3 x y 2 – 6 x 2 y = 3 x y (   y   –   2 x )

Lời giải:

Ta có:

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

a) \(2x^2+5x+2\)

\(=2x^2+4x+x+2\)

\(=2x\left(x+2\right)+\left(x+2\right)\)

\(=\left(x+2\right)\left(2x+1\right)\)

b) \(4x^2-4x-9y^2+12y-3\)

\(=\left(4x^2-4x+1\right)-\left(9y^2-12y+4\right)\)

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

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

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

c) \(x^4-2x^3-4x^2+4x-3\)

\(=x^4+x^3-x^2+x-3x^2-3x+3x-3\)

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

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

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

d) \(x^3-x+3x^2y+3xy^2+y^3-y\)

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

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

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

\(=\left(x+y\right)\left(x+y-1\right)\left(x+y+1\right)\)

a) 3xy² - 3x³ - 6xy + 3x 

=3x (y² - x² - 2y +1)

= 3x [ (y-1)² -x² ]

=3x (y-1-x)(y-1+x)

b) 3x² +11x+6

= 3 x² +9x +2x +6

=3x (x+3)+2(x+3)

= (x+3)(3x+2)

c) -x³ - 4xy² + 4x²y +16x

= -x (x² + 4y² - 4xy -16 )

= -x [(x -2y)² - 4² ]

= -x(x-2y-4)(x-2y+4) 

Câu 1: C

Câu 2: B

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

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

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

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

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

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

(15x³y² - 6x²y-3x²y²):6x²y

= \(\dfrac{15xy}{6}-1-\dfrac{y}{2}\)

= \(\dfrac{5}{2}xy-1-\dfrac{y}{2}\)