Tìm GTNN của biếu thức :Tìm GTLN của biểu thức -3x^2+3yz+3xz+2x-2y-3-3y^2-z^2
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LT
0
LT
0
VM
3
DD
Đoàn Đức Hà
Giáo viên
20 tháng 8 2021
\(4A=12x^2+12y^2+4z^2+20xy-12yz-12zx-8x-8y+12\)
\(=9x^2+9y^2+4z^2+18xy-12yz-12zx+2\left(x^2+y^2+4-4x-4y+2xy\right)+x^2+y^2-2xy+4\)
\(=\left(3x+3y-2z\right)^2+2\left(x+y-2\right)^2+\left(x-y\right)^2+4\ge4\)
Dấu \(=\)khi \(\hept{\begin{cases}3x+3y-2z=0\\x+y-2=0\\x-y=0\end{cases}}\Leftrightarrow\hept{\begin{cases}x=y=1\\z=3\end{cases}}\).
Vậy \(minA=1\)khi \(x=y=1,z=3\).
20 tháng 8 2021
\(A=3x^2+3y^2+z^2+5xy-3yz-3xz-2x-2y+3\)
\(=\left(z-\frac{3}{2}x-\frac{3}{2}y\right)^2+\frac{3}{4}\left(x^2y^2+\frac{2}{3}xy-\frac{8}{3}x-\frac{8}{3}y\right)+3\)
\(=\left(z-\frac{3}{2}x-\frac{3}{2}y\right)^2+\frac{3}{4}[\left(x+\frac{y}{3}-\frac{4}{3}\right)^2+\frac{8}{9}y^2-\frac{16}{9}y-\frac{16}{9}]\)
\(=\left(z-\frac{3}{2}x-\frac{3}{2}y\right)^2+\frac{3}{y}[\left(x+\frac{y}{3}-\frac{4}{3}\right)^2+\frac{8}{9}\left(y-1\right)^2-\frac{2y}{9}]+3\)
\(=\left(z-\frac{3}{2}x-\frac{3}{2}y\right)^2+\frac{3}{y}[\left(x+\frac{y}{3}-\frac{4}{3}\right)^2+\frac{8}{9}\left(y-1\right)^2]+1\)
\(\Leftrightarrow A\ge1\Leftrightarrow MinA=1\)
Dấu '' = '' xảy ra khi:
\(\hept{\begin{cases}z-\frac{3}{2}x-\frac{3}{2}y=0\\y-1=0\\x+\frac{y}{3}-\frac{4}{3}=0\end{cases}}\)
\(\Leftrightarrow\hept{\begin{cases}z=0\\y=1\\x=1\end{cases}}\)
LD
1
CG
1
28 tháng 9 2020
A = x2 - 2xy + 3y2 - 2x + 1997
= ( x2 - 2xy + y2 - 2x + 2y + 1 ) + ( 2y2 - 2y + 1/2 ) + 3991/2
= [ ( x2 - 2xy + y2 ) - ( 2x - 2y ) + 1 ] + 2( y2 - y + 1/4 ) + 3991/2
= [ ( x - y )2 - 2( x - y ) + 12 ] + 2( y - 1/2 )2 + 3991/2
= ( x - y - 1 )2 + 2( y - 1/2 )2 + 3991/2 ≥ 3991/2 ∀ x, y
Dấu "=" xảy ra <=> x = 3/2 ; y = 1/2
=> MinA = 3991/2 <=> x = 3/2 ; y = 1/2
23 tháng 7 2023
a) \(\left\{{}\begin{matrix}a=x\\b=2y\\c=3z\end{matrix}\right.\Rightarrow a+b+c=2;a,b,c>0\)
\(\Rightarrow S=\sqrt{\dfrac{\dfrac{ab}{2}}{\dfrac{ab}{2}+c}}+\sqrt{\dfrac{\dfrac{bc}{2}}{\dfrac{bc}{2}+a}}+\sqrt{\dfrac{ca}{ca+2b}}\)
\(=\sqrt{\dfrac{ab}{ab+2c}}+\sqrt{\dfrac{bc}{bc+2a}}+\sqrt{\dfrac{ca}{ca+2b}}\)
Vì a,b,c>0 nên áp dụng BĐT AM-GM, ta có:
\(\sqrt{\dfrac{ab}{ab+2c}}=\sqrt{\dfrac{ab}{ab+\left(a+b+c\right)c}}=\sqrt{\dfrac{ab}{c^2+bc+ca+ab}}=\sqrt{\dfrac{ab}{\left(a+c\right)\left(b+c\right)}}\)
\(=\sqrt{\dfrac{a}{a+c}}.\sqrt{\dfrac{b}{b+c}}\le\dfrac{1}{2}\left(\dfrac{a}{a+c}+\dfrac{b}{b+c}\right)\)
\(\sqrt{\dfrac{bc}{bc+2a}}=\sqrt{\dfrac{bc}{\left(b+a\right)\left(c+a\right)}}\le\dfrac{1}{2}\left(\dfrac{b}{a+b}+\dfrac{c}{a+c}\right)\)
\(\sqrt{\dfrac{ca}{ca+2b}}=\sqrt{\dfrac{ca}{\left(c+b\right)\left(a+b\right)}}\le\dfrac{1}{2}\left(\dfrac{c}{b+c}+\dfrac{a}{a+b}\right)\)
\(\Rightarrow S\le\dfrac{1}{2}\left(\dfrac{a}{a+b}+\dfrac{b}{a+b}\right)+\dfrac{1}{2}\left(\dfrac{b}{b+c}+\dfrac{c}{b+c}\right)+\dfrac{1}{2}\left(\dfrac{a}{a+c}+\dfrac{c}{a+c}\right)=\dfrac{1}{2}+\dfrac{1}{2}+\dfrac{1}{2}=\dfrac{3}{2}\)
Dấu "=" xảy ra khi và chỉ khi: a=b=c=2/3=>\(\left(x,y,z\right)=\left\{\dfrac{2}{3};\dfrac{1}{3};\dfrac{2}{9}\right\}\)
NV
Nguyễn Việt Lâm
Giáo viên
5 tháng 8 2021
Đặt \(\left(x;2y;3z\right)=\left(a;b;c\right)\Rightarrow a+b+c=2\)
\(S=\sqrt{\dfrac{ab}{ab+2c}}+\sqrt{\dfrac{bc}{bc+2a}}+\sqrt{\dfrac{ca}{ca+2b}}\)
\(S=\sqrt{\dfrac{ab}{ab+c\left(a+b+c\right)}}+\sqrt{\dfrac{bc}{bc+a\left(a+b+c\right)}}+\sqrt{\dfrac{ca}{ca+b\left(a+b+c\right)}}\)
\(S=\sqrt{\dfrac{ab}{\left(a+c\right)\left(b+c\right)}}+\sqrt{\dfrac{bc}{\left(a+b\right)\left(a+c\right)}}+\sqrt{\dfrac{ca}{\left(a+b\right)\left(b+c\right)}}\)
\(S\le\dfrac{1}{2}\left(\dfrac{a}{a+c}+\dfrac{b}{b+c}+\dfrac{b}{a+b}+\dfrac{c}{a+c}+\dfrac{a}{a+b}+\dfrac{c}{b+c}\right)=\dfrac{3}{2}\)
Dấu "=" xảy ra khi \(a=b=c=\dfrac{2}{3}\Rightarrow x;y;z\)