\(=\left(x^2+y^2-5\right)^2-4\left(xy-2\right)^2=\left(x^2+y^2-5+2xy-4\right)\left(x^2+y^2-5-2xy+4\right)\)
\(=\left(\left(x+y\right)^2-9\right)\left(\left(x-y\right)^2-1\right)=\left(x+y-3\right)\left(x+y+3\right)\left(x-y+1\right)\left(x-y-1\right)\)

(x^2+y^2-5)^2 - 4x^2y^2 - 16xy -16
= (x^2 + y^2 - 5)^2 - (4x^2y^2 + 16xy + 16)
= (x^2 + y^2 - 5)^2 - (2xy + 4)^2
= (x^2 + y^2 - 5 - 2xy - 4)(x^2 + y^2 - 5 + 2xy + 4)
= [(x - y)^2 - 9][(x + y)^2 - 1]
=(x-y-3)(x-y+3)(x+y-1)(x+y+1)

câu a nek

(4x^2 -7x-50)^2 -4x^2 (4x^2+14x+49/4)

= (4x^2 -7x-50)^2 -(2x)^2 (2x+7/2)^2

= (4x^2 -7x-50)^2 - (4x^2+7x)^2

= (4x^2 -7x-50 +4x^2+7x) (4x^2 -7x-50-4x^2-7x)

= (8x^2-50) (-14x-50)

=2(4x^2-25)* (-2)(7x+25)

=-4 (2x-5)(2x+5)(7x+25)

Bài 2:

1: \(\left(2x-1\right)^2-4\left(2x-1\right)=0\)

=>\(\left(2x-1\right)\left(2x-1-4\right)=0\)

=>(2x-1)(2x-5)=0

=>\(\left[{}\begin{matrix}2x-1=0\\2x-5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{2}\\x=\dfrac{5}{2}\end{matrix}\right.\)

2: \(9x^3-x=0\)

=>\(x\left(9x^2-1\right)=0\)

=>x(3x-1)(3x+1)=0

=>\(\left[{}\begin{matrix}x=0\\3x-1=0\\3x+1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{1}{3}\\x=-\dfrac{1}{3}\end{matrix}\right.\)

3: \(\left(3-2x\right)^2-2\left(2x-3\right)=0\)

=>\(\left(2x-3\right)^2-2\left(2x-3\right)=0\)

=>(2x-3)(2x-3-2)=0

=>(2x-3)(2x-5)=0

=>\(\left[{}\begin{matrix}2x-3=0\\2x-5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{3}{2}\\x=\dfrac{5}{2}\end{matrix}\right.\)

4: \(\left(2x-5\right)\left(x+5\right)-10x+25=0\)

=>\(2x^2+10x-5x-25-10x+25=0\)

=>\(2x^2-5x=0\)

=>\(x\left(2x-5\right)=0\)

=>\(\left[{}\begin{matrix}x=0\\2x-5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{5}{2}\end{matrix}\right.\)

Bài 1:

1: \(3x^3y^2-6xy\)

\(=3xy\cdot x^2y-3xy\cdot2\)

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

2: \(\left(x-2y\right)\left(x+3y\right)-2\left(x-2y\right)\)

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

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

3: \(\left(3x-1\right)\left(x-2y\right)-5x\left(2y-x\right)\)

\(=\left(3x-1\right)\left(x-2y\right)+5x\left(x-2y\right)\)

\(=(x-2y)(3x-1+5x)\)

\(=\left(x-2y\right)\left(8x-1\right)\)

4: \(x^2-y^2-6y-9\)

\(=x^2-\left(y^2+6y+9\right)\)

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

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

5: \(\left(3x-y\right)^2-4y^2\)

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

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

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

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

6: \(4x^2-9y^2-4x+1\)

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

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

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

8: \(x^2y-xy^2-2x+2y\)

\(=xy\left(x-y\right)-2\left(x-y\right)\)

\(=\left(x-y\right)\left(xy-2\right)\)

9: \(x^2-y^2-2x+2y\)

