\(9x^2y^2-18xy+9\)

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

\(=9\left[\left(xy\right)^2-2xy+1\right]\)

\(=9\left(xy-1\right)^2\)

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

\(9x^2y^2-18xy+9\\ =9\left(x^2y^2-2xy+1\right)\\ =9\left[\left(xy\right)^2-2xy.1^2+1\right]\\ =9\left(xy-1\right)^2\)

Lời giải:

Ta có:
$9x^2y^3-3x^4y^2-6x^3y^2+18xy^4$

$=3xy^2(3xy-x^3-2x^2+6y^2)

d)5.(x-y)-y(x-y)

=(x-y)(5-y)

e) y.(x-z)+7(z-x)

=y.(x-z)-7(x-z)

=(x-z)(y-7)

A=9x^2+18xy-12x+13y^2-24y+5

\(=\left(3x\right)^2+2.3.3xy-2.3x.2+9y^2+4y^2-12y-12y+4+9-8\)

\(=\left[\left(3x\right)^2+\left(3y\right)^2+2^2+2.3x.3y+2.3x.2+2.3y.2\right]+\left[\left(2y\right)^2-2.2y.3+9\right]-8\)

\(=\left(3x+3y+2\right)^2+\left(2y-3\right)^2-8\ge-8\)

Vậy \(MinA=-8\Leftrightarrow\hept{\begin{cases}\left(3x+3y+2\right)^2=0\\\left(2y-3\right)^2=0\end{cases}\Leftrightarrow\hept{\begin{cases}3x+3y+2=0\\2y-3=0\end{cases}\Leftrightarrow}\hept{\begin{cases}x=-6,5\\y=1,5\end{cases}}}\)

\(\left(y-3x\right)\left(3x+y\right)+18xy-9\left(x+y\right)^2\)

=\(\left(y-3x\right)\left(y+3x\right)+18xy-9\left(x+y\right)^2\)

\(=y^2-9x^2+18xy-9\left(x+y\right)^2\)

\(=y^2-9x^2+18xy-9\left(x^2+2xy+y^2\right)\)

\(=y^2-9x^2+18xy-9x^2-18xy-9y^2\)

\(=-8y^2\)

\(\text{A=9x^2+18xy-12x+13y^2-24y+5}\)

\(=\left[\left(3x\right)^2+\left(3y\right)^2+2^2-12x+18xy-12y\right]+\left[\left(2y\right)^2-2.2y.3+9\right]-8\)

\(=\left(3x+3y-2\right)^2+\left(2y-3\right)^2-8\ge-8\)

Vậy \(MinA=-8\Leftrightarrow\hept{\begin{cases}3x+3y-2=0\\2y-3=0\end{cases}\Leftrightarrow\hept{\begin{cases}x=-1,5\\y=1,5\end{cases}}}\)

\(12x^2y-18xy^2-30y^2=6y\left(2x^2-3xy-5y\right)\)

\(d,5\left(x-y\right)-y\left(x-y\right)=\left(5-y\right)\left(x-y\right)\)

1, \(3x^2y^2-6x^2y^3+9x^2y^2\)

\(\Leftrightarrow12x^2y^2-6x^2y^2\)

\(\Leftrightarrow3x^2y^2\left(4+2y\right)\)

5x^2y^3 - 25x^3y^4 + 10x^3y^3

\(\Leftrightarrow5x^2y^3\left(1-5xy+2x\right)\)

\(12x^2y-18xy^2+30y^2\)

\(\Leftrightarrow6y\left(2x^2-3xy-5y\right)\)