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

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

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

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

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

mấy bn ơi, giúp mk nhanh vs nha!!!!!!!!!!!

a/ A = 2x² + y² - 2xy - 2x + 3

= (x² - 2xy + y²) + (x² - 2x + 1) + 2

= (x - y)² + (x - 1)² + 2\(\ge2\)

Mấy con còn lại làm tương tự nhé bạn

Bài làm

a) A = x² + 2y² - 6x + 8y + 25

A = ( x² + 6x + 9 ) + 2( y² + 4y + 4 ) + 8 

A = ( x + 3 )² + 2( y + 2 )² + 8 > 8 

Dấu " = " xảy ra <=> x = -3 ; y = -2.

Vậy A_Min = 8 khi x = -3; y = -2

Mấy câu sau tương tự, tự giải theo, bh duyệt bài bên lazi đây, 

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 $x+y-2=x-1=y-1=0\Leftrightarrow x=y=1$

------------------

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

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

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

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

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

a.

\(A=\left(x^4+y^2+1-2x^2y+2x^2-2y\right)+2\left(y^2-2y+1\right)+2026\)

\(A=\left(x^2-y+1\right)^2+2\left(y-1\right)^2+2026\ge2026\)

\(A_{min}=2026\) khi \(\left(x;y\right)=\left(0;1\right)\)

b.

Đặt \(x-1=t\Rightarrow x=t+1\)

\(\Rightarrow A=\dfrac{3\left(t+1\right)^2-8\left(t+1\right)+6}{t^2}=\dfrac{3t^2-2t+1}{t^2}=\dfrac{1}{t^2}-\dfrac{2}{t}+3=\left(\dfrac{1}{t}-1\right)^2+2\ge2\)

\(A_{min}=2\) khi \(t=1\Rightarrow x=2\)

\(A=\dfrac{3x^2-8x+6}{x^2-2x+1}=\dfrac{3x^2-8x+6}{\left(x-1\right)^2}=\dfrac{2\left(x-1\right)^2+\left(x-2\right)^2}{\left(x-1\right)^2}=2+\dfrac{\left(x-2\right)^2}{\left(x-1\right)^2}\ge2\)

Dấu \("="\Leftrightarrow x=2\)