a)

\(A=4x-x^2+3=-\left(x^2-4x-3\right)=-\left(x^2-4x+4\right)+7=-\left(x-2\right)^2+7\le7\)

Daaus = xayr ra khi: x = 2

b) \(B=4x^2-12x+15=4\left(x^2-3x+9\right)-21=4\left(x-3\right)^2-21\ge-21\)

Dấu = xảy ra khi x = 3

c) \(C=4x^2+2y^2-4xy-4y+1=\left(4x^2-4xy+y^2\right)+\left(y^2-4y+4\right)-3=\left(2x-y\right)^2+\left(y-2\right)^2-3\ge-3\)

Dấu = xảy ra khi

2x = y và y = 2

=> x = 1 và y = 2

a) A = \(-x^2+4x+3=-\left(x-2\right)^2+7\le7\)

Dấu "=" <=> x = 2

b) \(4x^2-12x+15=\left(2x-3\right)^2+6\ge6\)

Dấu "=" xảy ra <=> \(x=\dfrac{3}{2}\)

c) \(4x^2+2y^2-4xy-4y+1\)

= \(\left(4x^2-4xy+y^2\right)+\left(y^2-4y+4\right)-3\)

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

Dấu "=" <=> \(\left\{{}\begin{matrix}x=1\\y=2\end{matrix}\right.\)

A = [y^2 +2y(x-2) + (x-2)^2] + (x^2-6x+9) + 1 

= (y+x-2)^2 + (x-3)^2 + 1 >=1 

Dấu = xảy ra khi  <=> y+x-2 = x-3=0

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

Giups mik vs

biet tong cua so thu nhat va so thu hai bang 5,8.Tong cua so thu hai va so thu ba bang 6,7.Tong so thu nhat va so thu ba bang 7,5.Tim moi so do?

H= (2x+y)^2 - 2(2x+y) + 1+ y^2 - 2y + 1 + 1

H= (2x+y+1)^2 + (y+2)^2 + 1 

Min h là 1 

= \(\left(9x^2+12xy+4y^2\right)+\left(x^2+6x+9\right)+2017\)

\(=\left(3x+2y\right)^2+\left(x+3\right)^2+2017\ge2017\)

=> \(MinP=2017\Leftrightarrow\left\{{}\begin{matrix}2y=-3x\\x=-3\end{matrix}\right.\)⇔\(\left\{{}\begin{matrix}x=-3\\y=\dfrac{9}{2}\end{matrix}\right.\)

Ô cho mình hỏi \(Min\) là gì ạ lớp 9 rồi mà chưa học bao giờ.

Để mik suy nghĩ đã sau đó mik trả lời giúp bạn nhé!

\(x^2-4xy+4y^2+3x^2-2x+\frac{1}{3}-\frac{1}{3}\\ =\left(x-2y\right)^2+3\left(x-\frac{1}{3}\right)^2-\frac{1}{3}\ge-\frac{1}{3}\)

khi \(x=\frac{1}{3},y=\frac{1}{6}\)