A=(5x-3y-2)² + (x+y+1)² + 4

Vậy giá trị nhỏ nhất của A là 4

a)  ... = (x^2 -2xy + y^2)+(x^2 -2x+1)+2014=(x-y)^2 + (x-1)^2 +2014 >= 2014 

Đăngt thức xay ra khi x=y=1

Lời giải:

$2Q=2x^2+2xy+2y^2-6x-6y+3998$

$=(x^2+2xy+y^2)+x^2+y^2-6x-6y+3998$

$=(x+y)^2-4(x+y)+(x^2-2x)+(y^2-2y)+3998$

$=(x+y)^2-4(x+y)+4+(x^2-2x+1)+(y^2-2y+1)+3992$

$=(x+y-2)^2+(x-1)^2+(y-1)^2+3992\geq 3992$

$\Rightarrow Q\geq 1996$

Vậy $Q_{\min}=1996$ khi $x+y-2=x-1=y-1=0\Leftrightarrow x=y=1$

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$R=(x^2+2xy+y^2)+x^2-2x+2y+15$

$=(x+y)^2+2(x+y)+x^2-4x+15$

$=(x+y)^2+2(x+y)+1+(x^2-4x+4)+10$

$=(x+y+1)^2+(x-2)^2+10\geq 10$
Vậy $R_{\min}=10$ khi $x+y+1=x-2=0$

$\Leftrightarrow x=2; y=-3$

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

\(D=x^2+2y^2-2xy-3y+2x-5\)

\(=\left(x^2-2xy+2x+y^2-2y+1\right)+\left(y^2-y-6\right)\)

\(=\left(x-y+1\right)^2+\left(y^2-y-\frac{24}{4}\right)\)

\(=\left(x-y+1\right)^2+\left(y^2-y+\frac{1}{4}\right)-\frac{25}{4}\)

\(=\left(x-y+1\right)^2+\left(y-\frac{1}{2}\right)^2-\frac{25}{4}\ge-\frac{25}{4}\forall x,y\)

Đẳng thức xảy ra khi \(\hept{\begin{cases}x-y+1=0\\y-\frac{1}{2}=0\end{cases}}\Rightarrow\hept{\begin{cases}x=-\frac{1}{2}\\y=\frac{1}{2}\end{cases}}\)

\(E=\left(3x-1\right)^2-4\left|3x-1\right|+5\)

\(=9x^2-6x+1-4\left|3x-1\right|+5\)

*)Xét \(x\ge\frac{1}{3}\Rightarrow3x-1\ge0\Rightarrow\left|3x-1\right|=3x-1\) thì:

\(E=9x^2-6x+1-4\left(3x-1\right)+5\)

\(=9x^2-6x+6-12x+4\)\(=9x^2-18x+10\)

\(=9x^2-18x+9+1=9\left(x^2-2x+1\right)+1\)

\(=9\left(x-1\right)^2+1\ge1\forall x\)

*)Xét \(x< \frac{1}{3}\Rightarrow3x-1< 0\Rightarrow\left|3x-1\right|=-3x+1\) thì:

\(E=9x^2-6x+1-4\left(-3x+1\right)+5\)

\(=9x^2-6x+6+12x-4=9x^2+6x+2\)

\(=9\left(x^2+\frac{2x}{3}+\frac{1}{9}\right)+1=9\left(x+\frac{1}{3}\right)^2+1\ge1\forall x\)

Ta thấy cả 2 trường hợp đều có Min=1 vậy ta chốt là Min=1 nhé

Đẳng thức xảy ra khi \(\orbr{\begin{cases}x=-\frac{1}{3}\\x=1\end{cases}}\)

câu b) hơi dài , tôi làm cách khác

đặt /3x-1/=t

ta có E=\(t^2-4t+5=\left(t^2-4t+4\right)+1=\left(t-2\right)^2+1>=1\)

=>Min E=1 dấu "=" xảy ra khi t-2=0<=>t=2=>/3x-1/=2=>\(\orbr{\begin{cases}3x-1=2\\3x-1=-2\end{cases}< =>\orbr{\begin{cases}x=1\\x=-\frac{1}{3}\end{cases}}}\)

\(...P=x^2-8x+16+x^2+2xy+y^2+2y^2-2y+2\)

\(P=\left(x-4\right)^2+\left(x+y\right)^2+2\left(y^2-y+1\right)\left(1\right)\)

Xét \(y^2-y+1=y^2-y+\dfrac{1}{4}-\dfrac{1}{4}+1=\left(y-\dfrac{1}{2}\right)^2+\dfrac{3}{4}\ge\dfrac{3}{4}\left(\left(y-\dfrac{1}{2}\right)^2\ge0\right)\)

\(\Rightarrow2\left(y^2-y+1\right)\ge2.\dfrac{3}{4}=\dfrac{3}{2}\)

mà \(\left(x-4\right)^2\ge0;\left(x+y\right)^2\ge0\)

\(\left(1\right)\Rightarrow P\ge\dfrac{3}{2}\Rightarrow Min\left(P\right)=\dfrac{3}{2}\)