a. 3xy( 4x + y - \(\dfrac{4}{3}\) )

b. 2x²( 3x + 1 )

c. (2x + 3 )( x - y )

d. xy( 1 - x )( x - 1 )

e. 6( 2x + 1 )( x + y )

\(a,=x\left(x-2\right)\\ b,=2b\left(x-3y\right)+a\left(x-3y\right)=\left(a+2b\right)\left(x-3y\right)\\ c,=x\left(x^2+2xy+y^2-4\right)=x\left[\left(x+y\right)^2-4\right]=x\left(x+y+2\right)\left(x+y-2\right)\\ d,=4-\left(x+y\right)^2=\left(2-x-y\right)\left(2+x+y\right)\\ đ,=5\left(x-y\right)\left(x+y\right)+3\left(x+y\right)^2=\left(x+y\right)\left(5x-5y+3x+3y\right)\\ =\left(x+y\right)\left(8x-2y\right)=2\left(4x-y\right)\left(x+y\right)\\ e,=3x\left(2xy-3\right)\\ b,=x\left(4x^2-4xy+y^2-4\right)=x\left[\left(2x-y\right)^2-4\right]=x\left(2x-y-2\right)\left(2x-y+2\right)\\ f,=\left(x+y\right)^2-z^2=\left(x+y-z\right)\left(x+y+z\right)\)

a,=x(x−2)b,=2b(x−3y)+a(x−3y)=(a+2b)(x−3y)c,=x(x2+2xy+y2−4)=x[(x+y)2−4]=x(x+y+2)(x+y−2)d,=4−(x+y)2=(2−x−y)(2+x+y)đ,=5(x−y)(x+y)+3(x+y)2=(x+y)(5x−5y+3x+3y)=(x+y)(8x−2y)=2(4x−y)(x+y)e,=3x(2xy−3)b,=x(4x2−4xy+y2−4)=x[(2x−y)2−4]=x(2x−y−2)(2x−y+2)f,=(x+y)2−z2=(x+y−z)(x+y+z)

Bài 2: 

a: \(x^2+5x-6=\left(x+6\right)\left(x-1\right)\)

b: \(5x^2+5xy-x-y\)

\(=5x\left(x+y\right)-\left(x+y\right)\)

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

c:\(-6x^2+7x-2\)

\(=-6x^2+3x+4x-2\)

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

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

1.

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

b) \(=\left(x+y\right)^3-\left(x+y\right)=\left(x+y\right)\left[\left(x+y\right)^2-1\right]\)

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

c) \(=5\left[\left(x^2-2xy+y^2\right)-4z^2\right]=5\left[\left(x-y\right)^2-4z^2\right]\)

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

2.

a) \(=x\left(x+2\right)+3\left(x+2\right)=\left(x+2\right)\left(x+3\right)\)

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

c) \(=-\left[3x\left(2x-1\right)-2\left(2x-1\right)\right]=-\left(2x-1\right)\left(3x-2\right)\)

3.

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

c) \(=-\left[5x\left(x-3\right)-1\left(x-3\right)\right]=-\left(x-3\right)\left(5x-1\right)\)

4.

a) \(\Rightarrow\left(x-1\right)\left(5x-1\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x=1\\x=\dfrac{1}{5}\end{matrix}\right.\)

b) \(\Rightarrow2\left(x+5\right)-x\left(x+5\right)=0\)

\(\Rightarrow\left(x+5\right)\left(2-x\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x=-5\\x=2\end{matrix}\right.\)

Lời giải:
a. $5x^2-10xy=5x(x-2y)$

b. $3x(x-y)-6(x-y)=(x-y)(3x-6)=3(x-y)(x-2)$
c. $2x(x-y)-4y(y-x)=2x(x-y)+4y(x-y)=(x-y)(2x+4y)=2(x-y)(x+2y)$

d. $9x^2-9y^2=9(x^2-y^2)=9(x-y)(x+y)$

e. $x^2-xy-x+y=(x^2-xy)-(x-y)=x(x-y)-(x-y)=(x-y)(x-1)$

f. $xy-xz-y+z=(xy-y)-(xz-z)=y(x-1)-z(x-1)=(x-1)(y-z)$

a) Cách 1.

Ta có 2xy + 3z + 6y + xz = (2xy + xz) + (3z + 6y)

= x(2 y + z)+3(z + 2 y) = (z + 2y)(x + 3).

Cách 2.

Ta có 2xy + 3z + 6y + xz = (2x1/ + 6y) + (3z + xz)

= 2y(x + 3) + z(3 + x) = (z + 2y)(x + 3).

b) Biến đổi được a 4   -   9 rt 3   +   a 2 -9a = (a- 9)a( a 2  +1).

c) Biến đổi được 3 x 2  + 5y - 3xy + (-5x) = (x - y)(3x - 5).

d) Biến đổi được  x 2  - (a + b)x + ab = (x- a)(x - b).

e) Ta có 4 x 2 - 4xy + y 2   –   9 t 2 =  ( 2 x   -   y ) 2   -   ( 3 t ) 2

= (2x - y - 3t )(2x - y + 31).

g) Ta có  x 3   -   3 x 2 y   +   3 xy 2   -   y 3   -   z 3

= ( x   -   y ) 3   -   z 3 = (x - y - z)( x 2   +   y 2   +   z 2  - 2xy + xz - yz).

h) Ta có x 2   -   y 2 + 8x + 6y+ 7 = ( x 2  +8x + 16) - ( y 2  - 6y+ 9)

= ( x   +   4 ) 2   - ( y - 3 ) 2  =(x-y + 7)(x + y + l).

