A=2x^2+9y^2-6xy-6x-12y+2024 
A = (x^2 -6xy +9y^2) + 4(x -3y) + x^2 - 10x + 2024
A = (x -3y)^2 +4(x -3y) + 4 + x^2 -10x +25 + 1995
A = (x -3y +2)^2 + (x -5)^2 + 1995 \geq 1995
Min A = 1995 
 x - 5 = 0 => x = 5
Và x - 3y + 2 = 0 hay 5 -3y +2 = 0 => -3y = -7 => y = 7/3 

\(K\)\(nha!~!\)

\(a,9x^2+y^2+2z^2-18x+4z-6y+20=0\\ \Leftrightarrow9\left(x-1\right)^2+\left(y-3\right)^2+2\left(z+1\right)^2=0\\ \Leftrightarrow\left\{{}\begin{matrix}x=1\\y=3\\z=-1\end{matrix}\right.\)

\(b,5x^2+5y^2+8xy+2y-2x+2=0\\ \Leftrightarrow4\left(x+y\right)^2+\left(x-1\right)^2+\left(y+1\right)^2=0\\ \Leftrightarrow\left\{{}\begin{matrix}x=-y\\x=1\\y=-1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=1\\y=-1\end{matrix}\right.\)

\(c,5x^2+2y^2+4xy-2x+4y+5=0\\ \Leftrightarrow\left(2x+y\right)^2+\left(x-1\right)^2+\left(y+2\right)^2=0\\ \Leftrightarrow\left\{{}\begin{matrix}2x=-y\\x=1\\y=-2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=1\\y=-2\end{matrix}\right.\)

\(d,x^2+4y^2+z^2=2x+12y-4z-14\\ \Leftrightarrow\left(x-1\right)^2+\left(2y-3\right)^2+\left(z+2\right)^2=0\\ \Leftrightarrow\left\{{}\begin{matrix}x=1\\y=\dfrac{3}{2}\\z=-2\end{matrix}\right.\)

\(e,x^2+y^2-6x+4y+2=0\\ \Leftrightarrow\left(x-3\right)^2+\left(y+2\right)^2=11\)

Pt vô nghiệm do ko có 2 bình phương số nguyên có tổng là 11

e: Ta có: \(x^2-6x+y^2+4y+2=0\)

\(\Leftrightarrow x^2-6x+9+y^2+4y+4-11=0\)

\(\Leftrightarrow\left(x-3\right)^2+\left(y+2\right)^2=11\)

Dấu '=' xảy ra khi x=3 và y=-2

Bài làm

a) xy + y² - x - y

= ( xy + y² ) - ( x + y )

= y( x + y ) - ( x + y )

= ( x + y )( y - 1 )

b) 25 - x² + 4xy - 4y²

= 25 - ( x² - 4xy + 4y² )

= 25 - ( x - 2y )²

= ( 5 - x + 2y )( 5 + x - 2y )

c) xy + xz - 2y - 2z

= ( xy + xz ) - ( 2y + 2z )

= x( y + z ) - 2( y + z )

= ( y + z )( x - 2 )

d) x² - 6xy + 9y² - 25z²

= ( x² - 6xy + 9y² ) - 25z²

= ( x - 3y )² - 25z²

= ( x - 3y - 5z )( z - 3y + 5z )

e) 3x² - 3y² - 12x + 12y

= 3( x - y )( x + y ) - 12( x - y )

= ( x - y )[ 3( x + y ) - 12 ]

f) 4x³ + 4xy² + 8x²y - 16x

= 4x( x² + y² + 2xy - 4 )

= 4x[ ( x + y)² - 4 ]

= 4x( x + y - 2 )( x + y + 2 )

g) x² - 5x + 4

= x² - x - 4x + 4

= x( x - 1 ) - 4( x - 1 )

= ( x - 1 )( x - 4 )

h) x⁴ + 5x² + 4

= x⁴ + x² + 4x² + 4

= x²( x² + 1 ) + 4( x² + 1 )

= ( x² + 1 )( x² + 4 )

i) 2x² + 3x - 5

= 2x² - 5x + 2x - 5

= 2x( x + 1 ) - 5( x + 1 )

= ( x + 1 )( 2x - 5 )

k) x³ - 2x² + 6x - 5 ( không biết làm )
l) x² - 4x + 3

= ( x² - 4x + 4 ) - 1

= ( x - 2 )² - 1

= ( x - 3 )( x - 1 )

# Học tốt #

d)  D = x⁴ - 6x² + 10

D = (X²)² - 2. x². 3 + 3² + 1

D = (x² - 3)² + 1

(x² - 3)²  >= 0 với mọi x

(x² - 3)² + 1 >=1 với moi5 x

Vậy GTNN của D là 1

\(M=2x^2+9y^2-6xy-6x-12y+2028\\ =3\left(x^2-2xy+y^2\right)-\left(x^2+6x+9\right)+6\left(y^2-2y+1\right)+2025\\ =\left(x-y\right)^2-\left(x-3\right)^2+6\left(y-1\right)^2+2025\ge2025\)

Dấu \("="\Leftrightarrow\left\{{}\begin{matrix}x=y\\x=3\\y=1\end{matrix}\right.\) (vô lí) nên dấu \("="\) ko thể xảy ra

\(N=x^2-4xy+5y^2+10x-22y+28\\ =\left(x^2+4y^2+25-4xy-20y+10x\right)+\left(y^2-2y+1\right)+2\\=\left(x-2y+5\right)^2+\left(y-1\right)^2+2\ge2\)

Dấu \("="\Leftrightarrow\left\{{}\begin{matrix}x-2y=5\\y=1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=7\\y=1\end{matrix}\right.\)

\(M=2x^2+9y^2-6xy-6x-12y+2028=\left(x+2\right)^2-6y\left(x+2\right)+9y^2+\left(x-5\right)^2+1999=\left(x+2-3y\right)^2+\left(x-5\right)^2+2019\ge1999\)

\(ĐTXR\Leftrightarrow\left\{{}\begin{matrix}x=5\\y=\dfrac{7}{3}\end{matrix}\right.\)

\(N=x^2-4xy+5y^2+10x-22y+28=\left(x+5\right)^2-4y\left(x+5\right)+4y^2+\left(y-1\right)^2+2=\left(x+5-2y\right)^2+\left(y-1\right)^2+2\ge2\)

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

b) \(\left(4x^2+4xy+y^2\right):\left(2x+y\right)=\dfrac{\left(2x+y\right)^2}{2x+y}=2x+y\)

c) \(\left(x^2-6xy+9y^2\right):\left(3y-x\right)=\dfrac{\left(3y-x\right)^2}{3y-x}=3y-x\)