\(32x-2x^3+4x^2y-2xy^2=2x\left(16-x^2+2xy-y^2\right)=2x\left(4+x-y\right)\left(4-x+y\right)\)

a) xy+3x-7y-21

=x(y+3)-7(x+3)

=(x-7)(y+3)

b)2xy-15-6x-5y

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

=(2x-5)(y-3)

c)2x^2y+2xy^2-2x-2y

=2x(xy-1)+2y(xy-1)

=(2x+2y)(xy-1)

x(x+3)-5x(x-5)-5(x+3)

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

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

Câu cuối mình bị nhầm dòng cuối phải là (x-5)(x+3+x-5)=(x-5)(2x-2)nha bạn

a) xy+3x-7y-21=(xy+3x)-(7y+21)= x(y+3)-7(y+3)=(y+3)(x-7)

b)2xy-15-6x+5y=(2xy-6x)+(5y-15)=2x(y-3)+5(y-3)=(y-3)(2x+5)

c)2x^2y+2xy^2-2x-2y=2xy(x+y)-2(x+y)=2(x+y)(xy-1)

d) x(x+3)-5x(x-5)-5(x+3)=[x(x+3)-5(x+3)]-5x(x-5)=(x+3)(x-5)-5x(x-5)=(x-5)(x+3-5x)=(x-5)(3-4x)

\(a,=5x^2-5x+3x-3=\left(x-1\right)\left(5x+3\right)\\ b,=2x^2-5x+2x-5=\left(2x-5\right)\left(x+1\right)\\ c,=x^2+5x-3x-15=\left(x+5\right)\left(x-3\right)\\ d,=7x^2-7x+x-1=\left(x-1\right)\left(7x+1\right)\)

c: =(x+5)(x-3)

d: =(x-1)(7x+1)

\(a,5x^2-2x-3=\left(5x^2-5x\right)+\left(3x-3\right)=5x\left(x-1\right)+3\left(x-1\right)=\left(x-1\right)\left(5x+3\right)\\ b,2x^2-3x-5=\left(2x^2+2x\right)-\left(5x+5\right)=2x\left(x+1\right)-5\left(x+1\right)=\left(x+1\right)\left(2x-5\right)\\ c,x^2+2x-15=\left(x^2-3x\right)+\left(5x-15\right)=x\left(x-3\right)+5\left(x-3\right)=\left(x-3\right)\left(x+5\right)\\ d,7x^2-6x-1=\left(7x^2-7x\right)+\left(x-1\right)=7x\left(x-1\right)+\left(x-1\right)=\left(x-1\right)\left(7x+1\right)\)

\(2x^3-3x^2+2x-1\)

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

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

Ta có:

\(\left(x-1\right)\left(x+2\right)\left(x+3\right)\left(x-6\right)+32x^2\)

\(=\left(x^2-7x+6\right)\left(x^2+5x+6\right)+32x^2\)

Đặt : \(x^2+6=a\left(a< 0\right)\). Khi đó pt trở thành:

\(\left(a-7x\right)\left(a+5x\right)+32x^2\)

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

\(=\left(x^2+x+6\right)\left(x^2-3x+6\right)\)

\(1,=x\left(x^2-2x+1-y^2\right)=x\left[\left(x-1\right)^2-y^2\right]=x\left(x-y-1\right)\left(x+y-1\right)\\ 2,=\left(x+y\right)^3\\ 3,=\left(2y-z\right)\left(4x+7y\right)\\ 4,=\left(x+2\right)^2\\ 5,Sửa:x\left(x-2\right)-x+2=0\\ \Leftrightarrow\left(x-2\right)\left(x-1\right)=0\Leftrightarrow\left[{}\begin{matrix}x=1\\x=2\end{matrix}\right.\)

Ta có \(x^4+10x^3+32x^2+40x+16=\left(x^4+2x^3\right)+\left(8x^3+16x^2\right)+\left(16x^2+32x\right)+\left(8x+16\right)\)

\(=x^3\left(x+2\right)+8x^2\left(x+2\right)+16x\left(x+2\right)+8\left(x+2\right)\)

\(=\left(x+2\right)\left(x^3+8x^2+16x+8\right)=\left(x+2\right)\left(x+2\right)\left(x^2+6x+4\right)\)

\(=\left(x+2\right)^2\left(x^2+6x+4\right)\)

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

\(\Leftrightarrow x^4+x^3-5x^3+x^2-5x^2+2x^2-5x+2x+2\)

\(\Leftrightarrow x^4+x^3+x^2-5x^3-5x^2-5x+2x^2+2x+2\)

\(\Leftrightarrow x^2\left(x^2+x+1\right)-5x\left(x^2+x+1\right)+2\left(x^2+x+1\right)\)

\(\Leftrightarrow\left(x^2-5x+2\right)\left(x^2+x+1\right)\)

Xin tick ạ !!!

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