P = x² + 2y² + 2xy – 6x – 8y + 2028

P = (x² + y² + 2xy) – 6(x + y) + 9 + y² – 2y + 1 + 2018

P = (x + y – 3)² + (y – 1)² + 2018 \(\ge\) 2018

=> Giá trị nhỏ nhất của P = 2018 khi x = 2; y = 1

P=x²+2y²+2xy-6x-8y+2028

=x²+2xy+y²+y²-8y+x²-6x-x²+2028

=(x²+2xy+y²)+(y²-8y+16)+(x²-6x+9)-x²+2028-16-9

=(x-y)²+(y-4)²+(x-3)²-x²+2003\(\ge2003\)

Vì \(\left\{{}\begin{matrix}\left(x-y\right)^2\ge0\\\left(y-4\right)^2\ge0\\\left(x-3\right)^2\ge0\\x^2\ge0\end{matrix}\right.\) nên:

Để P=2003 thì :

\(\Leftrightarrow\left\{{}\begin{matrix}\left(x-y\right)^2=0\\\left(x-3\right)^2=0\\\left(y-4\right)^2=0\\x^2=0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x-y=0\\x-3=0\\y-4=0\\x=0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x=y\\x=3\\y=4\\x=0\end{matrix}\right.\)

Vậy min P=2003\(\Leftrightarrow\left(x=y\right)\in\left\{0;4;3\right\}\)

P = x² + 2y² + 2xy – 6x – 8y + 2028

P = (x² + y² + 2xy) – 6(x + y) + 9 + y² – 2y + 1 + 2018

P = (x + y – 3)² + (y – 1)² + 2018  2018

=> Giá trị nhỏ nhất của P = 2018 khi x = 2; y = 1

Lời giải:

$P=(x^2+y^2+2xy)+y^2-6x-8y+2028$

$=(x+y)^2-6(x+y)+(y^2-2y)+2028$
$=(x+y)^2-6(x+y)+9+(y^2-2y+1)+2018$

$=(x+y-3)^2+(y-1)^2+2018\geq 0+0+2018=2018$

Vậy $P_{\min}=2018$

Giá trị này đạt tại $x+y-3=y-1=0$

$\Leftrightarrow y=1; x=2$

a

à lộn

thôi chịu

a)

Áp dụng BĐT Bunhiacopxki ta có:

\(\left(a+b+c\right)^2\le\left(a^2+b^2+c^2\right)\left(1^2+1^2+1^2\right)\)

\(\Rightarrow\left(a^2+b^2+c^2\right).3\ge\left(\dfrac{3}{2}\right)^2=\dfrac{9}{4}\)

\(\Rightarrow a^2+b^2+c^2\ge\dfrac{3}{4}\)

a/ chtt

b/ \(P=x^2+2y^2+2xy-6x-8y+2028\)

\(=\left(x^2+2xy+y^2\right)-6\left(x+y\right)+9+\left(y^2-2y+1\right)+2018\)

\(=\left(x+y\right)^2-6\left(x+y\right)+9+\left(y-1\right)^2+2018\)

\(=\left(x+y-3\right)^2+\left(y-1\right)^2+2018\ge2018\)

Dấu = xảy ra khi \(\left\{{}\begin{matrix}x=2\\y=1\end{matrix}\right.\)

Vậy....

a)

b) P = x2 + 2y2 + 2xy – 6x – 8y + 2028

P = (x2 + y2 + 2xy) – 6(x + y) + 9 + y2 – 2y + 1 + 2018

P = (x + y – 3)2 + (y – 1)2 + 2018 2018

=> Giá trị nhỏ nhất của P = 2018 khi x = 2; y = 1

Cách khác câu a

\(a^2+b^2+c^2\ge\dfrac{3}{4}\)

\(\Leftrightarrow a^2+b^2+c^2\ge\dfrac{\left(a+b+c\right)^2}{3}\)

\(\Leftrightarrow3a^2+3b^2+3c^2\ge a^2+b^2+c^2+2ab+2bc+2ca\)