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

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

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

Viết liền thế ai mà đọc được

1: \(6x^2y-9xy^2+3xy\)

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

2: \(\left(4-x\right)^2-16\)

\(=\left(4-x-4\right)\left(4-x+4\right)\)

\(=-x\cdot\left(8-x\right)\)

3: \(x^3+9x^2-4x-36\)

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

\(=\left(x+9\right)\left(x-2\right)\left(x+2\right)\)

1) \(6x^2y-9xy^2+3xy=3xy\left(2x-3y+1\right)\)

2) \(\left(4-x\right)^2-16=\left(4-x\right)^2-4^2=\left(4-x-4\right)\left(4-x+4\right)=-x\left(8-x\right)\)

3) \(x^3+9x^2-4x-36\\ =\left(x^3-2x^2\right)+\left(11x^2-22x\right)+\left(18x-36\right)\\ =x^2\left(x-2\right)+11x\left(x-2\right)+18\left(x-2\right)\\ =\left(x^2+11x+18\right)\left(x-2\right)\\ =\left[\left(x^2+2x\right)+\left(9x+18\right)\right]\left(x-2\right)\\ =\left[x\left(x+2\right)+9\left(x+2\right)\right]\left(x-2\right)\\ =\left(x+2\right)\left(x+9\right)\left(x-2\right)\)

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

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

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

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

h ) \(x^2-ax^2-y+ay+cx^2-cy\)

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

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

d)\(\left(x^2+y^2-z^2\right)^2-4x^2y^2\)

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

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

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

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

e)Đặt \(x^2+3x=a\)

Có: \(\left(x^2+3x+1\right)\left(x^2+3x-3\right)-5\)

\(=\left(a+1\right)\left(a-3\right)-5\)

\(=a^2-3a+a-3-5\)

\(=a^2-2a-8\)

\(=a^2+2x-4x-8\)

\(=a\left(a+2\right)-4\left(a+2\right)\)

\(=\left(a+2\right)\left(a-4\right)\)

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

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

\(=\left[x\left(x+1\right)+2\left(x+1\right)\right]\left[x\left(x-1\right)+4\left(x-1\right)\right]\)

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

\(d,\left(x^2+y^2-z^2\right)^2-4x^2y^2\)
\(=\left(x^2+y^2-z^2\right)^2-\left(2xy\right)^2\)
\(=\left(x^2+y^2-z^2-2xy\right)\left(x^2+y^2-z^2+2xy\right)\)
\(=\left[\left(x^2-2xy+y^2\right)-z^2\right]\left[\left(x^2+2xy+y^2\right)-z^z\right]\)
\(=\left[\left(x-y\right)^2-z^2\right]\left[\left(x+y\right)^2-z^2\right]\)
\(=\left(x-y-z\right)\left(x-y+z\right)\left(x+y-z\right)\left(x+y+z\right)\)
\(e,\left(x^2+3x+1\right)\left(x^2+3x-3\right)-5\left(1\right)\)
\(\text{Đặt }x^2+3x+\frac{1-3}{2}=t\)
\(\text{hay }x^2+3x-2=t\left(2\right)\)
\(\left(1\right)\Leftrightarrow\left(t+3\right)\left(t-1\right)-5\)
\(\Rightarrow t^2-t+3t-3-5\)
\(=t^2+2t-8\)
\(=t^2-2t+4t-8\)
\(=t\left(t-2\right)+4\left(t-2\right)\)
\(=\left(t-2\right)\left(t+4\right)\left(3\right)\)
\(\text{Thay (2) vào (3),ta được:}\)
\(\left(x^2+3x-2-2\right)\left(x^2+3x-2+4\right)\)
\(=\left(x^2+3x-4\right)\left(x^2+3x+2\right)\)
\(=\left(x^2-x+4x-4\right)\left(x^2+x+2x+2\right)\)
\(=\left[x\left(x-1\right)+4\left(x-1\right)\right]\left[x\left(x+1\right)+2\left(x+1\right)\right]\)

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