^2-5\)\(\ge-5\)
Dấu "="xảy ra khi:\(\hept{\begin{cases}\left(x+2y\right)^2=0\\\left(x+4\right)^2=0\\\left(y-2\right)^2=0\end{cases}}\Leftrightarrow\hept{\begin{cases}x=-4\\y=2\end{cases}}\)
Vậy...
b, \(C=2x^2+4xy+4y^2-3x-1\)
\(=\left(x+2y\right)^2+\left(x-\frac{3}{2}\right)^2-\frac{5}{4}\ge-\frac{5}{4}\)
sau đó giải tương tự câu a nhé