\(\Leftrightarrow2a^2+2b^2+2c^2-2ab-2bc-2ca\ge0\)

\(\Leftrightarrow\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2\ge0\)(luôn đúng)

=>đpcm

Câu b) P = x² + 2y² + 2xy - 6x - 8y + 2028

P = x² + 2xy + y² - 6x - 6y + 9 + y² - 2y + 1 + 2018

P = ( x + y)² - 6( x + y) + 9 + ( y - 1)² + 2018

P = ( x + y - 3)² + ( y - 1)² + 2018

Do : ( x + y - 3)² ≥ 0 ∀x,y

( y - 1)² ≥ 0 ∀x,y

⇒( x + y - 3)² + ( y - 1)² ≥ 0

⇒ ( x + y - 3)² + ( y - 1)² + 2018 ≥ 2018

⇒ P_MIN = 2018 ⇔ x = 2 ; y = 1

\(A=x^2+2x\left(y+1\right)+\left(y+1\right)^2-\left(y+1\right)^2+2y^2-4y+2028\)

\(=\left(x+y+1\right)^2-y^2-2x-1+2y^2-4y+2028\)

\(=\left(x+y+1\right)^2-6x+y^2+2027\)

\(=\left(x+y+1\right)+\left(y-3\right)^2+2018\ge2018\forall x;y\) (do...)

=> MinA = 2018 \(\Leftrightarrow\left\{{}\begin{matrix}x+y=-1\\y=3\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=-4\\y=3\end{matrix}\right.\)

Giải :(x²+2xy+y²)+y²-6x-8y+2024=(x+y)²-2(x+y)3+y²-2y+2024

=(x+y-3)²+(y²-2y+1)+2014=(x+y-3)²+(y-1)²+2014 >=2014

vì (x+y-3)²;(y-1)²>=0 với mọi x;y

nên Pmin=2014khi y=1;x=2

MinP=2024 nha!

2024 đó !đúng 100% luôn !

\(P=x^2+2y^2+2xy-6x-8y+2027\\ =\left(x^2+y^2+9+2xy-6x-6x\right)+\left(y^2-2y+1\right)+2017\\ =\left(x+y-3\right)^2+\left(y-1\right)^2+2017\)

Do \(\left(x+y-3\right)^2\ge0\forall x;y\)

\(\left(y-1\right)^2\ge0\forall x;y\)

\(\Rightarrow\left(x+y-3\right)^2+\left(y-1\right)^2\ge0\forall x;y\\ \Rightarrow\left(x+y-3\right)^2+\left(y-1\right)^2+2017\ge2017\forall x;y\)

Dấu "=" xảy ra khi: \(\left\{{}\begin{matrix}\left(x+y-3\right)^2=0\\\left(y-1\right)^2=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x+y-3=0\\y-1=0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x=3-y\\y=1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=2\\y=1\end{matrix}\right.\)

Vậy \(P_{\left(Min\right)}=2017\) khi \(\left\{{}\begin{matrix}x=2\\y=1\end{matrix}\right.\)

P = x²+2y² +2xy-6x-8y+2027

=x²+2xy+y²+y²-6x-6y-2y+1+9+2017

=(x²+2xy+y²)-(6x+6y)+9+(y²-2y+1)+2017

=(x+y)²-6(x+y)+9+(y-1)²+2017

=[(x+y)²-6(x+y)+9]+(y-1)² +2017

=(x+y-3)²+(y-1)²+2017

Do (x+y-3)² \(\ge0\forall x\)

(y-1)² \(\ge0\forall x\)

=>\(\left(x+y-3\right)^2+\left(y-1\right)^2\ge0\)

=>\(\left(x+y-3\right)^2+\left(y-1\right)^2+2017\ge2017\)=> P\(\ge2017\)

Min P=2017 khi

y-1=0

=> y=1

x+y-3=0

=>x+1-3=0

=> x=2

Vậy GTNN của P=2017 khi y=1 và x=2