Bài `1`

\(a,5x^2-10xy=5x\left(x-2y\right)\\ b,3x\left(x-y\right)-6\left(x-y\right)=\left(x-y\right)\left(3x-6\right)\\ =3\left(x-y\right)\left(x-2\right)\\ c,2x\left(x-y\right)-4y\left(y-x\right)=2x\left(x-y\right)+4y\left(x-y\right)\\ =\left(x-y\right)\left(2x+4y\right)=2\left(x-y\right)\left(x+2y\right)\\ d,9x^2-9y^2=\left(3x\right)^2-\left(3y\right)^2=\left(3x-3y\right)\left(3x+3y\right)\\ f,xy-xz-y+z=\left(xy-xz\right)-\left(y-z\right)\\ =x\left(y-z\right)-\left(y-z\right)=\left(y-z\right)\left(x-1\right)\)

Bài `3`

\(a,3x^2+8x=0\\ \Leftrightarrow x\left(3x+8\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\3x+8=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\3x=-8\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\x=-\dfrac{8}{3}\end{matrix}\right.\)

\(b,9x^2-25=0\\ \Leftrightarrow\left(3x\right)^2-5^2=0\\ \Leftrightarrow\left(3x-5\right)\left(3x+5\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}3x-5=0\\3x+5=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}3x=5\\3x=-5\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{5}{3}\\x=-\dfrac{5}{3}\end{matrix}\right.\)

\(c,x^3-16x=0\\ \Leftrightarrow x\left(x^2-16\right)=0\\ \Leftrightarrow x\left(x-4\right)\left(x+4\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\x-4=0\\x+4=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\x=4\\x=-4\end{matrix}\right.\)

\(d,x^3+x=0\\ \Leftrightarrow x\left(x^2+1\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x^2+1\in\varnothing\\x=0\end{matrix}\right.\Rightarrow x=0\)

a: \(2x^2+3xy-14y^2\)

\(=2x^2+7xy-4xy-14y^2\)

\(=\left(2x^2+7xy\right)-\left(4xy+14y^2\right)\)

\(=x\left(2x+7y\right)-2y\left(2x+7y\right)\)

\(=\left(2x+7y\right)\left(x-2y\right)\)

b: \(\left(x-7\right)\left(x-5\right)\left(x-3\right)\left(x-1\right)+7\)

\(=\left(x-7\right)\left(x-1\right)\left(x-5\right)\left(x-3\right)+7\)

\(=\left(x^2-8x+7\right)\left(x^2-8x+15\right)+7\)

\(=\left(x^2-8x\right)^2+15\left(x^2-8x\right)+7\left(x^2-8x\right)+105+7\)

\(=\left(x^2-8x\right)^2+22\left(x^2-8x\right)+112\)

\(=\left(x^2-8x\right)^2+8\left(x^2-8x\right)+14\left(x^2-8x\right)+112\)

\(=\left(x^2-8x\right)\left(x^2-8x+8\right)+14\left(x^2-8x+8\right)\)

\(=\left(x^2-8x+8\right)\left(x^2-8x+14\right)\)

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

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

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

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

\(=\left(7x-5\right)\left(-2x-2\right)\)

\(=-2\left(x+1\right)\left(7x-5\right)\)

d: \(xy\left(x-y\right)+yz\left(y-z\right)+zx\left(z-x\right)\)

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

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

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

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

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

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

\(=\left(x-z\right)\left[y\cdot\left(x-y\right)-z\left(x-y\right)\right]\)

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

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

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

b: \(=\left(x-4\right)\left(x+3\right)\)

e: =(x+3)(x-2)

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

b) \(=x\left(x-4\right)+3\left(x-4\right)=\left(x-4\right)\left(x+3\right)\)

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

d) \(=4xy\left(y-3x+2\right)\)

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

f) \(=x\left(x^2+2xy+y^2-4z^2\right)=x\left[\left(x+y\right)^2-4z^2\right]=x\left(x+y-2z\right)\left(x+y+2z\right)\)

g) \(=x\left(x^2-2xy+y^2-25\right)=x\left[\left(x-y\right)^2-25\right]=x\left(x-y-5\right)\left(x-y+5\right)\)

h) \(=x\left(x+1\right)-3\left(x+1\right)=\left(x+1\right)\left(x-3\right)\)

i) \(=x^2\left(x-3\right)-9\left(x-3\right)=\left(x-3\right)\left(x^2-9\right)=\left(x-3\right)^2\left(x+3\right)\)