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x² - y² + 2yz - z²

= x² - ( y² -2yz + z² )

= x² - ( y-z )²

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

x² - y² + 2yz - z²

= x² - ( y² - 2yz + z² )

= x² - ( y - z )²

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

x² - 6xy - 25z² + 9y²

= ( x² - 6xy + 9y² ) - 25z²

= ( x - 3y ) ² - ( 5z)²

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

x² - 6xy - 25z² + 9y²

= ( x² - 6xy + 9y² ) - 25z²

= ( x - 3y )² - ( 5z )²

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

#TNQ

3x² - 6xy + 3y² - 12z² 

 = 3(x² - 2xy + y² - 4z²) 

= 3[(x - y)²  - (2z)²]

= 3(x - y - 2z)(x - y + 2z) 

\(x^2-2xy+y^2-\left(2z\right)^2\)

\(=\left(x-y\right)^2-\left(2z\right)^2\)

\(=\left(x-y-2z\right)\times\left(x-y+2z\right)\)

XIN TiiCK

1, \(\left(\frac{x^2-3x}{x^2-9}-1\right):\left(\frac{9-x^2}{x^2+x-6}-\frac{x-3}{2-x}-\frac{x-2}{x+3}\right)\)ĐK : \(x\ne-3;2\)

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

\(=\left(\frac{3}{x+3}\right):\left(\frac{9-x^2+x^2-9-x^2+4x-4}{\left(x+3\right)\left(x-2\right)}\right)\)

\(=\frac{3}{x+3}:\frac{-x^2+4x-4}{\left(x+3\right)\left(x-2\right)}=\frac{3}{x+3}:\frac{2-x}{x+3}=\frac{3}{2-x}\)

\(A=-x^2+4x+5=-\left(x^2-4x+4-4\right)+5\)

\(=-\left(x-2\right)^2+9\le9\)Dấu ''='' xày ra khi x = 2

Vậy GTLN của A bằng 9 tại x = 2

b, \(B=-x^2-4y^2+2x-4y+3\)

\(=-\left(x^2-2x+1-1\right)-\left(4y^2+4y+1-1\right)+3\)

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

\(=-\left(x-1\right)^2-\left(2y+1\right)^2+4\le4\)

Dấu ''='' xảy ra khi x = 1 ; y = -1/2

Vậy GTNN của B bằng 4 tại x = 1 ; y = -1/2