a) x^2+2xy+y^2-16

=(x+y)²-16

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

b) 3x^2+5x-3xy-5y

=(3x²-3xy)+(5x-5y)

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

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

c) 4x^2-6x^3y-2x^2+8x

ko bik hoặc sai đề

d) x^2-4-2xy+y^2

=(x-y)²-4

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

e) x^3-4x^2-12x+27

=sai đề

g) 3x^2-18x+27

=3(x²-6x+9)

=3(x-3)²

h) x^2-y^2-z^2-2yz

=x²-(y²+z²+2yx)

=x²-(y+z)²

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

k) 4x^2(x-6)+9y^2(6-x)

=4x²(x-6)-9y²(x-6)

=(x-6)(4x²-9y²)

=(x-6)(2x-3y)(2x+3y)

l)6xy+5x-5y-3x^2-3y^2

=(5x-5y)+(-3x²+6xy-3y²)

=5(x-y)-3(x²-2xy+y²)

=5(x-y)-3(x-y)²

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

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

Ta có:

\(x^2+5y^2-4x-4xy+6y+5=0\\\Rightarrow[(x^2-4xy+4y^2)-(4x-8y)+4]+(y^2-2y+1)=0\\\Rightarrow[(x-2y)^2-4(x-2y)+4]+(y-1)^2=0\\\Rightarrow(x-2y-2)^2+(y-1)^2=0\)

Ta thấy: \(\left\{{}\begin{matrix}\left(x-2y-2\right)^2\ge0\forall x,y\\\left(y-1\right)^2\ge0\forall y\end{matrix}\right.\)

\(\Rightarrow\left(x-2y-2\right)^2+\left(y-1\right)^2\ge0\forall x,y\)

Mà: \(\left(x-2y-2\right)^2+\left(y-1\right)^2=0\)

nên: \(\left\{{}\begin{matrix}x-2y-2=0\\y-1=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=2y+2\\y=1\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}x=2\cdot1+2=4\\y=1\end{matrix}\right.\)

Thay \(x=4;y=1\) vào \(P\), ta được:

\(P=\left(4-3\right)^{2023}+\left(1-2\right)^{2023}+\left(4+1-5\right)^{2023}\)

\(=1^{2023}+\left(-1\right)^{2023}+0^{2023}\)

\(=1-1=0\)

Vậy \(P=0\) khi \(x=4;y=1\).

VV

9x² + y² + 2z² - 18x + 4z - 6y + 20 = 0

<=>9x²-18x+9+y²-6y+9+2z²+4z+2=0

<=>(3x-3)²+(y-3)²+2(z²+2z+1)=0

<=>(3x-3)²+(y-3)²+2(z+1)²=0

=>3x-3=0 và y-3=0 và z+1=0

<=>x=1 và y=3 và z=-1

\(9x^2+y^2+2z^2-18x+4z-6y+20=0\)

\(\Leftrightarrow\left(9x^2-18x+9\right)+\left(y^2-6y+9\right)+2\left(z^2+2z+1\right)=0\)

\(\Leftrightarrow\left(3x-3\right)^2+\left(y-3\right)^2+2\left(z+1\right)^2=0\)

Suy ra hoặc \(3x-3=0\Leftrightarrow x=1\)

            hoặc \(y-3=0\Leftrightarrow y=3\)

            hoặc \(z+1=0\Leftrightarrow z=-1\)