Đặt \(A=3x^2-4xy+2y^2-3x+2007\)

       \(A=2x^2-4xy+2y^2+x^2-3x+2007\)

      \(A=2\left(x-y\right)^2+x^2-2.\frac{3}{2}+\frac{9}{4}+\frac{8019}{4}\)

        \(A=2\left(x-y\right)^2+\left(x-\frac{3}{2}\right)^2+\frac{8019}{4}\ge\frac{8019}{4}\)

              Dấu = xảy ra khi \(\hept{\begin{cases}x-y=0\\x-\frac{3}{2}=0\end{cases}\Rightarrow}\hept{\begin{cases}x=y\\x=\frac{3}{2}\end{cases}}\Rightarrow\hept{\begin{cases}x=\frac{3}{2}\\y=\frac{3}{2}\end{cases}}\)

Vậy Min A = \(\frac{8019}{4}\) khi \(x=y=\frac{3}{2}\)

\(A=2\left(x^2-2xy+y^2\right)+\left(x^2-3x+\dfrac{9}{4}\right)+\dfrac{8067}{4}\)

\(A=2\left(x-y\right)^2+\left(x-\dfrac{3}{2}\right)^2+\dfrac{8067}{4}\ge\dfrac{8067}{4}\)

\(A_{min}=\dfrac{8067}{4}\) khi \(x=y=\dfrac{3}{2}\)

a: =2(x^2+2x+9/2)

=2(x^2+2x+1+7/2)

=2(x+1)^2+7>=7

Dấu = xảy ra khi x=-1

b: \(=2\left(\dfrac{3}{2}x^2-2xy+y^2-\dfrac{3}{2}x+\dfrac{2007}{2}\right)\)

\(=2\left(x^2-2xy+y^2+\dfrac{1}{2}x^2-\dfrac{3}{2}x+\dfrac{2007}{2}\right)\)

\(=2\left(x-y\right)^2+x^2-3x+2007\)

\(=2\left(x-y\right)^2+\left(x-\dfrac{3}{2}\right)^2+2004.75>=2004.75\)

Dấu = xảy ra khi x=y=3/2

Bạn có ghi nhầm đề không vậy? 

a) = 9(x² - 2.x/2.9 + 1/324) - 9/324 +5

GTNN A = 4,97

b) = (2x +y)² + y² + 2018

GTNN B = 2018 khi x=0;y=0

c) = -4(x² - 2.3x/ 4.2 + 9/16) +9/16 +10

GTLN C = 169/16

d) = -(x-y)² - (2x +1) +1 + 2016

GTLN D = 2017

(trg bn cho bài khó dữ z, làm hại cả não tui)

cảm ơn nhiều lắm đấy

đang cần gấp ạ

a) \(A=-x^2+2x=-\left(x^2-2x+1\right)+1=-\left(x-1\right)^2+1\le1\)

\(maxA=1\Leftrightarrow x=1\)

b) \(B=\left(2-3x\right)\left(3+2x\right)=-6x^2-5x+6=-6\left(x^2+\dfrac{5}{6}x+\dfrac{25}{144}\right)+\dfrac{169}{24}=-6\left(x+\dfrac{5}{12}\right)^2+\dfrac{169}{24}\le\dfrac{169}{24}\)

\(minB=\dfrac{169}{24}\Leftrightarrow x=-\dfrac{5}{12}\)

c) \(C=4xy-4x-2y-4x^2-2y^2-3=-\left[4x^2-4x\left(y-1\right)+\left(y-1\right)^2\right]+\left(y^2-4y+4\right)-6=\left(2x-y+1\right)^2+\left(y-2\right)^2-6\le-6\)

\(minC=-6\Leftrightarrow\)\(\left\{{}\begin{matrix}x=\dfrac{1}{2}\\y=2\end{matrix}\right.\)

A= (x²+4y²+9/4+4xy+3x+3y) + (y²+5x+95/4)

  = (x+2y+3/2)² + (y+5/2)² + 15

=> A min = 15

Dấu "=" xảy ra khi y=-5/2 ; x=7/2

\(A=x^2+5y^2+4xy+3x+8y+26\)

\(=\left(x^2+4xy+4y^2\right)+\left(3x+6y\right)+\frac{9}{4}+\left(y^2+2y+1\right)+\frac{91}{4}\)

\(=\left(x+2y\right)^2+3\left(x+2y\right)+\frac{9}{4}+\left(y+1\right)^2+\frac{91}{4}\)

\(=\left(x+2y+\frac{3}{2}\right)^2+\left(y+1\right)^2+\frac{91}{4}\ge\frac{91}{4}\forall x,y\)

Dấu"="xảy ra khi \(\orbr{\begin{cases}x+2y+\frac{3}{2}=0\\y+1=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x+2y=-\frac{3}{2}\\y=-1\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=\frac{1}{2}\\y=-1\end{cases}}}\)

Vậy .....